Physics-1.Unit-6-ROTATION MOTION AND MOMENT OF INERTIA

ROTATION MOTION AND MOMENT OF INERTIA SYLLABUS Centre of mass of a two-particle system. Centre of mass of a rigid body, general motion of a rigid body, nature of rotational motion, torque, angular momentum, its conservation and applications. Moment of Inertia, parallel and perpendicular axes theorem, expression of moment of inertia for ring, disc and sphere. 1. RIGID BODY A rigid body is a body with a definite and unchanged shape and size i.e. a body is said to be rigid if the distance between any two particles of the body remains invariant. 2. DEFINITION OF CENTRE OF MASS The concept of centre of mass is a pure mathematical concept. If n particle having mass m1, m2 ….m n and are placed in space (x1, y1, z1), (x2, y2, z2) ……….(x n, y n, z n) then centre of mass of system is defined as (X, Y, Z) where X = Y = and Z = where M = is the total mass of the system. Locate the point with coordinates (X, Y, Z). This point is called the centre of mass of the given collection of the particles. If the position vector of the ith particle is ri, the centre of mass is defined to have the position vector where = i x cm + j y cm + k (z cm) differentiating above equation Again differentiating above equation 3. MOMENTUM CONSERVATION AND CENTRE-OF-MASS MOTION We now consider the case when no net external force acts on the system. The particles of the system, however, move only under the influence of their mutual internal forces. In this case, setting Ftot= 0 MaCM=0 thus aCM=0 or VCM=0 or VCM=constant where VCM is the velocity of the centre of mass of the system. Thus in the absence of external forces (or if the external forces balance) the velocity of the centre of mass remains constant, i.e. RCM(t) = RCM(0) + VCMt and it moves uniformly in a straight line. This is Newton's first of motion. The case of an isolated system (i.e. a system on which no external forces act) is significant because, as will be evident form the following, the total linear momentum of such a system is conserved. If Ftot= 0, it implies that F1+ F2+ F3+. . . +FN= 0 or m1a1+ m2a2+ m3a3+. . .+ mNaN= 0 or (m1V1+ m2V2+ m3V3+. . . + mNVN)= 0 where V1, V2, V3, . . .VN are the velocities of the particles of the system. Thus m1V1+ m2V2+ m3V3+. . . + mNVN = constant or P=P1 + P2 + P3 + ... + PN= constant where P is the total linear momentum of the system. Now, Differentiating Eq.(5.1) with respect to time, we have or VCM= (m1V1+ m2V2+ m3V3+. . . + mNVN) where VCM is the velocity of the centre of mass and V1, V2, etc. are the velocities of the individual particles of the system. P = MVCM Which is an equivalent definition of the momentum of a system of particles. In other words states that the total linear momentum of a system of particles is equal to the product of the total mass of the system and the velocity of its centre of mass. States that if no net external force acts on the system, the total linear momentum of the system remains constant. This simple and general result is called the principle of conservation of linear momentum. 4. ANGULAR DISPLACEMENT Consider a rigid body undergoing rotation about an axis, perpendicular to the plane of the paper and passing through O. Suppose that A and B are any two particles of the rigid body at the position 1 while A′ and B′ are their subsequent locations when the body is at the position 2. Since the body undergoes rotation, OA = OA′ and OB = OB′ Further AB = A′B′, since the body is rigid. ⇒ ΔOAB ≅ΔOA′B′(congruent) i.e. ∠AOB = ∠A′OB′ Adding ∠AOB′ to both sides of the above equation, we get, ∠BOB′ = ∠AOA′ = θ (say) This implies that in a given interval of time the angular displacements of all particles of the rigid body undergoing rotation are identical. Therefore, a single variable, viz. angular displacement (θ) can be used to describe the rotational motion of the rigid body. Angular displacement is not a vector quantity. 