Plus One Physics Notes & Questions 2024-25 PDF

Chapter 1 Units and Measurement 1.1 Introduction Measurement of any physical quantity involves comparison with a certain basic, internationally accepted reference standard called unit. The result of a measurement of a physical quantity is expressed by a number accompanied by a un...

Chapter 1 Units and Measurement 1.1 Introduction Measurement of any physical quantity involves comparison with a certain basic, internationally accepted reference standard called unit. The result of a measurement of a physical quantity is expressed by a number accompanied by a unit. Fundamental and Derived Quantities. The quantities ,which can be measured directly or indirectly are called physical quantities. There are two types of physical quantities- Fundamental quantities(Base quantities) and Derived quantities. ▪ The physical quantities, which are independent of each other and cannot be expressed in terms of other physical quantities are called fundamental quantities. Eg: length, mass, time. ▪ The physical quantities , which can be expressed in terms of fundamental quantities are called derived quantities. Eg: volume, velocity, force Fundamental and Derived Units ▪ The units for the fundamental or base quantities are called fundamental or base units. The units of all other physical quantities can be expressed as combinations of the base units. ▪ The units of the derived quantities are called derived units. 1.2 The International System of Units A complete set of both the base and derived units, is known as the system of units. Three such systems, the CGS, the FPS (or British) system and the MKS system were in use extensively till recently. The base units for length, mass and time in these systems were as follows : ▪ CGS system - centimetre, gram and second. ▪ FPS system - foot, pound and second. ▪ MKS system - metre, kilogram and second. ---------------------------------------------------------------------------------- Seema Elizabeth, HSST Physics, MARM Govt HSS Santhipuram Downloaded from hssreporter.com  ▪ In 1971 the General Conference on Weights and Measures developed an internationally accepted system of units for measurement with standard scheme of symbols, units and abbreviations. ▪ This is the Système Internationale d’ Unites (French for International System of Units), abbreviated as SI system. ▪ SI system is now for international usage in scientific, technical, industrial and commercial work. In SI system there are seven base units and two supplementary units. Downloaded from hssreporter.com Multiples and Sub multiples of Units 1.3 Significant figures The result of measurement is a number that includes all digits in the number that are non reliable plus the first digit that is uncertain. The reliable digits plus the first uncertain digit in a measurement are known as significant digits or significant figures. If the period of oscillation of a symbol pendulum is 1.6 s, the digits 1 and 6 are reliable and certain, while the digit 2 is uncertain Downloaded from hssreporter.com Rule 6: The power of 10, in scientific notation is irrelevant to the determination of significant figures. Downloaded from hssreporter.com  4.700 m = 4.700 × 𝟏𝟎𝟐 cm = 4.700 × 𝟏𝟎𝟑 mm = 4.700 × 𝟏𝟎−𝟑 km All these numbers have 4 significant namaste figures. Rounding off the Uncertain Digits 1) If the insignificant digit to be dropped is more than 5, the preceding digit is raised by 1 A number 2.746 rounded off to three significant figures is 2.75 Here the insignificant digit , 6 > 5 and hence 1 is added to the preceeding digit 4.(4+1=5) 2) If the insignificant digit to be dropped less than 5, the preceding digit is left unchanged. A number 2.743 rounded off to three significant figures is be 2.74. Here the insignificant digit , 3< 5 and hence the preceeding number 4 does not change. 3) If the insignificant digit to be dropped is 5, Case i) If the preceding digit is even, the insignificant digit is simply dropped. A number 2.745 rounded off to three significant figures is 2.74. Here the preceding digit 4 is even and hence 5 is simply dropped. Case ii- ) If the preceding digit is odd, the preceding digit is raised by 1. A number 2.735 rounded off to three significant figures is 2.74 Here the preceding digit 3 ,is odd and hence 1 is added to It. (3+1=4) Rules for Arithmetic Operations with Significant Figures (1)In multiplication or division, the final result should retain as many significant figures as are there in the original number with the least significant figures. Downloaded from hssreporter.com Eg:If mass of an object is measured to be, 4.237 g (four significant figures) and its volume is measured to be 2.51cm3(3 significant figures), then find its density in appropriate significant figures. 𝑚𝑎𝑠𝑠 4.237 𝑔 Density = = = 1.688047 𝑣𝑜𝑙𝑢𝑚𝑒 2.51𝑐𝑚3 As per rule the final result should be rounded to 3 significant figures. So the answer is 1.69 g/ 𝒄𝒎𝟑 (2) In addition or subtraction, the final result should retain as many decimal places as are there in the number with the least decimal places. Eg:Find the sum of the numbers 436.32 g, 227.2 g and 0.301 g to appropriate significant figures. 436.32 g + (2 decimal places) 227.2 g + (1 decimal place) 0.301 g (3 decimal places) ______________ 663.821 g As per rule ,the final result should be rounded to 1 decimal place. So the answer 663.8 g 1.4 Dimensions of Physical Quantities The nature of a physical quantity is described by its dimensions. All the physical quantities represented by derived units can be expressed in terms of some combination of seven fundamental or base quantities. This base quantities are known as seven dimensions of physical world ,which are denoted with square brackets [ ]. Thus, length has the dimension [L], mass [M], time [T], electric current [A], thermodynamic temperature K], luminous intensity [cd] and amount of substance [mol]. The dimensions of a physical quantity are the powers (or exponents) to which the base quantities are raised to represent that quantity. Eg: The volume occupied by an object= lengthx breadth x thickness The dimensions of volume is represented as [V] Downloaded from hssreporter.com Thus volume has zero dimension in mass, zero dimension in time and three dimensions in length. 1.5 Dimensional Formulae and Dimensional Equations Unit of density =kg 𝒎−𝟑 Downloaded from hssreporter.com Downloaded from hssreporter.com Downloaded from hssreporter.com Downloaded from hssreporter.com 1.6 Dimensional Analysis and its Applications 1. Checking the dimensional consistency (correctness) of equatons. 2. Deducing relation among the physical quantities. Downloaded from hssreporter.com Downloaded from hssreporter.com Downloaded from hssreporter.com Since dimensions of all terms are the same for Equations (b) and (d) , these equations can be considered as the equation for kinetic energy. 9.The Van der waals equation of 'n' moles of a real gas is 𝒂 (P+ 𝟐 )(V−b)=nRT. Where P is the pressure, V is the volume, T is absolute 𝑽 temperature, R is molar gas constant and a, b, c are Van der waal constants. Find the dimensional formula for a and b. a (P+ 2 )(V−b)=nRT. V By principle of homegeneity, the quantities with same dimensions can be added or subtracted. a [P] =[ 2 ] V [a] =[PV 2 ] =ML−1 T −2 x L6 [a] = M𝐋𝟓 𝐓 −𝟐 [b] = [V] [b] =𝐋𝟑 Downloaded from hssreporter.com Downloaded from hssreporter.com Downloaded from hssreporter.com  Chapter 2 Motion in a Straight Line Introduction The study of motion of objects along a straight line is also known as rectilinear motion. Average Velocity and Average Speed Average Velocity Average velocity is defined as the ratio of total displacement to the total time interval. 𝐓𝐨𝐭𝐚𝐥 𝐝𝐢𝐬𝐩𝐥𝐚𝐜𝐞𝐦𝐞𝐧𝐭 Average velocity= 𝐓𝐨𝐭𝐚𝐥 𝐭𝐢𝐦𝐞 𝐢𝐧𝐭𝐞𝐫𝐯𝐚𝐥 𝐱 𝟐 −𝐱 𝟏 ∆𝐱 𝐯̅ = = 𝐭 𝟐 −𝐭 𝟏 ∆𝐭 where x1 and x2 are the positions of the object at time t1 and t2 Average speed Average speed is defined as the ratio of total path length (distance travelled) to the total time interval. 𝐓𝐨𝐭𝐚𝐥 𝐩𝐚𝐭𝐡 𝐥𝐞𝐧𝐠𝐭𝐡 Average speed= 𝐓𝐨𝐭𝐚𝐥 𝐭𝐢𝐦𝐞 𝐢𝐧𝐭𝐞𝐫𝐯𝐚𝐥 Uniform motion If an object moving along the straight line covers equal distances in equal intervals of time, it is said to be in uniform motion along a straight line. In uniform motion velocity of the object remains constant. 