Rotational Dynamics One Shot PDF

 Introduction Circular motion Statement: Circular motion is a movement of an object along the circumference of a circle or rotation along a circular path. Characteristics of Circular Motion 1) It is an accelerated motion: As the direction of velocity cha...

 Introduction Circular motion Statement: Circular motion is a movement of an object along the circumference of a circle or rotation along a circular path. Characteristics of Circular Motion 1) It is an accelerated motion: As the direction of velocity changes at every instant, it is an accelerated motion. 2) It is a periodic motion: During the motion, the particle repeats its path along the same trajectory. Thus, the motion is periodic. Kinematics of Circular Motion  Period Definition: Time taken by particle performing UCM to complete one revolution is called as period.  Frequency Definition: The numbers of revolutions performed by particle performing UCM in unit time is called as frequency of revolution  Uniform circular motion Definition: During circular motion if the speed of the particle remains constant, it is called UCM. Only the direction of its velocity changes at every instant in such a way that the velocity is always tangential to the path.  Centripetal Acceleration Definition: The acceleration of particle performing UCM which is directed towards the centre and along the radius of circular path is called as centripetal acceleration. Note: Only the direction of its velocity changes at every instant in such a way that the velocity is always tangential to the path.  Non-uniform circular motion Definition: Non-uniform circular motion is a circular motion when the speed of the body doesn't remain constant. Tangential Acceleration  Tangential Acceleration Definition: The acceleration responsible for changing the magnitude of velocity is directed along or opposite to the velocity, hence always tangential and is called as tangential acceleration 𝑎ԦT. Centripetal force  Centripetal force Q. Define centripetal force [March 2024] 1 Mark Explain the concept of centripetal force. [March 2017] 2 Marks Definition: A force that acts on a body moving in a circular path and is directed towards the centre of the circle. Ex: 1. Planets orbiting around the Sun 2. A Bike moving along a curve road Centrifugal force  Centrifugal force Q. Explain the concept of centrifugal force. Definition: Centrifugal force is a pseudo force in a circular motion which acts along the radius and is directed away from the center of the circle. Ex : 1.Weight of an object at the poles and on the equator 2. A bike making a turn. 3. Vehicle driving around a curve  Q. Distinguish between centripetal force and centrifugal force. [July 2017] 2 Marks Centripetal force Centrifugal force 1. It is the real force It is the pseudo force 2. Centripetal force is directed along the Centrifugal force is directed away from the radius of a circle towards the centre. centre of a circle along the radius. 3. 𝐹Ԧ = −𝑚𝜔2 𝑟Ԧ 𝐹Ԧ = +𝑚𝜔2 𝑟Ԧ 4. It is regarded in an inertial frame of It is regarded in a non-inertial frame of reference. reference. Applications of Uniform Circular Motion Vehicle Along a Horizontal Circular Track  Vehicle Along a Horizontal Circular Track Q. Derive an expression for maximum speed moving along a horizontal circular track. [MHT-CET 2024] Fig. shows vertical section of a car on a horizontal circular track of radius 𝑟. Plane of figure is a vertical plane, perpendicular to the track but includes only centre C of the track. Forces acting on the car (considered to be a particle) are (i) weight 𝑚𝑔, vertically downwards, (ii) normal reaction 𝑁, vertically upwards that balances the weight mg and (iii) force of static friction 𝑓s between road and the tyres. This is static friction because it prevents the vehicle from outward slipping or skidding. This is the resultant force which is centripetal. While working in the frame of reference attached to the vehicle, it balances the centrifugal force Well (or Wall) of Death  Well (or Wall) of Death Q. Derive an expression for minimum safest velocity of a body to move in well 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 (i) Normal reaction N