# ray optics - Tejas Engineers Academy

PHYSICS

ROTATIONAL MOTION

CLASS: XI

CENTRE OF MASS The point where the whole mass of the body can be supposed to be concentrated is known as centre of mass of the body. CENTRE OF MASS OF TWO PARTICLE SYSTEM Consider a system of two particles of masses m1 and m2 placed at points A and B having position   vector as r1 and r2 r2

B m2 C m1

r

A r1

 Let C be their centre of mass. r be the position vector of centre of mass. Suppose: f1 , f2 be the respective external forces acting on m1 and m2. F12 and F21 be the respective internal forces on A and B respectively. a1 and a2 be the acceleration of both particles, and acm be the acceleration of their centre of mass. According to Newton’s second law of motion. [Net force] = [Net mass of system]×[acceleration of C.M]       F12  F21  f1  f 2  (m1  m2 )acm Internal forces are equal and opposite, they cancel out in pair.     f1  f 2  (m1  m2 )acm     m1a1  m2 a2  (m1  m2 )acm   m1 a1  m2 a 2   acm  ---------------------------------------------(1) (m1  m2 ) We know that: acceleration = rate of change of velocity, thus:   dv1 a1  dt Putting in equation (1)   d (v1 ) d (v 2 )  m1  m2 d (vcm ) dt dt   dt (m1  m2 )   d  d  m1v1  m2 v2    (vcm )   dt dt  (m1  m2 )    m1v1  m2 v2   vcm  ------------------------------------------------(2) (m1  m2 ) Velocity is rate of change of displacement: dr i.e. v1  1 dt Putting in equation (2) 1

PHYSICS

ROTATIONAL MOTION

CLASS: XI

  d ( r ) d ( r 1 2)  d (r ) m1 dt  m2 dt   dt (m1  m2 )   d  d  m1 r1  m2 r2    (r )   dt dt  (m1  m2 )     m1 r1  m2 r2 r  (m1  m2 ) If there are n-particles in a system then we have:     m1r1  m2 r2      mn rn r  (m1  m2    mn ) in

m r

 r 

i i

i 1 i n

m i 1

---------------------------------------------------(3) i

COORDINATES OF CENTRE OF MASS Suppose all the particles in a system have coordinates in general as (xi, yi, zi) for i th particle. Then position vector for i th particle will be.  ri  xi iˆ  yi ˆj  zi kˆ and that of centre of mass:  r  x iˆ  y ˆj  z kˆ Substituting these values in equation-(3) i n

 xiˆ  yˆj  zkˆ 

 m ( x iˆ  y ˆj  z kˆ) i

i 1

i

i

i

i n

m i 1

i

Comparing values with same coefficient on LHS and RHS we get: i n

m x

x 

i

i 1 i n

m

i 1 i n

y 

; i

m y i 1 i n

i

m i 1

i

i

i

in

z 

m z i 1 i n

i

m i 1

i

i

2

PHYSICS

ROTATIONAL MOTION

CLASS: XI

Note: 1. Centre of mass of two particles always lies on the line joining their individual centre of masses. 2. If masses are equal their centre of mass lies on the mid of line joining them. 3. If one of the mass is heavier then centre of mass lies close to that mass. 4. If no external force acts on a system, then centre of mass moves with constant velocity. This can proves as follows: Consider a system of particles, Net force acting on system is:   dP F dt Where P is the linear momentum of system. Since no external force acts we have:  dP 0 dt   dP  0   Md (Vcm )  0   Vcm  cons tan t CONCEPT OF TORQUE Torque is defined as the turning effect of a given force.

Mathematically torque of a given force about any point is equal to product of force and perpendicular distance of force from that point. It is denoted by τ Thus τ = (Force applied)×(Perpendicular distance) Consider a particle at P at which a force F is acting as shown. This particle is turning about point O. Distance OP = r, and angle between F and r is φ Thus τ =(F)×(rsinφ)  τ = rFsinφ In vector form it may be written as      r F SI units: SI units of torque is Nm It is a vector quantity and its direction is perpendicular to plane containing r and F. 3

PHYSICS

ROTATIONAL MOTION

CLASS: XI

CONCEPT OF ANGULAR MOMENTUM Turning effect of linear momentum of a particle about any point is called angular momentum. Angular momentum is also called moment of momentum. Mathematically angular momentum of a particle about any point is equal to the product of its linear momentum and its perpendicular distance from that point.