2.2 Instantaneous Velocity and Speed Instantaneous velocity The average velocity tells us how fast an object has been moving over a given time interval but does not tell us how fast it moves at different instants of time during that interval. For this, we define instantaneous velocity or simply velocity v at an instant t. The velocity at an instant is called instantaneous velocity and is defined as the limit of the average velocity as the time interval Δt becomes infinitesimally small 𝚫𝐱 𝐝𝐱 v = 𝐥𝐢𝐦 = 𝚫𝐭→𝟎 𝚫𝐭 𝐝𝐭 𝐝𝐱 v = 𝐝𝐭 Downloaded from hssreporter.com dx is the differential coefficient of x with respect to t.It is the rate of dt change of position with respect to time. Determining velocity from position-time graph. Velocity at t = 4 s is the slope of the tangent to the graph at that instant. We see from Table 2.1 that as we decrease the value of ∆t from 2.0 s to 0.010 s, the value of the average velocity approaches the limiting value 𝐝𝐱 3.84 m s–1 which is the value of velocity at t = 4.0 s, i.e. the value of 𝐝𝐭 at t = 4.0 s. Instantaneous speed Instantaneous speed or simply speed is the magnitude of velocity. For example, a velocity of 24 m s –1 and a velocity of – 24 m s –1 — both have an associated speed of 24.0 m s -1. Downloaded from hssreporter.com Example The position of an object moving along x-axis is given by x = a + bt2 where a = 8.5 m, b = 2.5 m s–2 and t is measured in seconds. (a)What is its velocity at t = 0 s and t = 2.0 s. (b) What is the average velocity between t = 2.0 s and t = 4.0 s ? (a) x = a + bt 2 𝐝𝐱 𝐝 v= = (a + bt 2 ) = 2bt 𝐝𝐭 𝐝𝐭 At t = 0 , v = 0 At t = 2 , v = 2 x 2.5 x 2 = 10 m s-1 x2 −x1 x4 −x2 (b) v̅ = = t2 −t1 4−2 a+16b−a−4b = 2 12b 12 x 2.5 = = =15 m s-1 2 2 2.3 Acceleration Suppose the velocity itself is changing with time. In order to describe its effect on the motion of the particle, we require another physical quantity called acceleration. The rate of change of velocity of an object is called acceleration. Average Acceleration The average acceleration a over a time interval is defined as the change of velocity divided by the time interval. 𝐯𝟐 −𝐯𝟏 ∆𝐯 𝐚̅ = = 𝐭 𝟐 −𝐭 𝟏 ∆𝐭 ▪ Unit of acceleration is ms-2 , [a] =LT-2 ▪ Acceleration is a vector quantity. ▪ If velocity is increasing with time, acceleration is +ve. ▪ If velocity is decreasing with time, acceleration is -ve. ▪ -ve acceleration is called retardation or deceleration. Uniform acceleration If the velocity of an object changes by equal amounts in equal intervals of time, it has uniform acceleration. Downloaded from hssreporter.com Instantaneous acceleration The acceleration of a particle at any instant of its motion is called instantaneous acceleration. It is defined as the limit of the average acceleration as the time interval Δt becomes infinitesimally small. 𝚫𝐯 𝐝𝐯 a = 𝐥𝐢𝐦 = 𝐝𝐭 𝚫𝐭→𝟎 𝚫𝐭 𝐝𝐯 a = 𝐝𝐭 ⅆ𝟐 𝒙 or a = ⅆ𝒕𝟐 Position-time graph for motion with (a)positive acceleration (b) negative acceleration c)zero acceleration Velocity–time graph for motions with constant acceleration (a)Motion in positive direction with positive acceleration (b) Motion in positive direction with negative acceleration Downloaded from hssreporter.com (c)Motion in negative direction with negative acceleration (d)Motion of an object with negative acceleration that changes direction at time t1. (Between times 0 to t1, its moves in positive x - direction and between t1 and t2 it moves in the opposite direction.) Importance of Velocity - time graph for a moving object An interesting feature of velocity - time graph for any moving object is that the area under the velocity - time graph is equal to the displacement of the particle. Proof for this statement :- In uniform motion, velocity is the same at any instant of motion. Therefore, the velocity - time graph is a straight line parallel to the time axis. Area =uT = Displacement i.e.,the area under the velocity - time graph is equal to the displacement of the particle. Downloaded from hssreporter.com 2.4 Kinematic Equations for Uniformly Accelerated Motion Consider a body moving with uniform acceleration. The velocity – time graph is as shown in figure (1) Velocity – time relation From the graph , acceleration = slope BC a= AC v−u a= t v-u = at v = u +at -------------------- (1) or (v = v0 +at) (2) Position-time relation Displacement = Area under the graph s = Area of + Area of s = ut + ½ (v-u) t But from equation (1) v -u = at s = ut + ½ at x t s = ut + ½ at2 ------------------(2) or (s=v0 t +½ at2) (3)Position – velocity relation Displacement = Average velocity x time v+u v−u s=( )( ) 2 a v2 −u2 s=( ) 2a v 2 − u2 = 2as 𝐯 𝟐 = 𝐮𝟐 + 2as ----------------(3) Or (𝐯 𝟐 = 𝐯𝟎 𝟐 + 2as ) Downloaded from hssreporter.com Stopping distance of vehicles When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. 𝑣 2 = 𝑢2 + 2as 0 = 𝑢2 + 2as −𝑢2 = 2as −𝒖𝟐 s= 𝟐𝒂 Motion of an object under Free Fall Free fall is a case of motion with uniform acceleration. Since the acceleration due to gravity is always downward, a=–g =–9.8ms–2 The object is released from rest at y = 0. Therefore, u = 0 Then the equations of motion become v=0–gt = -9.8 t y = 0 – ½ g t2 = -4.9 t2 v 2 = 0 – 2 g y = -19.6 y (a)Variation of acceleration with time (b)Variation of velocity with time (c)Variation of distance with time Downloaded from hssreporter.com Example A ball is thrown vertically upwards with a velocity of 20 m s–1 from the top of a multistorey building. The height of the point from where the ball is thrown is 25.0 m from the ground. (a) How high will the ball rise ? and (b) how long will it be before the ball hits the ground? Take g = 10 m s–2 Example Galileo’s law of odd numbers :“The distances traversed, during equal intervals of time, by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity [namely, 1: 3: 5: 7…...].” Prove it. Downloaded from hssreporter.com Answer: Let us divide the time interval of motion of an object under free fall into many equal intervals τ and find out the distances traversed during successive intervals of time. Since initial velocity is zero, we have y = - ½ g t2 We find that the distances are in the simple ratio 1: 3: 5: 7: 9: 11… as shown in the last column. This law was established by Galileo Galilei. Downloaded from hssreporter.com  Chapter 3 Motion in a Plane 3.1 Introduction In one dimension only two directions are possible and we used positive and negative signs to represent the two directions. But in order to describe motion of an object in two dimensions (plane) or three dimensions (space) we need to use vectors to describe physical quantities like position, displacement, velocity ,acceleration etc Scalars and Vectors A scalar quantity has only magnitude and no direction. It is specified completely by a single number, along with the proper unit. Eg. distance ,mass , temperature, time. A vector quantity has both magnitude and direction and obeys the triangle law of addition or the parallelogram law of addition. A vector is specified by giving its magnitude by a number and its direction. Eg.displacement, velocity, acceleration and force. Representation of a Vector A vector is representedby a bold letter say A or an arrow by an arrow placed over a letter, say Ā. The magnitude of a vector is called its absolute value, indicated by |Ā|=A Graphically a vector is represented by a line segment with an arrow head. Q Q is the head of the vector Ā P is the tail of the vector P The length of line segment gives the magnitude of the vector and arrow mark gives its direction. Position and Displacement Vectors Downloaded from hssreporter.com Let P and P′ be the positions of the object at time t and t′ , respectively. OP is the position vector of the object at time t. OP = r. OP’ is the position vector of the object at time t’. OP’= r’ If the object moves from P to P′ , the vector PP′ is called the displacement vector. Displacement vector is the straight line joining the initial and final positions and does not depend on the actual path undertaken by the object between the two positions. Equality of Vectors Two vectors A and B are said to be equal if, and only if, they have the same magnitude and the same direction. (a) Two equal vectors A and