acting horizontally and towards the centre, (ii) Weight mg acting vertically downwards, and (iii) Force of static friction 𝐟s acting vertically upwards between vertical wall and the tyres. Vehicle on a Banked Road  Vehicle on a Banked Road As seen earlier, while taking a turn on a horizontal road, the force of static friction between the tyres of the vehicle and the road provides the necessary centripetal force (or balances the centrifugal force). However, the frictional force is having an upper limit. Also, its value is usually not constant as the road surface is not uniform. Thus, in real life, we should not depend upon it, as far as possible. For this purpose, the surfaces of curved roads are tilted with the horizontal with some angle 𝜃. This is called banking of a road or the road is said to be banked.  Vehicle on a Banked Road Banking of roads : The phenomenon in which the edges are raised for the curved roads above the inner edge to provide the necessary centripetal force to the vehicles so that they take a safe turn. 𝑚 = mass of the vehicle 𝑊 = 𝑚𝑔 = weight of vehicle 𝑁 = normal reaction 𝜃 = angle of banking Forces acting on the vehicle are i) Weight ( mg ), vertically downward. ii) Normal reaction (𝑁), normal to the surface. The normal reaction (𝑵) can be resolved into two components i) 𝑁cos 𝜃 vertically upward ii) 𝑁sin 𝜃 in the horizontal direction  Vehicle on a Banked Road Q. Draw a neat labelled diagram showing the various forces and their components acting on a vehicle moving along curved banked road. [July 2016] 2 Marks (a) Most safe speed  Vehicle on a Banked Road (b) Banking angle  (c) Speed limits BANKED ROAD : LOWER SPEED LIMIT  (c) Speed limits BANKED ROAD : UPPER SPEED LIMIT Q. Draw a diagram showing all components of forces acting on a vehicle moving on a curved banked road. Write the necessary equation for maximum safety, speed and state the significance of each term involved in it. [Oct 2014] 3 Marks  Conical Pendulum Q. Draw a neat labelled diagram of conical pendulum. State the expression for its periodic time in terms of length. [Oct 2015] 2 Marks  Conical Pendulum Definition: A simple pendulum in which the string moves along the surface of a cone and the point object performs a horizontal circular motion.  Derivation of Time Period of Conical Pendulum 𝐶 = center of the circle, 𝑟 = radius, 𝐿 = length of the string 𝑚 = mass of the bob, 𝑊 = 𝑚𝑔 = weight of bob 𝑇𝑜 = tension in the string, 𝜃 = semi vertical angle Forces acting on the bob are i) Weight ( mg), vertically downwards. ii) Tension 𝑇𝑜 , directed along the string towards point A. The tension 𝑇𝑜 can be resolved into two components, i) 𝑇𝑜 cos 𝜃 vertically upward. ii) 𝑇𝑜 sin 𝜃 in the horizontal direct Vertical Circular Motion  Vertical Circular Motion Two types of vertical circular motions are commonly observed in practice: (a) A controlled vertical circular motion such as a giant wheel or similar games. In this case the speed is either kept constant or NOT totally controlled by gravity.  Vertical Circular Motion (b) Vertical circular motion controlled only by gravity. In this case, we initially supply the necessary energy (mostly) at the lowest point. Then onwards, the entire kinetics is governed by the gravitational force. During the motion, there is interconversion of kinetic energy and gravitational potential energy. Roller coasters Water in a bucket is whirled in a vertical circle  Mass tied to a string Q. A particle of mass 𝑚, just completes the vertical circular motion. Derive the expression for the difference in tensions at the highest and the lowest points. [Mar 2013] 3 Marks OR Derive an expression for linear velocity at lowest position, midway position and the top-most position for a particle revolving in a vertical circle, if it has to just complete circular motion without string slackening at top. [Feb 2023] 4 Marks  Mass tied to a string Consider a bob of mass (m ) suspended from a light string of length (l) and rotated in a vertical circle of radius (r). The motion is non uniform. At any position of the bob, there are only two forces acting on the bob: (a) Weight ( W = mg ), vertically downwards, which is constant. (b) Tension (T) along the