5. ANGULAR VELOCITY () The rate of change of angular displacement with respect to time is known as angular velocity. Average angular velocity ω = Instantaneous angular velocity Instantaneous velocity means angular velocity at a particular dot instant. It is mathematically define as 6. ANGULAR ACCELERATIONS () Angular accelerations (): The rate of change of angular vel. with respect to time is known as angular acceleration. Average angular acceleration: Instantaneous angular acceleration: Instantaneous angular acceleration means angular acceleration. at a particular dot instant at t = t1 mathematically it is define as Direction of angular acceleration: If magnitude of ω increasing then direction of α will be same as direction of ω and vice–versa. 7. EQUATION OF ANGULAR MOTION 7.1 IF THE ANGULAR ACCELERATION IS CONSTANT, THE FOLLOWING RELATIONS HOLD ω(t) = ωo + αt θ(t) = θo + ωot + αt2 ω = Here ωo = magnitude of the initial angular velocity ω(t) = magnitude of the angular velocity after time t. θo = Initial angular position. θ(t) = Angular position after time t. Illustration 1: A disc starts rotating with constant angular acceleration of π radian/s2 about a fixed axis perpendicular to its plane and passing through its centre. Find (a) the angular velocity of the disc after 4 sec. (b) the angular displacement of the disc after 4 sec and (c) number of turns accomplished by the disc in 4 sec. Solution: Here =  rad/sec2 ω0 = 0 t = 4 sec (a) ω (4 sec) = 0 + (π rad/sec2) × 4 sec = 4π rad/sec. (b) θ (4 sec) = 0 + (π rad/sec2) × (16 sec2) = 8 πradian. (c) Let the number of turns be n ⇒ n 2 π rad = 8 π rad ⇒ n = 4 7.2 IF  IS NOT A CONSTANT If α is not constant then the following equation of motion will hold ω = α = Illustration 2: A particle is rotating in a circle and its angular position is varying according to law θ = radian. Find (a) Its instantaneous angular vel. and angular accel at t = 1 sec. (b) its average angular vel. and average angular accel. from t = 1 sec to t = 2 sec (c) its average angular vel. and angular accel. During who time of motion. Solution: θ = …….(1) ω = ……..(2) α = ……..(3) (a) at t = 1, ω = 1 rad/s α = –2 rad/s2. (b) at t = 1, θ1 = at t = 2, θ2 = at t1 = t1, ω1 = 1 t2 – t2 , ω2 = 0 (c) ω = 0 ⇒ t = 2 sec. So whole time of motion is 2 sec. t = 0 , θ0 = 6 t = 2, θ2 = t0 = 0, ω0 = 4 t2 = 2, ω2 = 0 8. RELATION BETWEEN LINEAR AND ANGULAR VARIABLES Consider a particle A of a rigid body undergoing rotation about a fixed axis- , the particle A describing an arc ABA′ of a circle with its centre O on the axis of rotation. Taking the origin at O, the position vector of A, ≡ . OA = OA′ = constant (radius of the circle) ∠A′OA = θ(t) (say) The arc length, ABA′, S = rθ The tangential velocity, vA = = rω The direction of the angular velocity vector be taken along the axis of rotation: = ω , being the unit vector along the axis of rotation. Then, , instantaneous velocity of A with respect to the axis of rotation, can be written as The acceleration of the point A with respect to the axis of rotation is ⇒ = ⇒ If is constant, then = 0 and, = × = − ω2 9. TORQUE Torque of a force about the axis of rotation Consider a force acting on a particle P. Choose an origin O and let be the position vector of the particle experiencing the force. We define the torque of the force F about O as .....(1) This is a vector quantity having its direction perpendicular to and according to the rule of cross product. Now consider a rigid body rotating about a given axis of rotation AB. Let F be a force acting on the particle O of the body. F may not be in the plane ABP. Take the origin O somewhere on the axis of rotation. Its unit is Newton-meter and dimensions is ML2T–2. Illustration 3: A particle of mass m is dropped at point A, find the torque about O. Solution: ⇒ r F sin = (r sin) F ⇒  = b mg The direction of torque is directed inward the paper or in other words, rotation about O is clockwise. 10. KINETIC ENERGY OF A KINGLY BODY WHICH IS PURLY ROTATING ABOUT AN AXIS AND MOMENT OF INERTIA Let a rigid body is purly rotating about an axis AB with angular vel. w consider a general particle Δm2 which is at a distance of r2 from axis of rotation. V2 = r2.