B. (b) Two vectors A ′ and B ′ are unequal eventhough they are of same length 3.3 Multiplication of Vectors by Real Numbers ▪ Multiplying a vector Ā with a positive number λ gives a vector whose magnitude is changed by the factor λ but direction is the same as that of Ā λxĀ =λĀ, if λ > 0 For example, if Ā is multiplied by 2, the resultant vector 2 Ā is in the same direction as Ā and has a magnitude twice of | Ā Downloaded from hssreporter.com  ▪ Multiplying a vector Ā by a negative number λ gives a vector λ Ā whose direction is opposite to the direction of Ā and whose magnitude is –λ times | Ā |. For example, multiplying a given vector A by negative numbers, say –1 and –1.5, gives vectors as 3.4 Addition and Subtraction of Vectors — Graphical Method Triangle law of vector addition If two vectors are represented in magnitude and direction by the two sides of a triangle taken in order ,then their resultant is given by the third side of the triangle taken in reverse order. This graphical method is called the head-to-tail method. If we find the resultant of B + A , the same vector R is obtained. ▪ Thus, vector addition is commutative: A+B=B+A ▪ The addition of vectors also obeys the associative law (A + B) + C = A + (B + C) Subtraction of vectors Subtraction of vectors can be defined in terms of addition of vectors. We define the difference of two vectors A and B as the sum of two vectors A and –B : A – B = A + (–B) Downloaded from hssreporter.com Parallelogram law of vector addition If two vectors are represented in magnitude and direction by the adjacent sides of a parallelogram ,then their resultant is given by the diagonal of the parallelogram. Example Rain is falling vertically with a speed of 35 m s –1. Winds starts blowing after sometime with a speed of 12 m s –1 in east to west direction. In which direction should a boy waiting at a bus stop hold his umbrella ? Unit vectors A unit vector is a vector of unit magnitude and points in a particular direction. It has no dimension and unit. It is used to specify a direction only. If we multiply a unit vector, say n by a scalar, the result is a vector. Downloaded from hssreporter.com In general, a vector A can be written as Ā =|𝐴̅|𝐴̂ ̅ ̂= 𝐀 𝐀 ̅| |𝑨 where  is the unit vector along Ā Unit vectors along the x-, y- and z-axes of a rectangular coordinate system are denoted by î , ĵ and k̂ , respectively. Since these are unit vectors, we have | î | = | ̂J | = | k̂ | = 1 These unit vectors are perpendicular to each other and are called orthogonal unit vectors 3.5 Resolution of a vector We can now resolve a vector A in terms of component vectors that lie along unit vectors î and ĵ. Downloaded from hssreporter.com 3.6 Vector Addition – Analytical Method Consider two vectors A and B in x-y plane Downloaded from hssreporter.com From the geometry of the figure, OS 2 = ON 2 + SN 2 but ON = OP + PN = A + B cos θ SN = B sin θ OS 2 = (A + B cos θ ) 2 + (B sin θ ) 2 R 2 = A 2 +2AB cos θ + B 2 cos 2θ +B 2sin2 θ R 2 = A 2 + B 2 + 2AB cos θ 𝐑 = √𝐀2 + 𝐁 2 + 2𝐀𝐁𝐜𝐨𝐬𝛉 This Equation gives the magnitude of the resultant of vectors A and B. From figure, These Equations gives the direction of the resultant of vectors A and B. Example A motorboat is racing towards north at 25 km/h and the water current in that region is 10 km/h in the direction of 60° east of south. Find the resultant velocity of the boat Downloaded from hssreporter.com 3.7 Motion in a Plane Position Vector The position vector r of a particle P at time t r = xî +y ĵ The position vector r of a particle P at time t’ r’ = x’î +y’ ĵ Displacement vector Δr = r’ -r Δr =( x’î +y’ ĵ ) - (xî +y ĵ ) Δr = ( x’- x) î + (y’ - y ) ĵ Δr = Δx î + Δy ĵ Velocity vector Δ𝐫 v= Δ𝐭 v = Δx î + Δy ĵ Δt v = Δx î + Δy ĵ Δt Δt v = vx î + v y ĵ Downloaded from hssreporter.com Instantaneous velocity 𝐝𝐫 v = 𝐝𝐭 v = vx î + vy ĵ where Acceleration a = ax î + a y ĵ Instantaneous Acceleration 𝐝𝐯 a= 𝐝𝐭 Downloaded from hssreporter.com 3.9 Projectile Motion ▪ An object that is in flight after being thrown or projected is called a projectile. ▪ The path (trajectory)of a projectile is a parabola. ▪ The components of initial velocity u are u cos θ along horizontal direction and u sin θ along vertical direction. Downloaded from hssreporter.com  ▪ The x-component of velocity(u cos θ ) remains constant throughout the motion and hence there is no acceleration in horizontal direction,i.e., ax = 0 ▪ The y- component of velocity (u sin θ ) changes throughout the motion. At the point of maximum height, u sin θ = 0. There is acceleration in vertical direction, ay = – g Equation of path of a projectile Displacement of the projectile after a time t x= ucosθ t x t =ucosθ 1 y= u sinθ t − g t 2 2 x 1 x 2 y= u sinθ (ucosθ) − 2 g (ucosθ) g y= tanθ x − x2 2 u2 cos2 θ This equation is of the form y = a x + b x 2 , in which a and b are constants. This is the equation of a parabola, i.e. the path of the projectile is a parabola. Time of Flight of a projectile (T) The total time T during which the projectile is in flight is called Time of Flight, T. Downloaded from hssreporter.com Consider the motion in vertical direction, s = ut +½ at2 s=0, u = u sin θ , a =-g , t=T 0 = u sin θ T - ½ gT2 u sin θ T = ½ gT2 𝟐 𝐮 𝐬𝐢𝐧 𝛉 T= 𝐠 Horizontal range of a projectile (R) The horizontal distance travelled by a projectile during its time of flight is called the horizontal range, R. Horizontal range = Horizontal component of velocity x Time of flight 𝟐 𝐮 𝐬𝐢𝐧 𝛉 R = u cos θ x 𝒈 𝒖𝟐 𝐱 𝟐 𝐬𝐢𝐧𝛉 cos θ R= 𝑔 𝐮𝟐 𝐬𝐢𝐧 𝟐𝛉 R= 𝐠 R is maximum when sin2θ is maximum, i.e., when θ = 45 0. 𝐮𝟐 Rmax = 𝐠 Show that for a given velocity of projection range will be same for angles 𝜽 and ( 90-𝜽 ) 𝐮𝟐 𝐬𝐢𝐧 𝟐𝛉 For angle θ, R= 𝐠 𝐮𝟐 𝐬𝐢𝐧 𝟐(𝟗𝟎−𝛉) For angle (90 -θ), R= 𝐠 𝐮𝟐 𝐬𝐢𝐧 (𝟏𝟖𝟎− 𝟐𝛉) R= 𝐠 sin (180 - 2θ) =sin 2θ 𝐮𝟐 𝐬𝐢𝐧 𝟐𝛉 R= 𝐠 for given velocity of projection range will be same for angles 𝜽 and ( 90-𝜽 ) Downloaded from hssreporter.com Maximum height of a projectile (H) It is the maximum height reached by the projectile. Consider the motion in vertical direction to the highest point v2 – u2 = 2as u = u sin θ, v=0, a = -g , s=H 0 - u2 sin 2θ = -2 g H 𝐮𝟐 𝐬𝐢𝐧𝟐 𝛉 H= 𝟐𝐠 Example A cricket ball is thrown at a speed of 28 m s –1 in a direction 30° above the horizontal. Calculate (a) the maximum height, (b) the time taken by the ball to return to the same level, and (c) the distance from the thrower to the point where the ball returns to the same level. u2 sin2 θ (a) H = 2g 282 sin2 30 H= = 10 m 2 x 9.8 2 u sin θ (b) T = g 2x9.8 sin 30 T= = 2.9 s 9.8 u2 sin 2θ (c) R = g 282 sin 60 R= = 69 m 9.8 3.10 Uniform Circular Motion When an object follows a circular path at a constant speed, the motion of the object is called uniform circular motion. The word “uniform” refers to the speed, which is uniform (constant) throughout the motion. Period The time taken by an object to make one revolution is known as its time period T Downloaded from hssreporter.com Frequency The number of revolutions made in one second is called its frequency. 𝟏 υ= 𝑻 unit - hertz (Hz) Angular velocity (ω ) angular velocity is the time rate of change of angular displacement 𝚫𝛉 ω= 𝚫𝒕 In the limit Δt tends to zero 𝐝𝛉 ω= ⅆ𝒕 Unit is rad/s During the time period T ,the angular displacement is 2π radian 2𝛑 ω= or ω = 2𝛑 υ 𝑻 Relation connecting angular velocity and linear velocity As the object moves from P to P′ in time Δt. Δθ is called angular displacement and Δ r is the linear diplacement arc angle = radius Δr Δθ= r Δr=r Δθ Δr Δθ =r Δ𝑡 Δ𝑡 v=rω Downloaded from hssreporter.com Angular Acceleration The rate of change of angular velocity is called angular acceleration. d𝛚 α = dt dθ But ω = 𝑑𝑡 d dθ α= ( 𝑑𝑡 ) dt 𝐝𝟐 𝛉 α= 𝐝𝐭 Centripetal acceleration A body in uniform circular motion experiences an acceleration , which is directed towards the centre along its radius.This is s called centripetal acceleration. Let r and r′ be the position vectors and v and v′ the velocities of the object when it is at point P and P ′ Δv Δr = v r vΔr Δv= r Δv vΔr = Δt r Δt v a= xv r v2 a= r If R is the radius of circular path, then centripetal acceleration. 