string, directed towards the centre. Uppermost position (A): The weight (mg) and tension TA acting in downward. vA is the speed at the uppermost point. Lowermost position (B): Force due to the tension, 𝑇𝐵 is vertically upwards, i.e., towards the centre, and opposite to 𝑚𝑔. vB is the speed at the lowermost point. Positions when the string is horizontal (C and D) Force due to the tension is the only force towards the centre as weight mg is perpendicular to the tension. Thus, force due to the tension is the centripetal force used to change the direction of the velocity and weight mg is used only to change the speed. Arbitrary positions: Force due to the tension and weight are neither along the same line, nor perpendicular. Tangential component of weight is used to change the speed. It decreases the speed while going up and increases it while coming down  Mass tied to a rod Consider a bob (point mass) tied to a (practically massless and rigid) rod and whirled along a vertical circle. The basic difference between the rod and the string is that the string needs some tension (T) to perform vertical circular motion. In the case of a rod tension is not necessary. Thus (practically) zero speed is possible at the uppermost point. Sphere of Death  Sphere of Death Q. Write a short note on sphere of death. This is a popular show in a circus. During this, two-wheeler rider (or riders) undergo rounds inside a hollow sphere. Starting with small horizontal circles, they eventually perform revolutions along vertical circles. The dynamics of this vertical circular motion is the 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. There are nearly horizontal circles. The linear speed is more for larger circles but angular speed (frequency) is more for smaller circles (while starting or stopping). This is as per the theory of conical pendulum.  Vehicle at the Top of a Convex Over Bridge Q. Explain the expression for upper limit on the speed when vehicle is at the top of a convex overbridge. m = mass of vehicle R = radius of convex bridge g = acceleration due to gravity Forces acting on the vehicle are (a) Weight mg (b) Normal reaction force N, both along the vertical line (topmost position). The resultant of these two must provide the necessary centripetal force (vertically downwards) if the vehicle is at the uppermost position. v is the speed at the uppermost point  Moment of Inertia as an Analogous Quantity for Mass If we open a door we give a certain angular displacement to it. The efforts needed for this depend not only upon the mass of the door, but also upon the (perpendicular) distance from the axis of rotation, where we apply the force. Thus, the quantity analogous to mass not only depends on the mass but also it is depends on the distance wise distribution of the mass about axis of rotation  Moment of Inertia as an Analogous Quantity for Mass Q. Define moment of inertia. State its SI unit and dimensions. [Oct. 2008, March 2017, Mar 2022 ] 2 Marks Statement: Moment of inertia of a rigid body about an axis of rotation is defined as the sum of product of the mass of each particle and the square of its perpendicular distance from the axis of rotation. Moment of Inertia as an Analogous Quantity for Mass  K.E. of System of N Particles Q. Derive an expression for the kinetic energy of a body rotating with a uniform angular speed. [Mar 2022] 2 Marks Let us consider the object consist of N no. of particles of masses m1 , m2 , … … mN respectively and 𝑟1 , 𝑟2 , …. 𝑟𝑁 be their perpendicular distances from the axis of rotation. As the object rotates, all these particles perform UCM with the same angular speed (𝜔), but with different linear speeds (v).  Significance of Moment of Inertia Q. What is the significance of moment of inertia (i) individual masses and (ii) the distribution of these masses about the given axis of rotation. For a different axis, it will again depend upon the mass distribution around that axis and will be different if there is no symmetry.  Moment of Inertia of a Uniform Ring Consider a ring of mass 𝑀 and radius 𝑅 is rotating about its own axis (line perpendicular to its plane and passing through its centre). 