ω So energy associated with this Δm2 is ΔK · E2 Δk · E2 = k·Etotal = k·Etotal = Where I is called as moment of inertial of body about an given axis of rotation. In this case I is about AB. Moment of inerter is also called as rotational mass of object. Illustration 4: Three point masses having mass 1, 2 and 3kg are placed at (1, 2, 3), (–4, 0, 5) and (–1, 2, –3) find moment of inerter of the system about z–axis. Solution: m1 = 1kg and (x1, y1, z1) = (1, 2, 3) So r1 = distance of m1 from t axis = M1 = 2kg and (x2, y2, z2) = (–4, 0, 5) and distance of m2 Distance of m2 from t axis = m3 = 3kg and (x3, y3, z3) = ( –1, 2, –3) 10.1 THEOREM OF PERPENDICULAR AXIS The moment of inertia of a plane lamina about an axis perpendicular to its plane is plane is equal to the sum of the moments of inertia of the lamina about two mutually perpendicular axes in its own plane intersecting each other at the point through which the perpendicular axis passes. Moment of inertia of the whole of lamina about X–axis, Ix = ∑mx2 Moment of inertia of the whole of lamina about Y–axis, Iy = ∑my2 Moment of inertia of the whole of lamina about Z –axis, Iz = ∑mr2 But r = x2 + y2 ∴ Iz = ∑m(x2 + y2) = ∑mx2 + ∑my2 or Iz = Iy + Ix. Illustration 5: The moment of inertia of a thin square plate ABCD (as shown in figure) of uniform thickness about an axis passing through the centre O and perpendicular to the Plane of plate is: (A) I1 + I2 (B) I3 + I4 (C) I1 + I3 (D) I1 + I2 + I3 + I4 Where I1, I2, I3 and I4 are moments of inertia about axis 1, 2, 3 and 4 rasp. Which are in the plane of plate? Solution: If I0 is moment of inertia of plane passing through the centre and ⊥ to plate, then according to the theorem of perpendicular axes. I0 = I1 + I2 = I3 + I4 From symmetry, I1 = I2 and I3 = I4 ∴ I0 = 2I2 = 2I3 i.e., I2 = I3 ∴ I0 = I1 + I2, I0 = I3 + I4 and I0 = I1 + I3 i.e., first three answers are correct. Ans. (A, B, C) 10.2 THEOREM OF PARALLEL AXES The moment of inertia of a body about any axis is equal to its moment of inertia about a parallel axis through its centre of gravity plus the product of the mass of the body and the square of the perpendicular distance between the two parallel axes. Moment of inertia of the whole of lamina about YY, I = ∑m(x + d)2 = ∑m(x2 + d2 + 2xd) or I = ∑mx2 + ∑md2 + ∑2mxd But ∑mx2 = Ig, ∑md2 = (∑m)d2 = Md2 where M ( = ∑m) is the mass of the lamina. Also, ∑mxd = 2d ∑mx ……….(1) I = Ig + md2 + 2d ∑mx ∑mgx = 0 or g ∑mx = 0 or ∑mx = 0 [ g ≠ 0] From equation n(1), I = Ig + Md2 Illustration 6: Calculate the moment of inertia of a uniform circular plate about its axis. Solution: Let the mass of the plate be M and its radius R. The centre is at O and the axis OX is perpendicular to the plane of the plate. its mass per unit area . Mass of the ring the moment of inertia of the elementary ring about OX is The moment of inertia of the plate about OX is 11. LAW OF ROTATION If a body rotates purly about an axis S with angular acceleration and net torque acting on the body about s is τs. τs = Iα Illustration 7: A triangular plate of uniform thickness and density is made to rotate about an axis perpendicular to the plane of the paper and (a) passing through A, (b) passing through B, by the application of the same force, F, at C (midpoint of AB) as shown in the figure. The angular acceleration in both the cases will be the same. Solution: τ = Iα ∴ τ = Force × perpendicular distance Torque is same in both the cases. But Since I will be different due to different mass distribution about the axis. ∴ α will be different. False. 