Rotational Dynamics PDF

# Rotational Dynamics ## Circular Motion - Motion of an object along the circumference of a circle is called circular motion. - **Examples:** - Motion of the earth around the sun - Motion of the moon around the sun - Motion of planets around the sun - Motion of an object tied at th...

# Rotational Dynamics ## Circular Motion - Motion of an object along the circumference of a circle is called circular motion. - **Examples:** - Motion of the earth around the sun - Motion of the moon around the sun - Motion of planets around the sun - Motion of an object tied at the end of a string and whirled in a circle. ## Characteristics of Circular Motion: - It is an accelerated motion. - It is a periodic motion. ## Kinematics of Circular Motion: - **Angular displacement:** \\(\theta = \frac{Arc}{Radius} = \frac{s}{r}\\) - It is a vector quantity. - Its SI unit is radian. - It is a dimensionless quantity. - **Angular velocity:** \\(\omega = \frac{Angular Displacement }{Time} = \lim_{\delta t \to 0} \frac {\delta \theta }{\delta t} or \omega = \frac{d \theta}{dt} \\) - It is a vector quantity. - Its SI unit is rad/s. - Its dimensions are [M^0 L^0 T^-1]. - **Angular acceleration:** \\(\alpha = \frac{Change in angular velocity}{Time} = \lim_{\delta t \to 0} \frac {\delta \omega }{\delta t} or \alpha = \frac{d \omega}{dt} \\) - It is a vector quantity. - Its SI unit is rad/s². - Its dimensions are [M^0 L^0 T^-2]. - **The quantities angular displacement, angular velocity, and angular acceleration are analogous to linear displacement, linear velocity, and linear acceleration of linear motion.** - **The direction of angular displacement, angular velocity, and angular acceleration can be determined by the right-hand rule.** ## Uniform Circular Motion - Motion of an object along the circumference of the circle with constant speed is called uniform circular motion. - **Examples:** - Motion of artificial satellites around the earth - Motion of electrons around its nucleus - Motion of blades of the windmills - Tip of second's hand of a watch with circular dial. ## Non-uniform Circular Motion - In circular motion if speed of an object is not constant then it is called as non-uniform circular motion. - **Examples:** - Oscillation of a pendulum - Motion of a train - A person jogging in the park ## Relation between Linear velocity & Angular velocity: - **Linear velocity = An Radius X Angular velocity.** - \\(v = r \omega \\) - **In vector form, \\(v = \omega * r\\)** ## Period and Frequency: - **Period (T) = Circumference / Linear velocity** - \\(T = \frac{2 \pi r}{v} = \frac{2 \pi}{\omega}\\) - SI unit of Period is second (S). - **Frequency (n) = 1 / Period** - \\( n = \frac{1}{T} \\) - SI unit of frequency is Hertz (Hz). ## Centripetal Force (CPF): - The force acting on a particle performing UCM, which is along the radius of the circle and directed towards the center of the circle. - **Magnitude of CPF = \\(mv^2/r = mr \omega^2\\)** - SI unit = N - Dimensions = [M^1 L^1 T^-2] ## Centrifugal Force (CCFF): - Centrifugal force is a pseudo force in UCM. It acts along the radius of circle and directed away from the center of the circle. - **Magnitude of CFF = \\(mv^2/r = mr \omega^2\\)** - SI unit = N - Dimensions = [M^1 L^1 T^-2] - **There are two ways of writing force egn for a CM. \\(mw^2 + (Read Force) = 0\\) or \\(mv^2/r + \frac{Read Force}{r} = 0\\)** - **Resultant force = -mw^2** ## Applications of UCM: - **Vehicle along a horizontal circular track:** - For a car moving on a Horizontal circular track of radius 'r', plane of figure is a vertical plane perpendicular to track. - **Forces acting on the car are:** - **Weight (mg)** vertically downwards - **Normal reaction (N)** vertically upwards that balance the weight (mg). - **Force of static friction (fs) between road & tires.