𝑑𝑚 - mass of each particle 𝑅 - perpendicular distance from axis of rotation.  Moment of Inertia of a Uniform Disc Q. State an expression for the moment of inertia of a solid uniform disc, rotating about an axis passing through it centre, perpendicular to its plane. Hence derive an expression for the moment of inertia and radius of gyration: (i) about a tangent in the plane of the disc, and (ii) about a tangent perpendicular to the plane of the disc. [Oct 2015] 4 Marks  Moment of Inertia of a Uniform Disc Consider a uniform disc of mass 𝑀 and radius 𝑅 rotating about its own axis, which is the line 𝑴 perpendicular to its plane and passing through its centre ∴ 𝝈 = 𝟐. 𝝅𝑹 As it is a uniform circular object, it can be considered to be consisting of a number of concentric rings of radii increasing from (practically) zero to 𝑅. One of such rings of mass 𝑑𝑚 is shown by shaded portion in the Fig. Width of this ring is 𝑑𝑟, which is so small that the entire ring can be considered to be of average radius 𝑟.  Expressions for M.I. and Radius of gyration of some bodies having regular shapes, about their axis of symmetry. No. Figure Description Axis of Rotation M.I. K Passing through its L Thin uniform rod of centre and ML2 1. mass 𝑀 and length L perpendicular to its 12 12 length Passing through its Thin uniform ring of centre and 2. mass M and radius R perpendicular to its MR2 R plane Passing through its centre and MR2 𝑅 Circular disc of mass 3. M and radius R perpendicular to its 2 2 plane  Expressions for M.I. and Radius of gyration of some bodies having regular shapes, about their axis of symmetry. No. Figure Description Axis of Rotation M.I. K MR2 𝑅 Solid Cylinder mass 𝑀 and Coinciding with the 4. radius R axis of the cylinder 2 2 Passing through its L2 R2 Solid Cylinder mass 𝑀 and L2 R2 + 5. radius R and length L centre and perpendicular M + 12 4 12 4 to its axis Hollow Cylinder mass Coinciding with the axis MR2 𝑅 6. 𝑀 and radius R of the cylinder  Expressions for M.I. and Radius of gyration of some bodies having regular shapes, about their axis of symmetry. No. Figure Description Axis of Rotation M.I. K 2 2 Solid Sphere mass Coinciding with the MR2 R 7. 𝑀 and radius R diameter of the sphere 5 5 2 2 Hollow Sphere mass Coinciding with any MR2 8. 𝑀 and radius R diameter of the sphere 3 R 3  Radius of Gyration As stated earlier, theoretical calculation of moment of inertia is possible only for mathematically integrable geometrical shapes. However, experimentally we can determine the moment of inertia of any object. It depends upon mass of that object and how that mass is distributed from or around the given axis of rotation. If we are interested in knowing only the mass distribution around the axis of rotation, we can express moment of inertia of any object as 𝐼 = 𝑀𝐾 2 , where 𝑀 is mass of that object. It means that the mass of that object is effectively at a distance 𝐾 from the given axis of rotation. In this case, 𝐾 is defined as the radius of gyration of the object about the given axis of rotation. In other words, if 𝐾 is radius of gyration for an object, 𝐼 = 𝑀𝐾 2 is the moment of inertia of that object. Larger the value of 𝐾, farther is the mass from the axis. MH-Board -Grade 12 - Physics - Content Credit: Text Book  Radius of Gyration Q. Define radius of gyration. Explain its physical significance. [Mar 2008, 2019; July 2018] 2 Marks Statement: Radius of gyration is generally defined as the distance from the axis of rotation to a point where total mass of any body is supposed to be concentrated, so that the moment of inertia about the axis may remain same. It is actually perpendicular distance from point mass to the axis of rotation. Physical Signification: It signifies distribution of mass about axis of rotation. MH-Board -Grade 12 - Physics - Content Credit: Text Book  Theorem of Parallel Axes In order to apply this theorem to any object, we need two axes parallel to each other with one of them passing through the centre of mass of the object. MH-Board -Grade 12 - Physics - Content Credit: Text Book  Theorem of Parallel Axes Statement: It states that, "The moment of inertia 𝐼O of an object about any axis is the sum of its moment of inertia 𝐼C about an axis parallel to the given