12. PROPERTY OF RIGID BODY PROPERTY 1: If a rigid body is moving in general motion i.e. neither pure rotation nor pure translation. Then if we consider two point on rigid body A and B. Then vel. of A along AB will be same as vel. of B along AB. ⇒ VA cosα = VB cosβ Illustration 8: A rod of length  is moving by the help of vertical and horizontal wall if at any particular time rod makes an angle θ with horizontal and vel. of A is V then what will be vel. of B. Solution: Vel. of A along BA = vel. of B along BA ⇒ V1 cos(90 – θ) = Vcosθ V1 = V1 = V cotθ PROPERTY 2: If a rigid body is performing a general motion then if we consider frame of reference of any point on rigid body then we will observe that the body is purly rotating about our self. Illustration 9: Two disc A and B (having mass, Radius and moment of inertia of disc are M1, R1, I1 and M2, R2, I2) body A is purely rotating and B is performing general motion frame all the eqn. To find linear and angular acceleration of A and B) Solution: F.B.D of A TR1 = I1 α1 ……….(1) F.B.D of B m2g – T = m2A2 ……….(2) F.B.D of B from frame of reference of centre of B TR = I2α2 …………. (3) accel. of A in down ward direction = accel. of B in downward direction = accel. of C in down ward ⇒ R1α1 = A2 – R2 α2 …………...(4) 13. MOTION OF ROUND OR SPHERICAL OBJECT ON HORIZONTAL SURFACE 13.1 CASE OF ROLLING AND SLIPPING: A disc at any time t = t was moving with linear vel. V0 and angular vel. ω0. If s is the point of contact then consider the vel. of point S. ; = (VO – RωO) i If vel. of point S is in forward direction with respect to contact surface then body is called is in forward rolling. If vel. of point S is in backward direction with respect to contact surface then it is called as in bock ward slipping. And if vel. of point S is 0 then the body is called as in rolling. So, VO – Rωo > 0 ⇒ VO > Rωo ⇒ forward slipping VO – Rωo < 0 ⇒ VO < Rωo ⇒ backward slipping VO – Rωo = 0 ⇒ Rolling. Point to be not iced Suppose a round or spherical object to moving on earth and it is rolling. If at any time t = t its linear acceleration is A and angular acceleration is α. Then for further rolling A = Rα 14. ROTATIONAL WORK AND ENERGY The rotational work done by a force about the fixed axis of rotation is defined as Wrot = Where is the torque produced by the force, and is the infinitesimally small angular displacement about the axis. The rotational kinetic energy of a body about a fixed rotational axis is defined as where I is the moment of inertia about the axis. Work–Energy Theorm In complete analog to the work energy theorem for the translatory motion, it can be stated for rotational motion as: Wrot = Krot The net rotational work done by the forces is equal to the change in rotational kinetic energy of the body. Conservation of Mechanical Energy In the absence of dissipative work done by non–conservative forces, the total mechanical energy of a system is conserved. or Kf + Uf = Ki + Ui 15. ANGULAR MOMENTUM OF A PARTICLE ABOUT A POINT AND ABOUT AN AXIS L0 = rp sin θ = r⊥ p where r⊥ = r sin θ = OA′ Lzz′ = where = unit vector along axis zz′. Where angular momentum of a particle of mass m about point 'O' 15.1 ANGULAR MOMENTUM OF A RIGID BODY ABOUT A POINT Suppose a body is rotating about centre of mass with angular vel. ω and centre of mass is moving with linear vel. . Then the angular momentum of body about o is defined as 15.2 RELATION BETWEEN TORQUE AND ANGULAR MOMENTUM  ⇒ = 0 + ⇒ This relation analogous to , is applied to rotation. 15.3 CONSERVATION OF ANGULAR MOMENTUM When there is no net external torque acting on a particle, then . = constant Therefore, the angular momentum of the particle remains invariant in the absence of any net external torque.   16. Expressions for moment of inertia of bodies of regular shapes about particular axes of rotation: Shape of body Axis of Rotation Expression for Moment of Inertia 1. Circular ring of mass M and Radius R (i) through centre, perpendi- cular to plane of ring MR2 (ii) any diameter (1/2) MR2 (iii) any tangent in the plane of ring (3/2) MR2 (iv) any tangent perpendicular to plane of ring 2 MR2 2. Circular disc of mass M and radius R (i) through centre, perpendi-cular to plane of disc (1/2) MR2 (ii) any diameter (1/4) MR2 (iii) tangent in the plane of the disc (5/4) MR2 (iv) tangent perpendicular to plane of disc (3/2) MR2 3. Sphere of mass M and Radius R (i) any diameter (2/5) MR2 (ii) any tangent plane (7/5) MR2 4. Cylinder of mass M, radius R and length L (i) own axis (1/2) MR2 (ii) through