𝐯𝟐 ac = 𝐑 Downloaded from hssreporter.com Centripetal acceleration can also be expressed as v=Rω v=Rω 𝐯𝟐 ac = R = v/ω 𝐑 v2 ac = R ac = R2 ω2 ac = v 2 R (v/ ω) a c = ω 2R ac = v ω Example An insect trapped in a circular groove of radius 12 cm moves along the groove steadily and completes 7 revolutions in 100 s. (a) What is the angular speed, and the linear speed of the motion? (b) Is the acceleration vector a constant vector ? What is its magnitude ? 100 Period, T= s 7 (a) The angular speed ω is given by 2π 2π 2π x7 ω= = 100 = =0.44 rad/s 𝑇 100 7 The linear speed , v =ω R = 0.44 × 0.12 = 5.3 x 10 -2 m s -1 (b) The direction of velocity v is along the tangent to the circle at every point. The acceleration is directed towards the centre of the circle. Since this direction changes continuously, acceleration here is not a constant vector. a = ω 2 R = (0.44 ) 2 x0.12 = 2.3x10 -2 m s -2 Downloaded from hssreporter.com  Chapter 4 Laws of Motion Galileo’s Law of Inertia If the net external force is zero, a body at rest continues to remain at rest and a body in motion continues to move with a uniform velocity. This property of the body is called inertia.Inertia means ‘resistance to change’. Suppose a person is standing in a stationary bus and the driver starts the bus suddenly. He gets thrown backward with a jerk. This is due to his inertia of rest. Similarly if a person is standing in a moving bus and if the bus suddenly stops he is thrown forward. This is due to his inertia of motion. Newton’s Laws of Motion Newton built on Galileo’s ideas and laid the foundation of mechanics in terms of three laws of motion. Galileo’s law of inertia was his starting point on which he formulated as the First Law of motion. Newton’s First Law of Motion (Law of inertia) Every body continues to be in its state of rest or of uniform motion in a straight line unless compelled by some external force to change that state. The state of rest or uniform linear motion both imply zero acceleration. If the net external force on a body is zero, its acceleration is zero. Acceleration can be non zero only if there is a net external force on the body. Momentum Momentum, P of a body is defined to be the product of its mass m and velocity v, and is denoted by p. p=mv Momentum is a vector quantity. Unit = kgm/s [p] = ML T −1 Downloaded from hssreporter.com  ▪ Suppose a light-weight vehicle car) and a heavy weight vehicle are parked on a horizontal road. A greater force is needed to push the truck than the car to bring them to the sam e speed in same time. Similarly, a greater opposing force is needed to stop a heavy body than a light body in the same time, if they are moving with the same speed. ▪ Speed is another important parameter to consider. A bullet fired by a gun can easily pierce human tissue before it stops, resulting in casualty. The same bullet fired with moderate speed will not cause much damage.Thus for a given mass, the greater the speed, the greater is the opposing force needed to stop the body in a certain time. Newton’s Second Law f Motion The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts. ∆𝐩 F∝ ∆𝐭 ∆p F =k ∆t For simplicity we choose k=1 In the limit ∆t →0 𝐝𝐩 F= 𝐝𝐭 Why a cricketer draws his hands backwards during a catch? By Newton's second law of motion , ∆𝐩 F= ∆𝐭 When he draws his hands backwards, the time interval (∆t) to stop the ball increases. Then force decreases and it does not hurt his hands. Force not only depends on the change in momentum but also on how fast the change is brought about. Downloaded from hssreporter.com Derivation of Equation of force from Newton's second law of motion By Newton's second law of motion , For a body of ixed mass m, p=mv = mv dt F = ma Force is a vector quantity Unit of force is kgm ( ) Definition of newton One newton is that which causes an acceleration of m s-2 to a mass of 1kg Important points about second law 1.The second Law is consistent with the first law. Zero acceleration implies the state of rest or uniform linear motion. i.e, when there is no external force , the body will remain in its state of rest or of uniform motion in a straight line. This is Newtons first law of motion. Downloaded from hssreporter.com 3. If a force makes some angle with velocity, the force changes only the component of velocity along the direction of force. 4. For a system of particles we can write 𝐅𝐧𝐞𝐭 = 𝐦𝐚 𝐅𝐧𝐞𝐭 refers to the total external force on the system and a refers to the acceleration of the system as a whole. 5. In equation F=ma Acceleration at any instant is determined by the force at that instant , not by any history of the motion of the particle. Example A bullet of mass 0.04 kg moving with a speed of 90 m/s enters a heavy wooden block and is stopped after a distance of 60 cm. What is the average resistive force exerted by the block on the bullet? Downloaded from hssreporter.com Example Impulse (I) There are some situations where a large force acts for a very short duration producing a finite change in momentum of the body. For example, when a ball hits a wall and bounces back, the force on the ball by the wall acts for a very short time when the two are in contact, yet the force is large enough to reverse the momentum of the ball. Impulse is the the product of force and time duration, which is the change in momentum of the body. Impulse = Force × time duration I=Fxt Unit = kg m s−1 [I] =M L T −1 Downloaded from hssreporter.com Impulsive force. A large force acting for a short time to produce a finite change in momentum is called an impulsive force. Impulse momentum Principle Impulse is equal to the change in momentum of the body. By Newton's second law of motion, dp F= dt F x dt = dp I = dp Impulse = change in momentum Example A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12 m s −1. If the mass of the ball is 0.15 kg, determine the impulse imparted to the ball. Impulse = change of momentum Change in momentum = final momentum – initial momentum Change in momentum = 0.15 × 12–(0.15×-12) Impulse = 3.6 N s Newton’s Third Law of Motion To every action, there is always an equal and opposite reaction. Important points about Third law 1. Forces always occur in pairs. Force on a body A by B is equal and opposite to the force on the body B by A. 2. There is no cause- effect relation implied in the third law. The force on A by B and the force on B by A act at the same instant. By the same reasoning, any one of them may be called action and the other reaction. 3. Action and reaction forces act on different bodies, not on the same body. So they do not cancel each other , eventhough they are equal and opposite. According to the third law, FAB = – FBA (force on A by B) = – (force on B by A) 4. However, if you are considering the system of two bodies as a whole, FAB and FBA are internal forces of the system (A + B). They add up to give a null force. Downloaded from hssreporter.com Weight of a body in a lift Weight of a body is the normal reaction exerted by the surface in contact with that body. Downloaded from hssreporter.com Example A man of mass 70 kg stands on a weighing scale in a lift which is moving, (a) upwards with a uniform speed of 10 m 𝐬 −𝟏 (b) downwards with a uniform acceleration of 5 m 𝐬−𝟐 (c) upwards with a uniform acceleration of 5 m 𝐬−𝟐 What would be the readings on the scale in each case? (d) What would be the reading if the lift mechanism failed and it falls down freely under gravity? Take g=10 m 𝐬−𝟐 (a)When lift moves with uniform speed , a=0 R= mg = 70 x 10=700 N Reading= 700/ 10= 70 kg (b)Acceleration a = 5m s-2 downwards R= m(g-a) =70 ( 10-5)= 70x5= 350N Reading = 350/ 10 = 35 kg (c) Acceleration a = 5m s-2 upwards R= m(g+a) = 70(10+5) = 70x 15= 1050N Reading = 1050/ 10 = 105 kg (d) when lift falls freely a=g R= m(g-g) = 0 Reading = 0 Law of Conservation of Momentum The total momentum of an isolated system of interacting particles is conserved. Or When there is no external force acting on a system of particles ,their total momentum remains constant. Proof of law of conservation of momentum dp By Newton's second law of motion , F= dt When F = 0 dp =0 dt dp = 0 , p=constant Thus when there is no external force acting on a system of particles, their total momentum remains constant. Downloaded from hssreporter.com Applications of law of conservation of linear momentum 1.Recoil of gun When a bullet is fired from a gun , the backward movement of gun is called recoil of the gun. If 𝐩𝐛 and 𝐩𝐠 are the momenta of the bullet and gun after firing 𝐩𝐛 + 𝐩𝐠 = 0 𝐩𝐛 = - 𝐩𝐠 The negative sign shows that the gun recoils to conserve momentum. Expression for Recoil velocity and muzzle velocity Momentum of bullet after firing , 𝐩𝐛 = 𝐦𝐯 Recoil momentum of the gun after firing , 𝐩𝐠 = 𝐌𝐕 𝐩𝐛 = - 𝐩𝐠 𝐦𝐯 = −𝐌𝐕 −𝐦𝐯 Recoil velocity of gun , V= 𝐌 −𝐌𝐕 Muzzle velocity of bullet , v= 𝐦 M= mass of gun, V= recoil velocity of bullet m= mass of bullet, v=muzzle velocity of bullet 2. The collision of two bodies Downloaded from hssreporter.com Total Final momentum = Total initial momentum i.e. , the total final momentum of the isolated system equals its total initial momentum. Equilibrium of a particle Equilibrium of a particle in mechanics refers to the situation when the net external force on the particle is zero. According to the first law, this means that, the particle is either at rest or in uniform motion. Downloaded from hssreporter.com If a number of forces 𝐅𝟏 , 𝐅𝟐 ,……..