axis, and passing through the centre of mass and the product of the mass of the object and the square of the distance between the two axes 𝑀ℎ2." Fig. shows an object of mass 𝑀. Axis MOP is any axis passing through point O. Axis ACB is passing through the centre of mass C of the object, parallel to the axis MOP, and at a distance ℎ from it ( ∴ ℎ = CO ). Consider a mass element 𝑑𝑚 located at point D. Perpendicular on OC (produced) from point D is DN. Moment of inertia of the object about the axis ACB is 𝑰𝑪 = ∫ (𝐃𝐂)𝟐 𝒅𝒎, and about the axis MOP it is 𝑰𝑶 = ∫ (𝐃𝐎)𝟐 𝒅𝒎. MH-Board -Grade 12 - Physics - Content Credit: Text Book  Theorem of Perpendicular Axes This theorem relates the moment of inertias of a laminar object about three mutually perpendicular and concurrent axes, two of them in the plane of the object and the third perpendicular to the object. Laminar object is like a leaf, or any two dimensional object, e.g., a ring, a disc, any plane sheet, etc. MH-Board -Grade 12 - Physics - Content Credit: Text Book  Theorem of Perpendicular Axes Q. State and prove the theorem of perpendicular axes about the M.I.. Statement: The theorem of perpendicular axes states that the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with the perpendicular axis and lying in the plane of the body. MH-Board -Grade 12 - Physics - Content Credit: Text Book Proof: i. Consider a rigid laminar object able to rotate about three mutually perpendicular axes X, Y and Z. ii. Axes X and Y are in the plane of the object while the Z axis is perpendicular to it, and all are concurrent at O. iii. Consider a mass element dm located at any point P. 𝐏𝐌 = 𝐲 and 𝐏𝐍 = 𝐱 are the perpendiculars drawn from P respectively on the X and Y axes. iv. The respective perpendicular distances of point P from X, Y and Z axes will then be y, x and 𝒚𝟐 + 𝒙𝟐. v. If 𝐼𝑥 , 𝐼𝑦 and 𝐼𝑧 are the respective moment of inertia of the body about 𝑥, 𝑦 and 𝑧 axes, then MH-Board -Grade 12 - Physics - Content Credit: Text Book MH-Board -Grade 12 - Physics - Content Credit: Text Book MH-Board -Grade 12 - Physics - Content Credit: Text Book  Angular Momentum or Moment of Linear Momentum The quantity in rotational mechanics, analogous to linear momentum is angular momentum or moment of linear momentum. It is similar to the torque being moment of a force. If 𝑝Ԧ is the instantaneous linear momentum of a particle undertaking a circular motion, its angular momentum at that instance is given by 𝑳 = 𝒓 × 𝒑, were 𝑟Ԧ is the position vector from the axis of rotation. In magnitude, it is the product of linear momentum & its perpendicular distance from the axis of rotation. ∴ 𝑳 = 𝑷 × 𝒓𝒔𝒊𝒏𝜽, where 𝜃 is the smaller angle between the directions of 𝑃 and 𝑟. Ԧ MH-Board -Grade 12 - Physics - Content Credit: Text Book  Expression for Angular Momentum in Terms of Moment of Inertia Q. Derive an expression that relates angular momentum with the angular velocity of a rigid body i. Consider a rigid object rotating with a constant angular speed ' 𝜔 ' about an axis perpendicular to the plane of paper. ii. Let us consider the object to be consisting of N number of particles of masses 𝒎𝟏 , 𝒎𝟐 , …. 𝒎𝑵 at respective perpendicular distances 𝒓𝟏 , 𝒓𝟐 , …. 𝒓𝑵 from the axis of rotation. iii. As the object rotates, all these particles perform UCM with same angular speed 𝜔, but with different linear speeds 𝒗𝟏 = 𝒓𝟏 𝝎, 𝒗𝟐 = 𝒓𝟐 𝝎, …. 𝒗𝑵 = 𝒓𝑵 𝝎. iv. Directions of individual velocities 𝑣1 , 𝑣Ԧ2 , … … vN , are along the tangents to the irrespective tracks.. MH-Board -Grade 12 - Physics - Content Credit: Text Book  Angular Momentum or Moment of Linear Momentum Q. Define the angular momentum of rotating body. State its units and dimensions. Angular momentum (L) : Angular momentum of rotating body about a given axis, is defined as the product of its moment of inertia about the given axis and its angular velocity. L = I𝜔 where 𝐼 = M.I. of body, 𝜔 = angular velocity. Units and Dimensions : S.I. system : kgm2 /s, C.G.S. system : gmcm2 /s [L] = M1 L2 T −1 MH-Board -Grade 12 - Physics - Content Credit: Text Book  Torque in Terms of Moment of Inertia MH-Board -Grade 12 - Physics - Content Credit: Text Book  Expression for Torque in Terms of Moment of Inertia Q. Obtain an expression for torque acting on a body rotating with uniform angular acceleration. [July 2016, 2017] 3 Marks i. Consider a rigid object rotating with a constant angular acceleration ' 𝛼 ' about an axis perpendicular to the plane of paper. ii. Let us consider the object to be consisting of N number of particles of masses 𝒎𝟏 , 𝒎𝟐 , …. 