It is denoted by L, thus L = (Linear momentum)×(perpendicular distance) Consider a particle having linear momentum p turning abut point O. Its distance from O is OP = r. and angle between r and p is say φ. By definition L = (p)×(OM) From right triangle: L = (p)×(rsinφ) L = rpsinφ. In vector form:    Lrp SI units: its units are kgm2s-1. It is a vector quantity and its direction is perpendicular to the plane containing r and p.

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PHYSICS

ROTATIONAL MOTION

CLASS: XI

KINETIC ENERGY OF ROTATION OF A BODY Consider a rigid body rotating about axis AB with angular velocity ω as shown. Suppose the body is made of large no. of particles m1, m2,----mn, having perpendicular distances r1, r2-------rn from the axis

Thus every particle of the body follows circular motion with same angular velocity ω, having radii of circular path as r1, r2,------rn respectively. Suppose v1, v2-----vn be the respective linear velocities of the particles. Total kinetic energy of the body is given as: 1 1 1 2 2 2 K .E  m1v1  m2v2      mnvn 2 2 2 putting values of velocities as v1 =r1ω, v2 = r2ω -------vn = rnω, we have K .E 

1 1 1 2 2 2 m1r1  2  m2 r2  2      mn rn  2 2 2 2

 K .E 

1 2 2 2 m1r1  m2 r2      mn rn  2 2

 K .E 

1 2 I 2

Where

I  m1r1  m2r2      mn rn is known as the moment of inertia of body. 2

2

2

MOMENT OF INERTIA Moment of inertia of a body rotating about a given axis is defined as the product of masses and perpendicular distances of all the particles of a body from the axis of rotation, mathematically is given as: 2 2 2 I  m1r1  m2r2      mn rn

5

PHYSICS

ROTATIONAL MOTION i n

I   mi r

or

CLASS: XI

2 i

i 1

2

SI units: SI units of moment of inertia are kgm . Its dimensions are [ML2T0] Moment of inertia of a body depends upon two factors: (1) Choice of axis of rotation. (2) Mass of body. RELATION BETWEEN ANGULAR MOMENTUM AND MOMENT OF INERTIA

Consider a rigid body rotating about axis AB. Suppose it is made of n-particles with masses m1, m2,-----mn rotating with angular velocity. Every particle of body follows circular motion with radii r1, r2,-----rn respectively. Angular momentum of body is given by: L = r1p1 + r2p2 + ---------+ rnpn.

 L = r1m1v1 + r2m2v2 + ------------+ rnmnvn. Putting values of velocities as v = rω, we have: L = r1m1r1 ω+ r2m2r2 ω+ ------------+ rnmnrnω

 L = (m1r12 + m2r22 + ----------+ mnrn2) ω  L = Iω Where I is the moment of inertia of body.

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PHYSICS

ROTATIONAL MOTION

CLASS: XI

RELATION BETWEEN MOMENT OF INERTIA AND TORQUE

Consider a rigid body rotating about axis AB. Suppose it is made of n-particles with masses m1, m2,-----mn rotating with angular acceleration α. Every particle of body follows circular motion with radii r1, r2,-----rn respectively. Net torque acting on body is: τ = F1r1 + F2r2+ -------+Fnrn. Putting force as product of mass and acceleration. τ = m1a1r1 + m2a2 r2 + -----------+mnanrn. using a=rα 2 τ = [m1r1 + m2r22+ -------+mnrn2]α Hence τ = Iα, Where I is the moment of inertia of body about given axis of rotation. RADIUS OF GYRATION

Radius of gyration of a body rotating about a given axis is defined as the shortest distance of a point from axis of rotation where whole mass of body may suppose to be concentrated. It is denoted by K. Consider a body consisting of n particles each of mass m. Its moment of inertia is: I = mr12+ mr22 + -----------------+mrn2 I = m(r12+ r22+ -----------+ rn2) 7

PHYSICS

ROTATIONAL MOTION

CLASS: XI

Multiplying and dividing by n I

mn 2 2 2 (r1  r2         rn ) n

(r  r2         rn ) ------------------------------------------(1) I M 1 n 2

2

2

Where M is the mass of body: MI of body is also given by: I = MK2 -------------------------------------------------------(2) From 1 and 2: (r1  r2        rn ) n 2