𝐅𝐧 , act on a particle 𝐅𝟏 + 𝐅𝟐 + … … …... +𝐅𝐧 = 𝟎 Example A mass of 6 kg is suspended by a rope of length 2 m from the ceiling. A force of 50 N in the horizontal direction is applied at the midpoint P of the rope, as shown. What is the angle the rope makes with the vertical in equilibrium ? (Take g = 10 m 𝐬−𝟐 ). Neglect the mass of the rope. Downloaded from hssreporter.com Common Forces in Mechanics There are two types of forces in mechanics- Contact forces and Non contact forces. Contact forces A contact force on an object arises due to contact with some other object: solid or fluid. Eg: Frictional force, viscous force, air resistance Non contact forces A non contact force can act at a distance without the need of any intervening medium. Eg: Gravitational force. Friction The force that opposes (impending or actual) relative motion between two surfaces in contact is called frictional force. There are two types of friction-Static and Kinetic friction Static friction 𝐟𝐬 ▪ Static friction is the frictional force that acts between two surfaces in contact before the actual relative motion starts. Or Static friction fs opposes impending relative motion. The maximum value of static friction is ( fs )max ▪ The limiting value of static friction ( fs )max , is independent of the area of contact. ▪ The limiting value of static friction ( fs )max , varies with the normal force(N) ( fs )max αN ( 𝐟𝐬 )𝐦𝐚𝐱 = 𝛍𝐬 𝐍 Where the constant 𝛍𝐬 is called the coefficient of static friction and depends only on the nature of the surfaces in contact. The Law of Static Friction The law of static friction may thus be written as , fs ≤ 𝛍𝐬 𝐍 0r ( 𝐟𝐬 )𝐦𝐚𝐱 = 𝛍𝐬 𝐍 Downloaded from hssreporter.com Note: If the applied force F exceeds ( fs )max , the body begins to slide on the surface. When relative motion has started, the frictional force decreases from the static maximum value ( fs )max Kinetic friction 𝐟𝐤 Frictional force that opposes (actual) relative motion between surfaces in contact is called kinetic or sliding friction and is denoted by fk.. ▪ Kinetic friction is independent of the area of contact. ▪ Kinetic friction is nearly independent of the velocity. ▪ Kinetic friction , fk varies with the normal force(N) fk αN 𝐟𝐤 = 𝛍𝐤 𝐍 where μk the coefficient of kinetic friction, depends only on the surfaces in contact. μk is less than μs The Law of Kinetic Friction The law of kinetic friction can be written as , 𝐟𝐤 = 𝛍𝐤 𝐍 where μk the coefficient of kinetic friction, Example Downloaded from hssreporter.com Body on an inclined surface The forces acting on a block of mass m When it just begins to slide are (i) the weight mg (ii) the normal force N (iii) the maximum static frictional force ( 𝐟𝐬 )𝐦𝐚𝐱 In equilibrium, the resultant of these forces must be zero. m g sin θ = ( fs )max But ( fs )max = μs N mg sin θ = μs N------------(1) m g cos θ = N-------------(2) (1) mg sin θ μs N Eqn(2) -------- = m g cos θ N 𝛍𝐬 = 𝐭𝐚𝐧 𝛉 𝛉 = 𝐭𝐚𝐧−𝟏 𝛍𝐬 This angle whose tangent gives the coefficient of friction is called angle of friction. μs = tan θ θ = 15°, μs = tan 15° = 0.27 Downloaded from hssreporter.com Applying second law to motion of the block 30 – T = 3a ---------------(1) Apply the second law to motion of the trolley T – fk = 20 a ------------(2) Nowfk = μk N, Here μk = 0.04, N = 20 x 10 = 200 N. Substituting in eq(2) T – 0.04 x 200 = 20 a T – 8 = 20a-------------(3) Solving eqns (1) and (3) 22 a= m s −2 = 0.96 m s−2 23 T = 27.1 N Rolling Friction It is the frictional force that acts between the surfaces in contact when one body rolls over the other. Rolling friction is much smaller than static or sliding friction Disadvantages of friction friction is undesirable in many situations, like in a machine with different moving parts, friction opposes relative motion and thereby dissipates power in the form of heat, etc. Downloaded from hssreporter.com Advantages of friction Friction is a necessary evil.In many practical situations friction is critically needed. Kinetic friction is made use of by brakes in machines and automobiles. We are able to walk because of static friction. It is impossible for a car to move on a very slippery road. On an ordinary road, the friction between the tyres and the road provides the necessary external force to accelerate the car. Methods to reduce friction (1)Lubricants are a way of reducing kinetic friction in a machine. (2)Another way is to use ball bearings between two moving parts of a machine. (3) A thin cushion of air maintained between solid surfaces in relative motion is another effective way of reducing friction. Circular Motion The acceleration of a body moving in a circular path is directed towards the centre and is called centripetal acceleration. 𝐯𝟐 a= 𝐑 The force f providing centripetal acceleration is called the centripetal force and is directed towards the centre of the circle. 𝐦𝐯 𝟐 𝐟𝐬 = 𝐑 where m is the mass of the body, R is the radius of circle. For a stone rotated in a circle by a string, the centripetal force is provided by the tension in the string. The centripetal force for motion of a planet around the sun is the gravitational force on the planet due to the sun. Motion of a car on a curved level road Downloaded from hssreporter.com Three forces act on the car. (i) The weight of the car, mg (ii) Normal reaction, N (iii) Frictional force, fs As there is no acceleration in the vertical direction N= mg The static friction provides the centripetal acceleration mv2 fs = R But , fs ≤ μs N mv2 ≤ μs mg (N=mg) R 𝟐 𝐯 ≤ 𝛍𝐬 𝐑𝐠 𝐯𝐦𝐚𝐱 = √𝛍𝐬 𝐑𝐠 This is the maximum safe speed of the car on a circular level road. Example A cyclist speeding at 18 km/h on a level road takes a sharp circular turn of radius 3 m without reducing the speed. The co-efficient of static friction between the tyres and the road is 0.1. Will the cyclist slip while taking the turn ? v 2 ≤ μs R g R = 3 m, g = 9.8 m s −2 , μs = 0.1. μs R g = 2.94 m2 s−2 v 2 ≤ 2.94 m2 s −2 Here v = 18 km/h = 5 m s−1 i.e., v 2 = 25 m2 s −2 The condition is not obeyed. The cyclist will slip while taking the circular turn. Downloaded from hssreporter.com Motion of a car on a banked road Raising the outer edge of a curved road above the inner edge is called banking of curved roads. Since there is no acceleration along the vertical direction, the net force along this direction must be zero. N cos θ = mg +f sin θ N cos θ - f sin θ = mg --------------(1) The centripetal force is provided by the horizontal components of N andfs. mv2 N sin θ + f cos θ = -------------(2) R Eqn(1) N cos θ − f sin θ mg ----- = mv2 Eqn(2) N sin θ + f cos θ R Dividing throughout by N cos θ f 1 −N tan θ Rg f = tan θ + N v2 f But , = μs for maximum speed N 1 −μs tan θ Rg = tan θ + μs v2 Rg(μs +tan θ ) v2 = 1 −μs tan θ 𝐑𝐠(𝛍𝐬 +𝐭𝐚𝐧 𝛉 ) 𝐯𝐦𝐚𝐱 = √ 𝟏 −𝛍𝐬 𝐭𝐚𝐧 𝛉 This is the maximum safe speed of a vehicle on a banked Curved road. If friction is absent, μs = 0 Then Optimum speed, 𝐯𝐨𝐩𝐭𝐢𝐦𝐮𝐦 = √𝐑𝐠 𝐭𝐚𝐧 𝛉 Downloaded from hssreporter.com Example A circular racetrack of ra dius 300 m is banked at an angle of 15°. If the coefficient of friction between the wheels of a race-car and the road is 0.2, what is the (a) optimum