centre perpendicular to length (iii) through end faces and ⊥ to length 5. One dimensional rod of mass M and length L (i) centre of rod and ⊥to length ML2/12 (ii) one end and ⊥to length ML2/3 6. Rectangular Lamina of mass M, length L and breadth B (i) length of lamina and in its plane MB2/3 (ii) breadth of lamina and in its plane ML2/3 (iii) centre of lamina and parallel to length or breadth in its plane (iv) centre of lamina and ⊥to its plane (v) centre of length and ⊥to its plane (vi) centre of breadth and ⊥to its plane 7. Rectangular block of mass M, length L, breadth B and height H (i) through centre or block and parallel to length or breadth or height of the block (ii) through end face and parallel to length or breadth or height of the block   17. ASSIGNMENT 1. A particle of mass m is projected with a velocity V making an angle of 45° with the horizontal. The magnitude of the angular momentum of the projectile about the projection when the particle is at its maximum height h is (A) zero (B) (C) (D) 2. The ratio of Earth’s orbital angular momentum (about the sun) to its mass is 4.4 × 1015 m2/s. The area enclosed by Earth’s orbit approximately ……..m2 (A) (B) (C) (D) 3. According to Kepler’s second law, the radius vector to a planet from the sun sweeps out equal areas in equal intervals of time. This law is a consequence of the conservation of …… (A) Angular momentum (B) linear momentum (C) Energy conservation (D) All are correct. 4. The moment of inertia of a ring about one of its diameters is I. What will be its moment of inertia about a tangent parallel to the diameter? (A) 4I (B) 2I (C) (D) 3I 5. A uniform cube of side a and mass m rests on a rough horizontal table. A horizontal force F is applied normal to one of the faces at a point that is directly above the centre of the face, at a height 3a/4 above the base. The minimum value of F for which the cube begins to topple about the edge is ………(Assume that the cube does not slide). (A) (B) (C) (D) 6. In the given example A pulley is purly rotating about O. Find the angular acceleration about O. Given that moment of inertia of pulley about O is I. (A) α = (B) α = (C) α = (D) α = 7. A disc of Radius R is pulley rotating about O with angular vel. ω about O. If B is the point on disc such that OB is parallel to rod which is as shown in the fig. B is connected by A (A is bar which is smoothly moving on the rod). If point B and bar A is connected by mass less rod and at a particular time AB makes an angle 30° with vertical rod and disc is rotating with angular vel. ω as shown in fig. Then what will be the vel. of A. (A) (B) (C) (D) 8. A disc of mass m and radius R is moving on a horizontal surface. If force F is acting on the disc. Then find the angular acceleration of disc. (A) (B) (C) (D) 9. A spool of thread (having mass M radius r and R ) is moving in forward direction by force F. If rolling is taking place find the acceleration. (A) Rα (B) Mα (C) rα (D) MAα 10. A disc of mass m has given a horizontal velocity Vo. And coefficient of friction between the disc and horizontal surface is . Find linear acceleration of disc when rolling starts. (A) Mg (B) μg (C) V0g (D) None of these. 11. A disc of mass m has given a horizontal velocity Vo. And coefficient of friction between the disc and horizontal surface is . Find angular acceleration of disc when rolling starts. (A) (B) (C) (D) 12. A disc of mass m has given a horizontal velocity Vo. And coefficient of friction between the disc and horizontal surface is . Find linear velocity of disc when rolling starts. (A) (B) (C) (D) 13. A disc of mass m has given a horizontal velocity Vo. And coefficient of friction between the disc and horizontal surface is . Find angular velocity of disc when rolling starts. (A) (B) (C) (D) CMP: A spool of thread was moving in horizontal direction by force F as shown in the figure. The coefficient of friction between spool and horizontal surface is . 