MK 2  M

2

(r1  r2         rn ) n 2

K

2

2

2

Thus radius of gyration of a body is the root mean square of all the perpendicular distances of constituting particles from axis of rotation. THEOREM OF PARALLEL AXES It states that moment of inertia of a body about an axis KL parallel to axis AB which passes through centre of mass of body is equal to the moment of inertia of body about axis AB plus product of mass of body and square of shortest distance between the two axes: i.e. IKL = IAB + Mh2

THEOREM OF PERPENDICULAR AXES It states that sum of moment of inertia of a plane lamina about any two axis mutually perpendicular to each other passing in the plane of body is equal to the moment of inertia of same lamina about an axis passing through common point of intersection and perpendicular to the plane of lamina. i.e. Iz = Ix + Iy

MOMENT OF INERTIA OF A THIN CIRCULAR RING 8

PHYSICS

ROTATIONAL MOTION

CLASS: XI

Consider a thin circular ring of mass M and radius R. To determine its moment of inertia about an axis passing through its centre and perpendicular to its plane. It mass per unit length = M ------------------------------------------------(1) 2R

Consider a smaller element of length dx and mass dm on its circumference. Small moment of inertia: dI = dm R2 -------------------------------------------------------(2) Mass of element will be dm = M dx 2R

putting this value in (2) dI 

M 2 R dx 2R

Integrating: MR 2 I 2R

2R

 dx 0

MR 2 2R  0 2R  I  MR2

I

MOMENT OF INERTIA OF THIN CIRCULAR DISC

Consider a thin circular disc of mass M and radius R. To determine its moment of inertia about an axis AB passing through its centre and perpendicular to its plane. Mass per unit area: = M2 R

Considering an elemental ring of mass dm, and radius x, suppose its thickness is dx. Its small moment of inertia will be: 9

PHYSICS

ROTATIONAL MOTION

CLASS: XI

dI = dm x2 -------------------------------------------------------(1) Mass of ring: dm = M2 (2xdx ) R

Where 2πxdx is the area of ring obtained by opening it: as shown. From 1: 2M dI  2 x 2 dx R Integrating: 2M R 2 I  2  x dx R 0 2M I  2 R I

 R4    0   4 

1 MR 2 2

MOMENT OF INERTIA OF A HOLLOW CYLINDER Consider a hollow cylinder of mass M and radius R. Consider a small elemental ring mass dm on it. It small moment of inertia: dI = dmR2 Integrating: M

I R

2

 dm

 I  MR 2

0

MOMENT OF INERTIA OF A SOLID CYLINDER

Consider a solid cylinder of mass M and radius R. Imagine there is a thin circular disc of mass dm. Small moment of inertia of the disc: dI = 1 dmR 2 2

Integrating: I

R2 2

M

1

 dm  I  2 MR

2

0

MOMENT OF INERTIA OF THIN ROD 10

PHYSICS

ROTATIONAL MOTION

CLASS: XI

Consider a rod of mass M and length l, rotating about axis AB passing through its mid point and perpendicular to its length. Imaging there is a small element of length dx and mass dm. Small moment of inertia of element: dI = dmx2 --------------------------------------------------(1) M Mass per unit length = -------------------------------------------------(2) l Mass of the element: M dm = dx l Putting in 1. M 2 dI  x dx l Integrating: l/2 M I x 2 dx  l l / 2 I

M l

 (l / 2) 3 (l / 2) 3     3   3 1  I  Ml 2 12

PRINCIPLE OF CONSERVATION OF ANGULAR MOMENTUM It states that if no external torque acts on a system then vector sum of angular momentum of all the particles present in system remains conserved and it do no changes with their mutual action or reaction. Torque is given by: dL  dt since no external torque acts on system thus: dL 0 dt  dL  0  L  cons tan t Angular momentum is given by L = Iω, thus Iω = constant. i.e. I1ω1 = I2ω2. 11

PHYSICS

ROTATIONAL MOTION

CLASS: XI

APPLICATIONS 1. When a diver contracts his body while diving into water, his spinning increases. On contracting the body, moment of inertia decreases. We have Iω = constant, i.e. 1  , I Hence ω increases. 2. While jumping from one roof to the other a cat expands her arms. On expanding her arms she increases her moment of inertia. Thus ω decreases. 3. A dancing girl contracts her arms while performing feats. This causes decrease in moment of inertia of her body due to which ω increases. ACCELERATION OF A BODY DOWN THE ROUGH INCLINED SURFACE