speed of the racecar to avoid wear and tear on its tyres, and (b) maximum permissible speed to avoid slipping ? Downloaded from hssreporter.com  Chapter 5 Work ,Energy and Power The Scalar Product or Dot Product The scalar product or dot product of any two vectors ⃗A and ⃗B, denoted as ⃗A.B ⃗ (read A dot B) is defined as 𝐀⃗ ⋅ ⃗𝐁 ⃗ = 𝐀𝐁 𝐜𝐨𝐬𝛉 where θ is the angle between the two vectors Since A, B and cos θ are scalars, the dot product of A and B is a scalar quantity. Each vector, A and B, has a direction but their scalar product does not have a direction. ⃗ ⋅B A ⃗ = A(B cosθ ) ⃗A ⋅ ⃗B = magnitude of Ax projection of B onto A ⃗A ⋅ ⃗B = (Acosθ )B ⃗A ⋅ ⃗B = magnitude of Bx projection of Aonto B Properties of scalar product ▪ The scalar product follows the commutative law ⃗A ⋅ ⃗B = ⃗B. ⃗A ▪ Scalar product obeys the distributive law ⃗ ⋅ (B A ⃗ +C ⃗ )=A ⃗ ⋅B ⃗ +A ⃗.C ⃗ ⃗A. (λ ⃗B) = λ ( ⃗A ⋅ ⃗B) where λ is a real number. ▪ For unit vectors 𝑖̂, 𝑗̂, 𝑘̂ we have 𝒊̂ ⋅ 𝒊̂ = 𝒋̂ ⋅ 𝒋̂ = 𝒌̂⋅𝒌 ̂=1 ̂=𝒌 𝒊̂ ⋅ 𝒋̂ = 𝒋̂ ⋅ 𝒌 ̂ ⋅ 𝒊̂ = 0 ▪ For two vectors A ̅ = Ax î+ Ay ĵ + Az k̂ ̅ = Bx î+ By ĵ + Bz k̂ B Downloaded from hssreporter.com  ⃗A ⋅ ⃗B =(Ax î+ Ay ĵ + Az k̂). (Bx î+ By ĵ + Bz k̂) ⃗ ⋅ ⃗𝐁 𝐀 ⃗ = 𝐀 𝐱 𝐁 𝐱 + 𝐀 𝐲 𝐁 𝐲 + 𝐀 𝐳 𝐁𝐳 ▪ ⃗A ⋅ ⃗A = Ax Ax + Ay Ay + Az Az ⃗A ⋅ ⃗A = Ax 2 + Ay 2 + Az 2 = A2 ▪ ⃗A ⋅ ⃗A = A A cos 0 = A2 ▪ If ⃗A and ⃗B are perpendicular ⃗A ⋅ ⃗B = A B cos 90 = 0 Example Find the angle between force ⃗F = (3î+ 4ĵ - 5k̂) unit and displacement ⃗d = (5î+ 4ĵ + 3k̂) unit. Also find the projection of F on d. ⃗F. ⃗d = 𝐹𝑑𝑐𝑜𝑠𝜃 ⃗ ⃗F.d cosθ = ----------(1) Fd ⃗F. ⃗d = 𝐹𝑥 𝑑𝑥 + 𝐹𝑦 𝑑𝑦 + 𝐹𝑧 𝑑𝑧 = (3x5 )+ (4x4 ) + (-5 x3) 𝐅. 𝐝 = 16 unit F =√Fx 2 + Fy 2 + Fz 2 =√32 + 42 + (−5)2 = √9 + 16 + 25 F = √50 unit d =√dx 2 + dy 2 + dz 2 =√52 + 42 + 32 = √25 + 16 + 9 d = √50unit Substituting the values in eq(1) 16 16 cos𝜃 = = = 0.32 √50 √50 50 𝜽 = 𝐜𝐨𝐬 −1 0.32 The projection of F on d = F cosθ=√50 x 0.32 =2.26 Downloaded from hssreporter.com Work Consider a constant force F acting on an object of mass m. The object undergoes a displacement d in the positive x-direction The work done by the force is defined to be the product of component of the force in the direction of the displacement and the magnitude of this displacement. W = (F cos θ )d W = F d cos θ W = ⃗⃗⃗ 𝐅 ⋅𝐝 Work can be zero, positive or negative. Zero Work The work can be zero,if (i)the displacement is zero. When you push hard against a rigid brick wall, the force you exert on the wall does no work. A weightlifter holding a 150 kg mass steadily on his shoulder for 30 s does no work on the load during this time. (ii) the force is zero. A block moving on a smooth horizontal table is not acted upon by a horizontal force (since there is no friction), but may undergo a large displacement. (iii) the force and displacement are mutually perpendicular Here θ = 90 o , cos (90) = 0. For the block moving on a smooth horizontal table, the gravitational force mg does no work since it acts at right angles to the displacement. Positive Work If θ is between 0 o and 90 o , cos θ is positive and work positive. Eg: Workdone by Gravitational force on a freely falling body is positive Negative work If θ is between 90 o and 180 o , cos θ is negative and work negative. Eg: the frictional force opposes displacement and θ = 180 o. Then the work done by friction is negative (cos 180 o = –1). Downloaded from hssreporter.com Units of Work and Energy ▪ Work and Energy are scalar quantities. ▪ Work and energy have the same dimensions, [ML 2 T –2 ]. ▪ The SI unit is kgm2s-2 or joule (J), named after the famous British physicist James Prescott Joule. Alternative Units of Work/Energy in J Example Work done by a Variable Force If the displacement Δx is small, we can take the force F (x) as approximately constant and the work done is then ΔW =F(x) Δx Downloaded from hssreporter.com  x2 W = ∫ F(x) Δx x1 In the limit Δx tends to zero 𝐱𝟐 W = ∫ 𝐅(𝐱) 𝐝𝐱 𝐱𝟏 Kinetic Energy The kinetic energy is the energy possessed by a body by virtue of its motion. If an object of mass m has velocity v, its kinetic energy K is 𝟏 𝟏 K = 𝐦𝐯̅ ⋅ 𝐯̅ = 𝐦𝐯 𝟐 𝟐 𝟐 Kinetic energy is a scalar quantity. Example In a ballistics demonstration a police officer fires a bullet of mass 50.0 g with speed 200 m s-1 on soft plywood of thickness 2.00 cm. The bullet emerges with only 10% of its initial kinetic energy. What is the emergent speed of the bullet ? The Work-Energy Theorem The work-energy theorem can be stated as :The change in kinetic energy of a particle is equal to the work done on it by the net force. Proof For uniformly accelerated motion v 2 − u 2 = 2 as 1 Multiplying both sides by 𝑚, we have 2 1 1 mv − mu2 = mas = Fs 2 2 2 Kf -Ki = W Change in KE = Work Downloaded from hssreporter.com Potential Energy Potential energy is the ‘stored energy’ by virtue of the position or configuration of a body. ▪ A body at a height h above the surface of earth possesses potential energy due to its position. ▪ A Stretched or compressed spring possesses potential energy due to its state of strain. Gravitational potential energy of a body of mass m at a height h above the surface of earth is mgh. Gravitational Potential Energy , V =mgh Show that gravitational potential energy of the object at height h, is equal to the kinetic energy of the object on reaching the ground, when the object is released. PE at a height h, V = mgh----------(1) When the object is released from a height it gains KE K = ½ mv 2 v 2 = u2 + 2as u=0, a=g , s=h 2 v = 2gh K = ½ m x 2gh K= mgh----------(2) From eq(1) and (2) Kinetic energy= Potential energy Conservative Force A force is said to be conservative, if it can be derived from a scalar quantity. − dV F= where V is a scalar dx Eg: Gravitational force, Spring force. ▪ The work done by a conservative force depends only upon initial and final positions of the body ▪ The work done by a conservative force in a cyclic process is zero Note: Frictional force , air resistance are non conservative forces. The Conservation of Mechanical Energy The principle of conservation of total mechanical energy can be stated as, The total mechanical energy of a system is conserved if the forces, doing work on it, are conservative. Downloaded from hssreporter.com Conservation of Mechanical Energy for a Freely Falling Body Consider a body of mass m falling freely from a height h At Point A PE = mgh KE = 0 (since v=0) TE = PE + KE = mgh + 0 TE = mgh-----------(1) At Point B PE = mg (h-x) KE = ½ mv 2 v 2 = u2 + 2as u=0, a=g , s= x 2 v = 2gx KE = ½ m x 2gx KE= mgx TE = PE + KE TE = mg (h-x) + mgx TE = mgh--------------(2) At Po int C PE = 0 (Since h=0) KE = ½ mv 2 v 2 = u2 + 2as u=0, a=g , s= h 2 v =2gh KE = ½ m x 2gh KE= mgh TE = PE + KE TE = 0 + mgh TE = mgh--------------(3) From eqns (1), (2) and (3), it is clear that the total mechanical energy is conserved during the free fall. Graphical variation of KE and PE with height from ground Downloaded from hssreporter.com Hooke’s Law Hooke’s law states that ,for an ideal spring, the spring force F is proportional displacement x of the block from the equilibrium position. F = − kx The displacement could be either positive or negative. The constant k is called the spring constant. Its unit is N𝑚−1 The spring is said to be stiff if k is large and soft if k is small. The Potential Energy of a Spring Consider a block of mass m attached to a spring and resting on a smooth horizontal surface. The other end of the spring is attached to a rigid wall. Let the spring be pulled through a distance x. Then the spring force F = − kx The work done by the spring force is x W = ∫0 F dx x W = − ∫0 kx dx 1 W = − kx 2 2 This work is stored as potential energy of spring 𝟏 𝐏𝐄 = 𝐤𝐱 𝟐 𝟐 Conservation of Mechanical Energy of an Oscillating Spring Consider a spring oscillating between −𝑥𝑚 and 𝑥𝑚. Downloaded from hssreporter.com At any point x between −𝑥𝑚 𝑎𝑛𝑑 𝑥𝑚 , the total mechanical energy of the spring TE = PE +KE 1 1 1 k𝐱 𝐦 2 = kx 2 + mv 2 2 2 2 At equilibrium position x=0, 1 1 k𝐱 𝐦 2 = mv 2 2 2 k v=√ xm m ▪ At equilibrium position PE is zero and KE is max. ▪ At extreme ends, the PE is maximum and KE is zero. ▪ The kinetic energy gets converted to potential energy and vice versa, however, the total mechanical energy remains constant. Graphical variation of kinetic Energy and potential of a spring Power Power is defined as the time rate at which work is done or energy is transferred. The average power of a force is defined as the ratio of the work, W, to the total time t taken. 