** This is static friction (fs) because it prevents the vehicle from outward slipping or skidding. This is the resultant force which is centripetal. *Diagram of a car on a horizontal road with forces acting on it:* [Diagram of a car on a horizontal road with forces acting on it] - **While working in the frame of reference attached to the vehicle, it balances the centrifugal force.** - \\(N = mg\\) - \\(f_s = \frac{mv^2}{r}\\) - **Dividing ② by ①** - \\(f_s/N = v^2/rg\\) - **v = √(rg f_s/N)** - **Vmax = √(urg)** as \\(f_s = \mu_s N\\) - **Well (or wall) of Death:** - This is a vertical cylindrical wall of radius 'r' inside which a vehicle is driven in horizontal circles. This can be seen while performing stunts. - **The forces acting on the vehicle are:** - **Normal reaction (N)** acting horizontally towards the center - **Weight (mg)** acting vertically downwards - **Force of static friction (fs) acting vertically upwards between walls & tires.** It has to prevent the downward slipping. Its magnitude is equal to mg in upward direction. *Diagram of a car inside the wall of death with forces acting on it:* [Diagram of a car inside the wall of death with forces acting on it] - **N = \\(mv^2/r\\) & mg = f_s** - **Force of static friction (f_s) is always less than or equal to μs N.** - \\(f_s ≤ μ N\\) - \\(mg ≤ μs (mv^2/r)\\) - \\(g ≤ μs v^2/r\\) - \\(v^2 ≥ \frac{gr}{\mu_s}\\) - **Vmin = √(gr/μ_s)** - **Vehicle on a Banked Road:** - **Banking of roads:** - The arrangement of keeping the outer edge of the road surface inclined with the horizontal is called banking of roads. - **Angle of banking:** - The angle made by the inclined road surface with the horizontal is called angle of banking. *Diagram of a car on a banked road:* [Diagram of a car on a banked road] - **Fig shows the vertical section of vehicle on curved road of radius 'r' banked at an angle θ with the horizontal** - Consider, the vehicle to be a point and ignoring friction and non-conservative forces like air resistance. - **There are two forces acting on a vehicle:** - **Weight (mg)** vertically downwards - **Normal reaction (N) perpendicular to surface of road.** - **N resolve into two components:** - **Nsino - horizontal component being resultant force, must be the necessary CP force.** - **Ncoso - vertical component balances weight (mg).** - **Nsino = \\(mv^2/r\\)** - **Ncoso = mg** - Dividing eqn ② by ①, - **tanθ = \\(v^2/rg\\)** - **a)** **Most safe speed**: - For a particular road of radius (r) and angle of banking (θ) are fixed. - From egn ③, **\\(v = √(rg tanθ)\\)** - **b)** **Banking angle:** - From egn ③, **θ = tan⁻¹(\\(v^2/rg\\))** - **c)** **Speed limits** *Diagram of a car on a banked road with forces acting on it:* [Diagram of a car on a banked road with forces acting on it] - **Fig shows vertical section of a vehicle on a rough curved road of radius 'r' banked at an angle θ.** - **If the vehicle is running exactly at the speed: \\(v_s = √(rg tanθ)\\)** - **The forces acting on the vehicle are:** - **Weight mg** vertically downwards - **Normal reaction N.