𝒎𝑵 at respective perpendicular distances 𝒓𝟏 , 𝒓𝟐 , ….. 𝒓𝑵 from the axis of rotation. iii. As the object rotates, all these particles perform circular motion with same angular acceleration 𝛼, but with different linear (tangential) accelerations 𝒂𝟏 = 𝒓𝟏 𝜶, 𝒂𝟐 = 𝒓𝟐 𝜶, … , 𝒂𝑵 = 𝒓𝑵 𝜶,etc. iv. Force experienced by the first particle is, 𝒇𝟏 = 𝒎𝟏 𝒂𝟏 = 𝒎𝟏 𝒓𝟏 𝜶 v. As these forces are tangential, the irrespective perpendicular distances from the axis are 𝑟1 , 𝑟2 , ⋯ ⋯ rN. MH-Board -Grade 12 - Physics - Content Credit: Text Book MH-Board -Grade 12 - Physics - Content Credit: Text Book MH-Board -Grade 12 - Physics - Content Credit: Text Book  Conservation of Angular Momentum In the article 4.7 of XIth Std. we have seen the conservation of linear momentum which says that linear momentum of an isolated system is conserved in the absence of an external unbalanced force. As seen earlier, torque and angular momentum are the respective analogous quantities to force and linear momentum in rotational dynamics. With suitable changes this can be transformed into the conservation of angular momentum.  Conservation of Angular Momentum MH-Board -Grade 12 - Physics - Content Credit: Text Book  Q. State and prove the principle of conservation of angular momentum. [ Mar 2023 ] Statement: The angular momentum of a body remains constant if the resultant external torque acting on the body is zero. Examples of conservation of angular momentum (i) Ballet dancers: During ice ballet, the dancers have to undertake rounds of smaller and larger radii. The dancers come together while taking the rounds of smaller radius (near the centre). In this case, the moment of inertia of their system becomes minimum and the frequency increases, to make it thrilling. While outer rounds, the dancers outstretch their legs and arms. This increases their moment of inertia that reduces the angular speed and hence the linear speed. This is essential to prevent slipping.  Examples of conservation of angular momentum (ii) Diving in a swimming pool (during competition): While on the diving board, the divers stretch their body so as to increase the moment of inertia. Immediately after leaving the board, they fold their body. This reduces the moment inertia considerably. As a result, the frequency increases and they can complete more rounds in air to make the show attractive. Again, while entering into water they stretch their body into a streamline shape. This allows them a smooth entry into the water.  Q. State the law of conservation of angular momentum and explain with a suitable example. [Oct 2014; July 2019, July 2022] 2 Marks Statement: The angular momentum of a body remains constant if the resultant external torque acting on the body is zero. Example: A ballet dancer makes use of the law of conservation of angular momentum to vary her angular speed. Explanation: The torque acting on her body is zero. By the law of conservation of angular momentum, 𝐿 = constant I𝜔 = constant ⇒ 𝜔 ∝ 1/I When she suddenly folds her arms and brings the stretched leg close to the body, her angular velocity increases on account of the decrease in moment of inertia. This helps in rotating safely on her legs.  Rolling Motion Definition: A motion is said to be rolling motion if a body possess both linear motion and rotational motion  Rolling Motion Q. Derive an expression for kinetic energy, when a rigid body is rolling on a horizontal surface without slipping. Hence find kinetic energy for a solid sphere. [Mar 2013] 3 Marks Consider an object of moment of inertia 𝐼, rolling uniformly. Following quantities can be related. v = Linear speed of the centre of mass 𝑅 = Radius of the body 𝜔 = Angular speed of rotation of the body ∴ 𝜔 = V𝑅 for any particle 𝑀 = Mass of the body 𝐾 = Radius of gyration of the body ∴ 𝐼 = 𝑀𝐾2 Linear Acceleration and Speed While Pure Rolling Down an Inclined Plane