𝐖 𝐏𝐚𝐯 = 𝐭 The instantaneous power The instantaneous power is defined as the limiting value of the average power as time interval approaches zero. 𝐝𝐖 P= 𝐝𝐭 The work done, dW = F. dr. dr P=F. dt P= F. v where v is the instantaneous velocity when the force is F. ▪ Power, like work and energy, is a scalar quantity. ▪ Its dimensions are ML2 T −3. Downloaded from hssreporter.com  ▪ SI unit of power is called a watt (W). 1W = 1 J/s ▪ The unit of power is named after James Watt. ▪ Another unit of power is the horse-power (hp) 1 hp = 746 W This unit is still used to describe the output of automobiles, motorbikes, etc kilowatt hour Electrical energy is measured in kilowatt hour (kWh). 1kWh = 3.6 × 𝟏𝟎𝟔 J Note: A 100 watt bulb which is on for 10 hours uses 1 kilowatt hour (kWh) of energy. Energy = Power x Time =100 (watt) × 10 (hour) = 1000 watt hour = =1 kilowatt hour (kWh) = 103 (W) × 3600 (s) = 3.6 × 106 J Problem An elevator can carry a maximum load of 1800 kg (elevator + passengers) is moving up with a constant speed of 2 m s–1. The frictional force opposing the motion is 4000 N. Determine the minimum power delivered by the motor to the elevator in watts as well as in horse power. The downward force on the elevator is F = m g + Frictional Force = (1800 × 10) + 4000 = 22000 N Power, P = F. v = 22000 × 2 = 44000 W In horse power, power = 44000/746 =59 hp Collisions In all collisions the total linear momentum is conserved; the initial momentum of the system is equal to the final momentum of the system. There are two types of collisions Elastic and Inelastic. Downloaded from hssreporter.com Elastic Collisions The collisions in which both linear momentum and kinetic energy are conserved are called elastic collisions. Eg: Collision between sub atomic particles Inelastic Collisions The collisions in which linear momentum is conserved, but kinetic energy is not conserved are called inelastic collisions.. Part of the initial kinetic energy is transformed into other forms of energy such as heat,sound etc.. Eg: Collision between macroscopic objects A collision in which the two particles move together after the collision is a perfectly inelastic collision. Elastic Collisions in One Dimension If the initial velocities and final velocities of both the bodies are along the same straight line, then it is called a one-dimensional collision, or head-on collision. Consider two masses m1 and m2 making elastic collision in one dimension. By the conservation of momentum m1 u1 + m2 u2 = m1 v1 + m2 v2 --------------(1) m1 u1 − m1 v1 = m2 v2 − m2 u2 m1 (u1 − v1 ) = m2 (v2 − u2 )----------------(2) By the conservation of kinetic energy 1 1 1 1 m1 u12 + m2 u22 = m1 v12 + m2 v22 -----------(3) 2 2 2 2 1 1 1 1 m1 u12 − m1 v12 = m2 v22 − m2 u22 2 2 2 2 1 1 m1 (u12 − v12 ) = m2 (v22 − u22 ) 2 2 m1 (u12 − v12 ) = m2 (v22 − u22 ) -------------(4) (4) m1 (u21 −v21 ) m2 (v22 −u22 ) Eqn ------------ = (2) m1 (u1 −v1 ) m2 (v2 −u2 ) Downloaded from hssreporter.com  (u1 +v1 ) (u1 −v1 ) (v2 +u2 )(v2 −u2 ) = (u1 −v1 ) (v2 −u2 ) u1 + v1 = v2 + u2 -------------(5) 𝐮1 − 𝐮2 = −(𝐯1 − 𝐯2 )--------(6) i.e., relative velocity before collision is numerically equal to relative velocity after collision. From eqn(5), v2 = u1 + v1 − u2 Substituting in eqn (1) m1 u1 + m2 u2 = m1 v1 + m2 (u1 + v1 − u2 ) m1 u1 + m2 u2 = m1 v1 + m2 u1 + m2 v1 − m2 u2 m1 u1 + m2 u2 − m2 u1 + m2 u2 = m1 v1 + m2 v1 (m1 − m2 )u1 + 2m2 u2 = (m1 + m2 )v1 (𝐦1 −𝐦2 )𝐮1 2𝐦2 𝐮2 𝐯1 = + -------- (7) 𝐦1 +𝐦2 𝐦1 +𝐦2 (𝐦2 −𝐦1 )𝐮2 2𝐦1 𝐮1 Similarly, 𝐯2 = + ------- (8) 𝐦1 +𝐦2 𝐦1 +𝐦2 Case 1 -If two masses are equal, 𝐦1 = 𝐦2 = 𝒎 Substituting in eqns (7) and (8) 2mu2 v1 = = u2 2m 2mu1 v2 = =u1 2m ie.,the bodies will exchange their velocities Case 2- If one mass dominates, 𝐦2 > > 𝐦1 and 𝐮2 = 0 m1 + m2 = m2 𝑎𝑛𝑑 m1 − m2 = −m2 ( m1 −m2 )u1 m2 u 1 v1 = =− = -u1 m1 +m2 m2 2m1 u1 2 x 0 x u1 v2 = = =0 m1 +m2 m2 (since m1 is very small , it can be neglected) The heavier mass comes to rest while the lighter mass reverses its velocity. Downloaded from hssreporter.com Elastic Collisions in Two Dimensions Consider the elastic collision of a moving mass m1 with the stationary mass m2. Since momentum is a vector ,it has 2 equations in x and y directions. Equation for conservation of momentum in x direction 𝐦1 𝐮1 = 𝐦1 𝐯1 𝐜𝐨𝐬𝛉1 + 𝐦2 𝐯2 𝐜𝐨𝐬𝛉2 Equation for conservation of momentum in y direction 0 = 𝐦1 𝐯1 𝐬𝐢𝐧𝛉1 − 𝐦2 𝐯2 𝐬𝐢𝐧 𝛉2 Equation for conservation of kinetic energy,(KE is a scalar quantity) 𝟏 𝟏 𝟏 𝐦𝟏 𝐮𝟐𝟏 = 𝐦𝟏 𝐯𝟏𝟐 + 𝐦𝟐 𝐯𝟐𝟐 𝟐 𝟐 𝟐 Downloaded from hssreporter.com  Chapter 6 Systems of Particles and Rotational Motion Rigid Body Ideally a rigid body is a body with a perfectly definite and unchanging shape. The distances between different pairs of such a body do not change. Motion of a rigid body The motion of a rigid body which is not pivoted or fixed in some way is either a pure translation or a combination of translation and rotation. The motion of a rigid body which is pivoted or fixed in some way is rotation. 1)Pure Translational Motion In pure translational motion at any instant of time every particle of the body has the same velocity. Eg: A block moving down an inclined plane. Any point like P1 or P2 of the block moves with the same velocity at any instant of time. 2)Pure Rotational Motion In pure rotational motion at any instant of time every point in the rotating rigid body has the same angular velocity,but different linear velocity. i) Rotation about a fixed axis Eg: A ceiling fan A potter’s wheel. The line along which the body is fixed is termed as its axis of rotation. In rotation of a rigid body about a fixed axis, every particle of the body moves in a circle, which lies in a plane perpendicular to the axis and has its centre on the axis. Downloaded from hssreporter.com ii) Rotation about an axis which is not fixed Eg: A spinning top An oscillating table fan 3)Rolling Motion It is a combination of translational and rotational motion. Eg A solid cylinder moving down an inclined plane. Points P1, P2, P3 and P4 have different velocities at any instant of time. In fact, the velocity of the point of contact P3 is zero at any instant, if the cylinder rolls without slipping. Centre Of Mass The centre of is a hypothetical point where the entire mass of an object may be assumed to be concentrated to visualise its motion. Consider a two particle system. Let C be the centre of mass which is at a distancev X from origin. ⃗⃗ = 𝐦𝟏 𝐫𝟏+𝐦𝟐𝐫𝟐 𝐑 𝒎𝟏 +𝒎𝟐 ⃗⃗ = 𝐦𝟏𝐫𝟏+𝐦𝟐𝐫𝟐 𝐑 where M=𝒎𝟏 + 𝒎𝟐 𝐌 Downloaded from hssreporter.com  𝑚1 𝑥1 +𝑚2 𝑥2 x coordinate of centre of mass 𝑋= 𝑚1 +𝑚2 𝑚1 𝑦1 +𝑚2 𝑦2 y coordinate of centre of mass 𝑌= 𝑚1 +𝑚2 𝑚1 𝑧1 +𝑚2 𝑧2 z coordinate of centre of mass 𝑍= 𝑚1 +𝑚2 If we have n particles of masses 𝑚1 , 𝑚2 ,... 𝑚𝑛 ⃗ = 𝐦𝟏𝐫𝟏 +𝐦𝟐𝐫𝟐 +⋯……..+𝐦𝐧 𝐫𝐧 ⃗𝐑 --------(1) 𝐌 ⃗⃗ = ∑𝐦𝐢 𝐫𝐢 𝐑 where M =m1 + m2 +…….+mn 𝐌 If the origin is chosen to be the centre of mass then𝑅⃗ = 0 ∑mi r⃗i 0= M ∑mi ri = 0 Example Find the centre of mass of three particles at the vertices of an equilateral triangle. The masses of the particles are 100g, 150g, and 200g respectively. Each side of the equilateral triangle is 0.5m long. 𝑚1 𝑦1 +𝑚2 𝑦2 +𝑚3 𝑦3 𝑌= 𝑚1 +𝑚2 +𝑚3 Downloaded from hssreporter.com Motion of Centre of Mass Position vector of centre of mass ⃗ = ∑mi r⃗i R M 𝐦𝟏 𝐫𝟏 +𝐦𝟐 𝐫𝟐 +⋯……..+𝐦𝐧 𝐫𝐧 ⃗𝐑 ⃗ = --------(1) 𝐌 where M =m1 + m2 +…….+mn Velocity of centre of mass Differentiating d ⃗R = d { m1 r⃗1 +m2 r⃗2 +⋯……..+mn r⃗n } dt dt M d d d m1 r1 + m2 r2 + ⋯ ….. +mn rn ⃗V = dt dt dt M 𝐦𝟏 𝐯⃗𝟏 +𝐦𝟐 𝐯⃗𝟐 +⋯……..+𝐦𝐧 𝐯⃗𝐧 ⃗ = 𝐕 --------------(2) 𝐌 Acceleration of centre of mass Differentiating d ⃗ 1 +m2 v d m1 v ⃗ 2 +⋯……..+mn v ⃗n ⃗V = { } dt dt M d d d m1 v ⃗ 1 + m2 v ⃗ 2 + ⋯ ….. +mn v ⃗n ⃗ = dt dt dt A M 𝐦𝟏 𝐚⃗𝟏 +𝐦𝟐 𝐚⃗𝟐 +⋯……..