** perpendicular to the surface of road - **N resolve into two components Ncoso & Nsing** - **In practice, vehicles never travel exactly with this speed. For speeds other than this the components of force of static friction between roads and tires help us upto a certain limit.** - **For speed \\(v_i < √(rg tanθ)\\)** - **\\(mv^2/r < N sino\\)** - **From fig, ** - \\(mg = f_s sino + Ncosθ\\)** - **\\(mv^2/r = Nsino - f_s cosθ\\)** - **Dividing ④ by ③** - **\\(v^2 _ s - Nsing-Fs cosθ\\). ** - **\\(rg Fs sinθ + Ncos θ\\) ** - **Putting \\(f_s = μsN\\)** - **\\(v_s^2 = rg(\frac{Nsino - μsNcos θ}{μs sinθ + Ncos θ})\\)** - **\\(v_s^2 = rg(\frac{tanθ - μs}{1+μs tanθ})\\)** - **For μs = tanθ, \\(v_min = 0\\)** - **This is true for rough surface for speeds \\(v_s < √(rg tanθ)\\)** - **In this case, the direction of force \\(f_s\\) is along the inclination of the road as shown in fig. below.** *Diagram of a car on a banked road with forces acting on it:* [Diagram of a car on a banked road with forces acting on it] - **From fig, ** - **mg = N cos θ - f_s sino** - **\\(mv_s^2/r = Nsinθ + f_s cosθ\\)** - **Dividing ⑥ by ⑦** - **\\(v_s^2 = rg(\frac{tanθ + μs}{1- μs tanθ})\\)** - **For \\(μs = 0\\) both eqns give \\(v = √(rg tanθ)\\)** ## conical Pendulum: - **Conical Pendulum:** - A simple pendulum which gives such a motion that bob describes a horizontal circle. The string describes a cone. Its construction is similar to an ordinary pendulum. Instead of swinging back and forth, the bob of a conical pendulum moves at a constant speed in a circle with the string tracing out a cone. *Diagram of a conical pendulum:* [Diagram of a conical pendulum] - **Fig shows the vertical section of a conical pendulum having bob of mass m & string of length 'l'.** - **In given position B, the forces acting on the bob are -** - **Weight mg.** vertically downwards - **The force To - along string** - **This To resolved into two components** - **To cosθ - balances the weight mg** - **To sino - becomes the resultant force which is the CP force.** - **To sino = \\(mrw^2\\)** - **To cosθ = mg** - **Dividing eqn ① by ②, ** - **\\(w^2 = \frac{gsino}{r cos θ}\\)** - **\\(w = \sqrt{\frac{gsino}{r cos θ}}\\)** - **But, T = \\(2π/w\\)** - **From egn ③ T= \\(2π \sqrt{\frac{r cos θ}{gsin θ}}\\)** - **From fig (b), r = Lsin θ** - **T= \\(2π \sqrt{\frac{L}{g}}\\)** - **Frequency (n) = 1/T** - **n= \\(\frac{1}{2π \sqrt{\frac{L}{g}}}\\)** - **n= \\(\frac{1}{2π} \sqrt{\frac{g}{Lcosθ}}\\)** ## Vertical Circular Motion (VCM): - **A body revolves in a vertical circle such that its motion at different points is different, then the motion of the body is said to be vertical circular motion.** - **Ex. 1) roller coasters** - **2) water buckets on a string** - **3) cars travelling on hilly roads.** - **Two types of VCM are observed in practice** - **a) A controlled VCM such as a giant wheel not totally controlled only by gravity.** - **b) VCM controlled only by gravity.** - **During the motion, there is inter conversion of KE and gravitational PE.** ## Point mass undergoing vertical circular motion under gravity: - **Case 1: Mass tied to a string** *Diagram of a point mass tied to a string performing vertical circular motion:* [Diagram of a point mass tied to a string performing vertical circular motion] - **Fig shows a bob tied to a massless and inextensible string. it is whirled along vertical circle so that bob perform VCM.