Title 1A. Rotational Dynamics (Rigid body dynamics) ( 1 - 24 ).pdf ✅

NISHITH Multimedia India (Pvt.) Ltd., 1 JEE ADVANCED - VOL-III ROTATIONAL DYNAMICS NISHITH Multimedia India (Pvt.) Ltd., SYNOPSIS Pure Rolling Motion When a rigid body moves on a surface. (i) Its motion is said to be pure rolling if there is no slipping between the points of contact of two (ii) Its motion is said to be rolling with slipping. If there is slipping between the points of contact of two surfaces. Consider a uniform sphere which is rolling on a rough plank as shown and let P and P’ be the points of contact of sphere and plank, respectively. Now, for pure rolling. V V ', P P    and p(t) p'(t) a a    p.c c p' v v v      and 0 a R a    0     R( i) vi v i ˆ ˆ ˆ 0 v R v    If surface is stationary, 0 v = 0 and 0 a 0  v R   and a R   If P P v v ',  then motion is said to be rolling with forward slipping. If P P v v ',  then motion is said to be rolling with backward slipping. If v and a are the velocity and acceleration of C.M. of a rolling body relative to the surface, then v R   and a R   are applicable for pure rolling motion even when the surface is moving. 1. Impure Rolling Motion: (a) In impure rolling motion, the point of contact of the body with the platform is not relatively at rest wrt platform on which it is performing rolling motion, as a result sliding occurs at the point of contact. (b) For impure rolling motion, 0 AB v  i,e., 0 v R  ω v . If platform is stationary i,e., 0 v  0 , then condition for impure rolling motion is v R  ω . V V0  B . V0 V R  .  A (c) Here as v R  ω, so a R  α . (d)As in impure rolling motion, velocity of point of contact is not zero relative to the platform, kinetic friction comes into the existence and kinetic friction is given by k μ N . (e) In this case, friction is opposing the relative motion i.e., rolling motion and is dissipative in nature i.e., work done by friction force is non-zero. (f) Direction of friction force is decided by the fact that it opposes the relative motion i.e., it is opposite to AB v  . (g)Consider a body which is performing rolling motion as shown in the figure. As here v R  ω, so impure rolling v R  ω 0i.e., velocity of point of contact wrt surface is in forward direction, so, friction force acts in backward direction. The free body ROTATIONAL DYNAMICS
Jr Chemistry E/M ROTATIONAL DYNAMICS 2 NISHITH Multimedia India (Pvt.) Ltd., JEE ADVANCED - VOL-III NISHITH Multimedia India (Pvt.) Ltd., diagram of the body is as shown in figure f . .  a m R I . . V V R /2 Rough surface Coefficent of friction  f ma   , R I  α , f  μmg Here, remember a R  α . Solving above equations we can analyse the situation. [When it is not clear that whether the body is sliding or not in rolling motion, assume that the body is performing pure rolling motion and find the magnitude of friction force. If s f  μ N , then solve the question from scratch by considering that the object is performing impure rolling motion.] Equilibrium of a Rigid Body and Toppling: (a)A body (which is in translational equilibrium) is in rotational equilibrium if net external torque acting on the body about any point is zero i.e., Σ 0 τ   If the body is rotational equilibrium but not in translational equilibrium then net external torque would be zero only about centre of mass of body. (b) Equilibrium is stable if PE of system is minimum and equilibrium is unstable if PEof system is maximum. When PE of system is constant (may be zero also), the equilibrium of system is termed as neutral equilibrium. (c)Concept of equilibrium can be used to determine the unknown force acting on a system if it is in equilibrium. Consider a block of mass M, height h, width w is placed on a rough horizontal floor as shown in the figure. f W N 0 h X . h Mg (a)For vertical translational equilibrium: If a force F is applied on one face of the block as shown in figure, then for equilibrium of block, F f  [ f is the static friction force] N Mg  (b)For rotational equilibrium: 0 0 2 2 h h F h f nx           0 x Fh Mg   (c)For sliding to occuor, l f F   F    μN F μMg (d) For no sliding, F  μMg (e)For not toppling, 2 2 0 w wMg x F h    (f) For toppling to occur first i.e., before sliding, 2 0 wMg μMg h  , i.e., 2 w h μ  (g) For sliding to occur before toppling, 0 . ., 2 2 wMg w μMg i e h h μ   Spinning ball falls on a rough horizontal floor: Velocity of ball just before collision with the surface is