+𝐦𝐧 𝐚⃗𝐧 ⃗⃗⃗𝐀 = ………………(3) 𝐌 Force on centre of mass Acceleration of centre of mass m a⃗ + m2 a⃗2 + ⋯ … ….. +mn a⃗n ⃗A = 1 1 M MA⃗ = m1 a⃗1 + m2 a⃗2 + ⋯ … ….. +mn a⃗n 𝐅𝐞𝐱𝐭 = m1 a⃗1 + m2 a⃗2 + ⋯ … ….. +mn a⃗n 𝐅𝐞𝐱𝐭 = 𝐅𝟏 + 𝐅𝟐 + ⋯ … ….. +𝐅𝐧 ⃗ 𝐅𝐞𝐱𝐭 = MA The centre of mass of a system of particles moves as if all the mass of the system was concentrated at the centre of mass and all the external forces were applied at that point. Downloaded from hssreporter.com The centre of mass of the fragments of the projectile continues along the same parabolic path which it would have followed if there were no explosion. Linear Momentum of centre of mass Velocity of centre of mass ⃗ 1 +m2 v m1 v ⃗ 2 +⋯……..+mn v ⃗n ⃗V = M ⃗ = m1 v MV ⃗ 1 + m2 v ⃗ 2 + ⋯ … ….. +mn v ⃗n ⃗ = 𝐩 𝐏 ⃗𝟏+𝐩 ⃗ 𝟐 + ⋯ … ….. +𝐩 ⃗𝐧 Law of Conservation of Momentum for a System of Particles If Newton’s second law is extended to a system of particles, d⃗P ⃗Fext = dt When the sum of external forces acting on a system of particles is zero ⃗Fext = 0 d⃗P =0 dt ⃗ = constant P Thus, when the total external force acting on a system of particles is zero, the total linear momentum of the system is constant. This is the law of conservation of the total linear momentum of a system of particles. But ⃗P = MV⃗ MV⃗ = constant ⃗V = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 When the total external force on the system is zero the velocity of the centre of mass remains constant or the CM of the system is in uniform motion. Downloaded from hssreporter.com Vector Product or Cross product of Two Vectors Vector product of two vectors A ⃗ and B⃗ is defined as ⃗ x ⃗𝐁 𝐀 ⃗ = AB 𝐬𝐢𝐧 𝜽 𝐧 ̂ where A and B are magnitudes of ⃗A and ⃗B 𝜽 is the angle between ⃗A and ⃗B 𝑛̂ is the unit vector perpendicular to the plane containing A ⃗ and B ⃗ The direction of ⃗A x ⃗B is given by right hand screw rule or right hand rule Right hand screw rule If we turn the head of screw in the direction from ⃗A to ⃗B ,then the tip of the screw advances in the direction ofA⃗ x ⃗B Right hand rule if the fingers of right hand are curled up in the direction from ⃗A to ⃗B, then the stretched thumb points in the direction of ⃗A x ⃗B ▪ The vector product is not commutative ⃗ x ⃗𝐁 𝐀 ⃗ ≠ ⃗𝐁 ⃗ x 𝐀 ⃗ ▪ Vector product obeys distributive law ⃗ x ⃗⃗⃗⃗ 𝐀 (𝐁 + 𝐂 )= 𝐀⃗ x ⃗𝐁 ⃗ +𝐀 ⃗ 𝐱𝐂 ⃗ = ⃗𝟎 ⃗ x𝐀 ▪ 𝐀 ▪ 𝒊̂ × 𝒊̂ = 𝟎 , 𝒋̂ × 𝒋̂ = 𝟎 , ̂×𝒌 𝒌 ̂=𝟎 ̂, ▪ 𝒊̂ × 𝒋̂ = 𝒌 ̂ = 𝒊̂, 𝒋̂ × 𝒌 ̂ × 𝒊̂ = 𝒋̂ 𝒌 ̂, ▪ 𝒋̂ × 𝒊̂ = −𝒌 ̂ × 𝒋̂ = −𝒊̂, 𝒌 ̂ = −𝒋̂ 𝒊̂ × 𝒌 Downloaded from hssreporter.com Angular Velocity and its Relation with Linear Velocity ⃗⃗⃗ is directed along the fixed axis The angular velocity is a vector quantity. 𝝎 as shown. The linear velocity of the particle is ⃗⃗⃗ 𝒗 =𝝎⃗⃗⃗ × 𝒓 ⃗ ⃗⃗⃗ and 𝒓 It is perpendicular to both 𝝎 ⃗ and is directed along the tangent to the circle described by the particle. Figure shows the direction of angular velocity when the body rotates in clockwise and anti clockwise direction. For rotation about a fixed axis, the direction of the vector ω does not change with time. Its magnitude may change from instant to instant. For the more general rotation, both the magnitude and the direction of ω may change from instant to instant. Angular acceleration Angular acceleration α ⃗ is defined as the time rate of change of angular velocity. ⃗⃗⃗ 𝐝𝛚 ⃗ = 𝛂 𝐝𝐭 If the axis of rotation is fixed, the direction of ω and hence, that of α is fixed. In this case the vector equation reduces to a scalar equation dω α= dt Downloaded from hssreporter.com Torque or Moment of Force The rotational analogue of force is torque or moment of force. If a force ⃗⃗⃗ 𝐅 acts on a single particle at a point P whose position with respect to the origin O is 𝒓⃗ ,then torque about origin o is ⃗ = r F sinθ 𝝉 ⃗ x ⃗⃗⃗ ⃗𝝉 = 𝒓 𝐅 ▪ Torque has dimensions M L2 T −2 ▪ Its dimensions are the same as those of work or energy. ▪ It is a very different physical quantity than work. ▪ Moment of a force is a vector, while work is a scalar. ▪ The SI unit of moment of force is Newton-metre (Nm) The magnitude of the moment of force may be written τ = (r sin θ ) F = 𝒓⊥ F τ = r (F sin θ ) = r 𝐅⊥ where r⊥ = r sin θ is the perpendicular distance of the line of action of F form the origin and F⊥ = Fsin θ is the component of F in the direction perpendicular to r. Angular momentum of a particle Angular momentum is the rotational analogue of linear momentum. Angular momentum is a vector quantity. It could also be referred to as moment of (linear) momentum. 𝒍=𝒓 ⃗ ×𝒑⃗ 𝒍 = 𝒓𝒑𝐬𝐢𝐧 𝛉 Downloaded from hssreporter.com Relation connecting Torque and Angular momentum 𝑙 =𝑟×𝑝 Differentiating 𝑑𝑙 d = (r×p ⃗ ) 𝑑𝑡 dt 𝑑𝑙 dr⃗ ⃗ dp = ×p ⃗ + rx 𝑑𝑡 dt dt dr⃗ ⃗ dp ⃗ = mv p ⃗ , =v ⃗ , = ⃗⃗F dt dt 𝑑𝑙 =v ⃗ + r x ⃗⃗F ⃗ × mv 𝑑𝑡 ⃗ = 0 , (r x ⃗⃗F =τ⃗ ) ⃗ ×v v 𝑑𝑙 =0 + 𝜏 𝑑𝑡 ⅆ𝒍 ⃗ = 𝝉 ⅆ𝒕 Thus, the time rate of change of the angular momentum of a particle is equal to the torque acting on it. ⃗ dp This is the rotational analogue of the equation ⃗⃗F = , which expresses dt Newton’s second law for the translational motion of a single particle. Relation connecting Torque and Angular momentum for a system of particles ⃗ ⅆ𝑳 ⃗𝝉 = ⅆ𝒕 ⃗ where 𝐿 = 𝑙1 + 𝑙2 + ⋯ ⋅ +𝑙𝑛 Law of Conservation of Angular momentum For a system of particles ⃗ dL τ⃗ext = dt If external torque τ⃗ext = 0 , ⃗ dL =0 dt 𝐋 = 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 If the total external torque on a system of particles is zero, then the total angular momentum of the system is conserved i.e, remains constant. Downloaded from hssreporter.com Example Find the torque of a force 𝟕𝐢̂ + 𝟑𝐉̂ − 𝟓𝐤̂ about the origin. The force acts on a ̂. particle whose position vector is 𝐢̂ − 𝐉̂ + 𝐤 𝜏 = 𝑟 x ⃗⃗F 𝜏 = (î − Ĵ + k̂) x(7î + 3Ĵ − 5k̂) + - + î 𝐽̂ 𝑘̂ 𝜏 =|1 −1 1 | 7 3 −5 ⃗τ = î [(−1 x − 5) −(3 x 1) ] - Ĵ [(1x-5)- (7x1)] + k̂ [(1x3) -(7x-1)] τ⃗ = î [5 - 3] - Ĵ [-5 – 7] + k̂ [3 – -7] τ⃗ = 2î +12 Ĵ + 10k̂ Equilibrium of a Rigid Body A rigid body is said to be in mechanical equilibrium, if it is in both translational equilibrium and rotational equilibrium. i.e, for a body in mechanical equilibrium its linear momentum and angular momentum are not changing with time. Translational Equilibrium When the total external force on the rigid body is zero, then the total linear momentum of the body does not change with time and the body will be in translational equilibrium. Rotational Equilibrium When the total external torque on the rigid body is zero, the total angular momentum of the body does not change with time and the body will be in rotational equilibrium. Partial equilibrium A body may be in partial equilibrium, i.e., it may be in translational equilibrium and not in rotational equilibrium, or it may be in rotational equilibrium and not in translational equilibrium. Downloaded from hssreporter.com Here net torque is zero and the body is in rotational equilibrium. Net force is not zero and the body is not in traslational equilibrium Here net torque is not zero and the body will not be rotational equilibrium. Net force is zero and the body will be in traslational equilibrium. Couple A pair of equal and opposite forces with different lines of action is known as a couple. A couple produces rotation without translation. Our fingers apply a couple to turn the lid The Earth’s magnetic field exerts equal and opposite forces on the poles of a compass needle. These two forces form a couple. Principles of Moments The lever is a system in mechanical equilibrium. For rotational equilibrium the sum of moments must be zero, 𝐝𝟏 𝐅𝟏 − 𝐝𝟐 𝐅𝟐 = 0 The equation for the principle of moments for a lever is 𝐝𝟏 𝐅𝟏 = 𝐝𝟐 𝐅𝟐 load arm × load = effort arm × effort 𝐹1 𝑑2 Mechanical Advantage MA = 𝐹2 = 𝑑1 Downloaded from hssreporter.com Centre of gravity The Centre of gravity of a body is the point where the total gravitational torque on the body is zero. ▪ The centre of gravity of the body coincides with the centre of mass. For a body is small, g does not vary from one point of the body to the other. Then the centre of gravity of the body coincides with the centre of mass. ▪ If the body is so extended that g varies from part to part of the body, then the centre of gravity and centre of mass will not coincide. Moment of Inertia Moment of Inertia is the rotational analogue of mass. Moment of inertia is a measure of rotational inertia The moment of inertia of a particle of mass m rotating about an axis is I =m𝐫 𝟐 The moment of inertia of a rigid body is The moment of inertia of a rigid body depends on The mass of the body, its shape and size Di