** - **Let, m be the mass of an object and r be the radius of vertical circle.** - **Let, \\(V_A\\) be the velocity of an object at highest point which is minimum.** - **\\(V_B\\) be the velocity of an object at lowest point which is maximum.** - **\\(V_c\\) be the velocity at point C.** - **The forces acting on object at point A are:** - **Tension \\(T_A\\)** acting in downward direction - **Weight mg** acting vertically downward. - **At point A: Centripetal force = Weight + Tension** - **\\(m V_A^2 /r = mg + T_A\\)** - **\\(T_A = \frac{mV_A^2}{r} - mg\\)** - **Now, the forces acting at point B are:** - **Tension \\(T_B\\)** acting vertically upwards - **Weight mg** acting vertically downwards - **At point B: \\(mV_B^2/r = T_B - mg\\)** - **\\(T_B = \frac{mV_B^2}{r} + mg\\)** - **a)** **Linear velocity at highest point A:** - **At point A string becomes slack** - **\\(T_A = 0\\)** - **... Egn ① becomes, \\(mV_A^2/r = mg \\)** - **\\(V_A = √(gr)\\)** - **b)** **Linear velocity at lowest point B:** - **As an object moves from point A to B these is decrease in PE while KE increases according to law of conservation of energy.** - **...(T.E.) at A = (T.E.) at B** - **...(K.E. + P.E.) at A = (K.E. + P.E.) at B** - **\\(\frac{1}{2}mV_A^2 + mgh_A = \frac{1}{2}mV_B² + mgh_B\\)** - **Putting \\(V_A = √(gr), h_p = 2r & h_p = 0\\)** - **\\(\frac{1}{2} m(gr) + 2mgr = \frac{1}{2} mV_B² + 0\\)** - **\\(\frac{1}{2}gr + 2gr = \frac{1}{2} V_B^2\\)** - **\\(\frac{5}{2}gr = \frac{1}{2} V_B^2\\)** - ** \\(V_B = √(5gr)\\)** - **c)** **Linear velocity at midpoint C:** - **According to law of conservation of energy** - **(T.E.) at B = (T.E.) at C** - **(K.E. + P.E.) at B =(K.E. + P.E.) at C** - **\\(\frac{1}{2}mV_B^2 + mgh_B = \frac{1}{2}mV_c^2 + mgh_C\\)** - **Putting \\(V_B = √(gr), h_p = 0 & h_c = r \\)** - **\\(\frac{1}{2}m x\sqrt{gr} + 0 = \frac{1}{2}mV_c^2 + mgr\\)** - **\\(\frac{5}{2}gr = V_c^2 + 2gr\\)** - **\\(V_c = √(3gr)\\)** - **\\(V_c = √(3gr)\\)** - **Case II: Mass tied to a rod:** - **i) Consider a bob tied to a massless and rigid rod and whirled along a vertical circle.** - **ii) The basic difference between the rod and the string is that the string needs some tension at all the points including the uppermost point.** - **iii) Thus, a certain minimum speed is necessary at the uppermost point in the case of a string. In the case of a rod, such a condition is not necessary.** - **iv) Thus i.e. \\(V_A = 0\\) is possible at the uppermost point.** - **Linear velocity at lowest point:** - **According to law of conservation of energy** - **(T.E.) B = (T.E.) A** - **\\(\frac{1}{2}mV_B^2 + mgh_B = \frac{1}{2}mV_A² + mgh_A\\)** - **\\(\frac{1}{2}mV_B^2 + 0 = 0 + mg(2r)\\)** - **\\(V_B = √(4gr)\\)** - **\\(V_B = 2√(gr)\\)** - **Similarly s-velocity at horizontal position is minimum velocity = √(2gr).** - **Also, \\(T_B - T_A = 6mg \\)** - **Sphere of Death:** - This is a popular show in a circus. It is similar to the wall of death, but in this act riders can loop vertically as well as horizontally. - During this, two-wheeler rider undergo rounds inside a hollow sphere starting with small horizontal circles they eventually perform revolutions along vertical circles. - This VCM is same as that of the point mass tied to the string, except that the force due to tension T is replaced by the normal reaction force N - The linear speed is more for larger circles but angular speed (frequency) is more for smaller circles (while starting or stopping). - **Vehicle at the top of Convex over-bridge:** *Diagram of a car on a convex over bridge with forces acting on it:* [Diagram of a car on a convex over bridge with forces acting on it] - **Fig. shows a vehicle at the top of a convex over-bridge. During its motion forces acting on the vehicle are.