NISHITH Multimedia India (Pvt.) Ltd., 3 JEE ADVANCED - VOL-III ROTATIONAL DYNAMICS NISHITH Multimedia India (Pvt.) Ltd., fk F   . 0 mg m F A 0 υ gh  2 After collision the velocity of rebound in vertical direction, y 0 υ eυ  ..... (i) Angular velocity of the ball when it strikes the ground is ω0 . As the floor is rough and the ball is spinning about the horizontal diametere, therefore it slips over the surfaface and as a result a kinetic frictional force acts. Impulse of kinetic friction imparts horizontal velocity. Frictional force is dependent on impulse of normal reaction; so impulse and angular impulse due to this force have to be taken into account. N dt m    υ υ 0 y   . 0 V1 V Vy . N dt m    υ eυ 0 0   ..... (ii) In horizontal direction, friction force exerts an impulse  f dt m k x   υ  μ N dt mυ     x ....... (iii) Dividing equation (iii) by equation (ii), we get   0 1 x υ e υ μ  x 1  0 υ e υ   μ ........ (iv) The torque of friction creates an angular impulse. From angular impulse-angular momentum equation we have  k  0  f rdt I   ω Iω μ N dt r Iω Iω     0  f k  μN From equation (iii), we get m x 0 υ r Iω Iω   Thus we can solve x υ , y υ and ω after collision. Since after collision the ball acquires both the vertical and the horizontal components of velocities therefore it follows the parabolci path after collision. A rolling ball strikes a vertical wall: A ball rolling without slipping on a horizontal surface strikes a vertical wall. The vertical wall is smooth. During collision the impulse due to normal reaction of wall changes the direction of its velcoity. Due to the absence of any tangential force there is no torque and the angular velocity is uncahnged. 0 V0 0 eV 0 . V 0 eV 0 Just Before colision Just after colision . . mg V0 F 0 Ve V FA V Wall is rough
Jr Chemistry E/M ROTATIONAL DYNAMICS 4 NISHITH Multimedia India (Pvt.) Ltd., JEE ADVANCED - VOL-III NISHITH Multimedia India (Pvt.) Ltd., If the wall is rough, impulse of F change momentum in x-direction and impulse of μF changes momentum in y  direction. Angular impulse of friction change angular velocity. If the wall is smooth, there will not be any change in angular velocity. F dt me 0 0   υ mυ  ........(i) μ F dt mυ   0  ..........(ii)    μ Fr dt I ω I ω  C C 0 .......(iii) A spinning ball falls on a rough plane: There is some friction between ball and plank. (i) If 0 u sin θ ω r  , then there will not be any friction force between the ball and the plank 0  r m2 Frictionless At rest m1 0 u sin u sin u cos  u At rest C .   u cos u sin 0 (ii) If 0 u sin θ ω r  1 1 F dt m eu   cos cos θ m u θ  ...( i ) 1 1 μ F dt m υ m μ θ   sin  ...( ii ) eu cos V C  F  F F  k F F k   V '   μ F dt m υ  2 ..........( iii )    μr F dt I ω I ω  C C 0 .... ( iv ) (iii) If μ θ ωr sin  1 1 Fdt m eu   cos cos θ m u θ  ....( i ) 1 1 μ Fdt m υ m u θ   sin  .....( ii ) '    μ Fdt m υ  2 ..... ( iii ) eu cos V C  F  F F  k F F k   V    μr Fdt I ω I ω  C C 0 ...... ( iv ) Instantaneous axis of rotation(IAOR) The position of ICR can be found out in the following cases. Case I: If the velocity of a point on the body and angular velocity are given. Draw a line perpendicular to v  , the instantaneous centre must be lying on this line at a distance ‘r’ given by r = v/w The position of ICR can be found out in the following cases. Case I: If the velocity of a point on the body and angular velocity are given. Draw a line perpendicular to v  , the instantaneous centre must be lying on this line at a distance ‘r’ given by r = v/w Case II: If the lines of action of two non-parallel velocities of two points of the rigid body are given. Draw the normals on the two non-parallel