14. The value of angle θ for which spool will not move and not rotate. (A) θ = tan–1 (r/R) (B) θ = cot–1 (r/R) (C) θ = cos–1 (r/R) (D) θ = sin–1 (r/R) 15. For above angle of θ, the maximum value of force F for which it will remains in equilibrium (A) (B) (C) (D) 16. Let at θ = θ1 body is in equilibrium then if θ > θ1 then direction and magnitude of friction on ground will be (if rolling is taking place I = mrR/2 about centre) (A) bark ward (B) bark ward (C) bark ward (D) bark ward 17. for θ = 180, where K = 1Kg/N. Then direction and magnitude of friction force on contact surface will be (A) F/15 (B) F/25 (C) F/20 (D) none of these 18. and . Then the friction force, linear and angular acceleration of spool will be (moment of inertia of spool about centre = I) (A) 0, F/M, Fr/I (B) 0, 0, Fr/I (C) 0, 0, 0 (D) none of these 19. In part (iv) the value of linear acceleration of spool will be (A) (B) (C) (D) none of these 20. In part (iv) if μ = F/(125m) k, where K = 1Kg/N. The friction is equal to (A) (B) (C) (D) none of these 21. A cubical block of side L rests on a rough horizontal surface with coefficient of friction μ. A horizontal force F is applied on the block as shown. If the coefficient of friction is sufficiently high so that the block does not slide before toppling, the minimum force required to topple the block is (A) infinitesimal (B) mg/4 (C) mg/2 (D) mg(1–μ) 22. Consider a body, shown in figure, consisting of two identical balls, each of mass M connected by a light rigid rod. If an impulse J = MV is imparted to the body at one of its ends, what would be its angular velocity? (A) V/L (B) 2V/L (C) V/3L (D) V/4L 23. The torque τ on a body about a given point is found to be equal to A × L where A is constant vector, and L is the angular momentum of the body about that point. From this it follows that (A) is perpendicular to L at all instants of time (B) the component of L in the direction of A does not change with time. (C) the magnitude of L does not change with time. (D) L does not change with time (A) A, B, C (B) A, C, D (C) B, C, D (D) A, B, C, D 24. A particle is confined to rotate in a circular path decreasing linear speed, then which of the following is correct? (A) (angular momentum ) is conserved about the centre (B) only direction of angular momentum is conserved (C) It spirals towards the centre (D) Its acceleration is towards the centre 25. A thin circular ring of mass ‘M and radius r is rotating about its axis with a constant angular velocity , Two objects, each of mass m, are attached gently to the opposite ends of a diameter of the ring. The wheel now rotates with an angular velocity (A) (B) (C) (D) 26. Two point masses of 0.3 kg and 0.7kg are fixed at the ends of a rod of length 1.4 m and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum is located at a distance of (A) 0.42 m from mass of 0.3 kg (B) 0.70 m from mass of 0.7 kg (C) 0.98 m from mass of 0.3 kg (D) 0.98 m from mass of 0.7 kg 27. A mass m is moving with a constant velocity along a line parallel to the x–axis, away from the origin. Its angular momentum with respect to the origin (A) is zero (B) remains constant (C) goes on increasing (D) goes on decreasing. 28. A smooth sphere A is moving on a frictionless horizontal plane with angular speed ω and centre of mass velocity υ. It collides elastically and head on with an identical sphere B at rest. Neglect friction everywhere. After the collision, their angular speeds are ωA and ωB, respectively. Then (A) ωA < ωB (B) ωA = ωB (C) ωA = ω (D) ωB = ω 29. A disc of mass M and radius R is rolling with angular speed ω on a horizontal plane as shown in Figure. The magnitude of angular momentum of the disc about th origin O is (A) (1/2)MR2ω (B) MR2ω (C) (3/2)MR2ω (D) 2MR2ω 30. A cubical block of side a is moving with velocity V on a horizontal smooth plane as shown in Figure. It hits a ridge at point O. The angular speed of the block after it hits O is (A) 3V/(4a) (B) 3V/(2a) (C) (D) zero   18. ANWER SHEET 1. B 11. C 21. C 2. A 12. D 22. A 3. A 13. A 23. A 4. D 14. C 24. B 5. B 15. D 25. C 6. C 16. D 26. C 7. A 17. B 27. B 8. C 18. B 28. C 9. A 19. B 29. C 10. B 20. A 30. A