** - **Weight (mg)** in downward direction - **Normal reaction force (N)** in upward direction. - **The resultant of these two must provide the necessary centripetal force.** - ** mg - N = \\(mv^2/r\\)** - **\\(N =mg - \frac{mv^2}{r}\\)** - **As speed is increased, N goes on decreasing. Thus for just maintaining contact N=0 mg = \\(mv^2/r\\)** - **\\(V = √(gr)\\)** ## Moments of Inertia as an analogous quantity for mass: - **In expression of linear momentum, force & KE, mass is a common term. In order to have their rotational analogues we need a replacement for mass. To know this let us derive an expression for rotational KE.** *Diagram of a body with n particles rotating about an axis:* [Diagram of a body with n particles rotating about an axis] - **Consider, a body of n particles of a rigid body rotating with constant angular speed (w) about an axis perpendicular to the plane of paper.** - **Suppose that, the body consists of n particles of masses m1, m2, m3… mn situated at distances \\(r_1, r_2, r_3… r_n\\) respectively from the axis of rotation.** - **Let, \\(v_1, v_2, v_3… v_n\\) be their linear velocities. Consider, particle 1 of mass m1. When the body rotates, this particle revolves in a circular orbit with radius \\(r_1\\) and velocity \\(v_1\\).** - **K.E. of 1st particle \\(E_1\\) is = \\(\frac{1}{2}\\)m1v1² - ** = \\(\frac{1}{2}\\)m1r1² w² (As \\(v_1 = r_1 w\\))** - **Similarly, K.E. of 2nd particle is \\(E_2 = \frac{1}{2}\\) m2r2² w²** - **K.E. of 3rd particle is \\(E_3 = \frac{1}{2}\\)m3r3² w²** - **and so on…** - **Total KE. of rotating body is, E= \\(E_1 + E_2 + E_3… + E_n\\)** - **E = \\(\frac{1}{2}\\) Im1r1²w² + \\(\frac{1}{2}\\) m2r2²w² + … \\(\frac{1}{2}\\) mnrn²w²** - **E = \\(\frac{1}{2}\\)w²(m1r1² + m2r2² + … + mnrn²)** - **E = \\(\frac{1}{2}\\)w²[∑(miri²)]** - **E =\\(\frac{1}{2}\\)Iw² ①** - **where, I = ∑(miri²)** - **I= \\(\frac{1}{2}\\)∑(miri²)** - **Eqn ① is analogous to translational K.E. = \\(\frac{1}{2}\\)mv²** - **Thus, I is defined to be the moment of inertia. I= \\(\frac{1}{2}\\)∑(miri²)** - **Also, I can be expressed in terms of Integration I = ∫r²dm** ## Moment of Inertia of a uniform ring: - **Consider a uniform ring of mass 'M' and radius 'R' rotating about an axis passing through centre of ring and perpendicular to the plane of the ring.** - **M.I of uniform ring rotating about an axis passing through point 'O' is I = \\(MR^2\\).** - **I = ∫r²dm** *Diagram of a ring rotating about an axis:* [Diagram of a ring rotating about an axis] ## Moment of Inertia of a uniform disc: - **Consider a uniform disc of mass 'M' and radius 'R' rotating about an axis which is perpendicular to its plane and passing through its centre.** - **Surface density (σ) = Mass/Area = M/πR²** - **As it is a uniform circular object it can be considered to be consisting of a number of concentric rings of radii increasing from zero to R. One of such ring of mass dm is shown by shaded portion.** *Diagram of a disc rotating about an axis:* [Diagram of a disc rotating about an axis] - **Let, r & dr be the radius and width of the ring** - **Area of this ring A= 2πr.dr** - **Mass = dm** - **= σ area** - **= 2πσr.dr** - **By defn of moment of inertia (with limit 0 to R)** - **I = ∫r²dm** - **I= ∫r².2πσr.dr** - **I = 2πσ ∫r³dr** - **I = 2πσ [ \\(\frac{r^4}{4}\\) ]_{0}^R** - **I = 2πσ(\\(\frac{R^4}{4}\\))** - **I = 2π x M / πR² x \\(\frac{R^4}{4}\\)** - **I = MR²/2** ## Radius of Gyration: - **The radius of gyration of a body about a given axis of rotation is defined as the distance between the axis of rotation and a point at which the whole mass of the body can be supposed to be concentrated so as to possess the same moment of inertia as that of the body.** - **I = MK²** - **i.e. K² = I/M** - **K= √(I/M)** - **The distance 'K' is called radius of gyration.** - **SI unit of K = m** - **Dimensions of K = [M^0 L^1 T^0]** *Diagram of the radius of gyration of a body:* [Diagram of the radius of gyration of a body] ## Theorem of Parallel Axes - **Statement:** - **The moment of inertia (I0) of a rigid body about any axis is equal to the sum of:** - **i) Its moment of inertia (I_c) about an axis parallel to the given axis passing through the centre of mass and** - **ii) the product of the mass of the object and the square of the distance between the two axes (Mh²).** - **.. I_0 = I_c + Mh²** *Diagram of the theorem of parallel axes:* [Diagram of the theorem of parallel axes] - **Proof:** - **Consider, the rigid body of mass M rotating about an axis passing through ‘o’. Let ‘c’ be the centre of mass of the body.** - **I_0 be the M.I. of the body about an axis passing through o** - **I_c be the M.I. of the body about parallel axis passing through c** - **oc = h be the distance between two parallel axes.** - **Consider, a particle of mass dm at point P. Join op and cp. Draw perpendicular pe to oc.** - **In A OPQ, ** - **(OP)² = (OQ)² + (PQ)²** - **(OP)² = (oc + CQ)² + (PQ)²** - **(op)² = (oc)² + 2oc.CQ + (CQ)² + (PQ)²** - **= (oc)² + 2oc.CQ + (cp)²** - **Multiplying by dm and integrating we get, ** - **∫(OP)²dm = ∫(oc)²dm + ∫2oc.CQdm + ∫(CP)²dm** - **By defn, ** - **I_0 = ∫(OP)²dm and putting oc = h, I_c = ∫(CP)²dm and ∫dm = M = total mass** - **Eqn ① becomes, ** - **I_0 = Mh² + 2h∫cQdm + I_c** - **I_0 = I_c + Mh²** - **:as ∫cQdm = 0 Hence the proof.** ## Theorem of Perpendicular Axes - **Statement:** - **The moment of inertia of a plane lamina about an axis perpendicular to its plane is equal to the sum of its moment of inertias about two mutually perpendicular axes in the plane of the lamina and intersecting at the point where the perpendicular axis cuts the lamina.** - **I_z = I_x + I_y** *Diagram of the theorem of perpendicular axes:* [Diagram of the theorem of perpendicular axes] - **Proof:** - **Let, ox & oy be the two mutually perpendicular axes in the plane of the lamina. Let, oz be the axis passing through their point of intersection ‘o’ perpendicular to their plane.** - **Let, I_x, I_y, and I_z be the moment of inertia of the lamina about ox, oy, and oz axes respectively.** - **Consider a particle of mass dm situated at point P with co-ordinates (x,y). Join op. Draw PM ⊥ to X-axis and PN ⊥ to Y-axis. Let, PN = OM = x and PM = ON = y.** *Diagram of a plane lamina with axes:* [Diagram of a plane lamina with axes] - **From fig. (OP)² = (OM)² + (PM)²** - **r² = x² + y²** - **Multiplying by dm and integrating we get, ** - **∫r²dm = ∫x²dm + ∫y²dm ①** - **By defn of moment of inertia about X,Y, and Z axes** - **I_x = ∫y²dm** - **I_y = ∫x²dm** - **I_z = ∫r²dm** - **Eqn ① becomes, ** - **I_z = I_x + I_y** - **Hence, the proof.** - **Angular momentum or moment of linear momentum:** - **The quantity angular momentum is analogous to linear momentum.** - **If P is the instantaneous linear momentum of a particle undertaking circular motion, its angular momentum is given by:** - **L = r*P** - **L= r *P sin θ** - **Where θ is the smaller angle beth P and r.** - **Expression for angular momentum in terms of moment of inertia:** *Diagram of a body with n particles rotating about an axis:* [Diagram of a body with n particles rotating about an axis] - **Consider a rigid body rotating with a uniform angular velocity (ω) about a fixed axis passing through point O.** - **Suppose that, the body consists of n particles of masses m1, m2, m3… mn situated at distances \\(r_1, r_2, r_3… r_n\\) respectively from axis of rotation. As the body rotates all the particles perform UCM with same angular velocity ω.** - **Consider, 1st particle of mass m1. Its linear momentum is, P1 = m1v1 = m1r1 ω** - **v=rω** - **Angular momentum of first particle is, L1= P1r1 = m1r1ω * r1** - **i.e. L1 = m1r1²ω** - **Similarly for 2nd particle angular momentum is, L2= m2r2²ω** - **For 3rd particle it is, L3 = m3r3²ω and so on…** - **Total angular momentum (L) of rotating body is, L = L1 + L2 + L3 + … +Ln** - **L= m1r1²ω + m2r2²ω + m3r3²ω + … + mnrn²ω** - **L = [m1r1² + m2r2² + … + mnrn²] ω** - **L = (∑m,r,²) ω** - **L = Iω** - **where, I = ∑(m,r,²) = moment of inertia** ## Expression for torque in terms of moment of inertia: - **Consider a rigid body rotating about a fixed axis passing through point 'O' with uniform angular accn (α).** *Diagram of a body with n particles rotating about an axis:* [Diagram of a body with n particles rotating about an axis] - **Suppose that the body consists of n particles of masses m1, m2, m3… mn situated at distances \\(r_1, r_2, r_3… r_n\\) respectively from the axis of rotation. ** - **Let, \\(f_1, f_2, f_3… f_n\\) be the forces acting on the particles.** - **Consider, first particle of mass m1. The linear acch of this particle is \\(a_1 = r_1α\\)** - **Force acting on 1st particle is, \\(f_1 = m_1a_1 \\)** - **\\(f_1 = m_1r_1α\\)** - **Torque (T1) = \\(f_1 r_1\\)** - **\\(T_1 = m_1r_1²α\\)** - **Similarly torque acting on 2nd particle is, \\(T_2= m_2r_2²α\\)** - **For 3rd particle, \\(T_3 = m_3r_3²α\\) so on.** - **Total torque acting on body is, T = T1 + T2 + T3 + … +Tn** - **T = m1r1²α + m2r2²α + m3r3²α + … + mnrn²α** - **T = (m1r1² + m2r2² + … + mnrn²) α** - **T = (Σm,r,²) α** - **T = Iα** - **where, I = ∑(m,r,²) = moment of inertia.** - **The relation T = Iα is analogous to F = ma for the translational motion.** - **Moment of inertia (I) is analogous to mass which is its physical significance.** ## Conservation of angular momentum: - **Statement:** - **If no external torque acts on a body then angular momentum of the body remains constant.** - **Proof:** - **Angular momentum of system is given by, L = r*P** - **Differentiating w.r.t. time we get, ** - **\\(\frac{dL}{dt} = \frac{d(r*P)}{dt}\\)** - **\\(\frac{dL}{dt} = r\frac{dP}{dt} + \frac{dr}{dt}*P \\)** - **But, we know that, \\(\frac{dr}{dt} = v\\) & \\(\frac{dP}{dt} = F \\)** - **... \\(\frac