Consider a body of mass M rolling down without slipping with linear acceleration a and angular acceleration α. Frictional force be f. it is in opposite direction to a. Force equation of body is: Mgsinθ – f = Ma---------------------------------------------------(1) Torque experienced by the body: τ = Iα  Rfsin900 = I(a/R) Ia  f  2 ------------------------------------------------------(2) R From 1 and 2, Ia Mg sin   2  Ma R Mg sin  a I M 2 R If body is cylinder put I = (1/2)MR2. we have: 12

PHYSICS

ROTATIONAL MOTION

CLASS: XI

Mg sin  (1 / 2) MR 2 M R2 2  a  g sin  3 ACCLERATION OF A POINT MASS TIED WITH A HORIZONTAL CYLINDER a

Consider a horizontal cylinder on which string is wounded. Suppose a point mass m is tied at one end. Suppose downward acceleration of mass is a and that the angular acceleration of cylinder is α. mg – T = ma-------------------------------------------------(1) Torque experienced by the body: τ = Iα RTsin900 = I(a/R)  Ia  T  2 ------------------------------------------------------(2) R From 1 and 2, Ia mg  2  ma R mg a I m 2 R If body is cylinder put I = (1/2)MR2. we have: mg a (1 / 2) MR 2 m R2 mg a M m 2

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PHYSICS

ROTATIONAL MOTION

CLASS: XI

ASSIGNMENT CENTRE OF MASS 1. Give two examples where the centre of mass lies outside the body mass. 2. Show that the distance of the two particles from their centre of mass is inversely proportional to their masses 3. Three equal masses having Coordinate of their positions are (1,1), (2,2) and (3,3) respectively find the position of their centre of mass. [(2,2)] 4. In the HCl molecule the separation between the nuclei of hydrogen and chlorine atoms is 1.27 A. If the mass of chlorine atom is 35.5 times that of a hydrogen atom, find the distance of the centre of mass of HCl molecule from hydrogen atom.[1.235 A from H] 5. In the CO molecule the centre of mass is at a distance of 0.068 nm from the carbon atom. Find the separation between the nuclei of carbon atom and oxygen atom. 6. A child is standing at one end of a long trolley moving with a speed v on a smooth horizontal track. If the child start running towards the other end of the trolley with a speed u, then find with which velocity the centre of mass of trolley and child will move. [NCERT] 7. Four particles of masses m, m, 2m and 2m are placed at the four corners of a square of side a as shown in fig-2. Find the (x,y) coordinate of the centre of mass. [a/2, 2a/3] 8. A circular plate of uniform thickness has a diameter of 56 cm. A circular portion of diameter 42 cm is removed from one edge of the plate. Find the position of the centre of mass of the remaining portion. 9. Two bodies of masses 1 kg and 2 kg are located at (1,2) and (-1,3) respectively. Calculate the coordinates of the centre of mass. [-1/3,8/3] 10. Two bodies of masses 0.5 kg and 1 kg are lying in XY plane at (-1,2), and (3,4) respectively. What are the coordinates of the centre of mass?[5/3,10/3] 11. Three point masses of 1 kg , 2kg and 3 kg lie at (1,2), (0,-1) and (2,-3) respectively. Calculate the coordinates of the centre of mass of the system. [7/6, -3/2] 12. Three identical spheres each of radius r and mass m are placed touching each other on a horizontal floor. Locate the position of centre of mass of the system. [r,r/31/2] 13. Two particles of mass 2 kg and 1 kg are moving along the same line with speeds 2 m/s and 5 m/s respectively. What is the speed of the centre of mass of the system if both the particles are moving (a) in same direction (b) in opposite direction? [3 m/s, 1/3 m/s in the direction of the motion of 1 kg] 14. Give the location of the centre of mass of a (i) sphere (ii) cylinder (iii) ring and (iv) cube each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body? [NCERT] 15. A system consists of three particles located at the corners of a right triangle. Find the centre of mass from the origin. 16. The centre of mass of a body 14

PHYSICS

ROTATIONAL MOTION

CLASS: XI

(a) lies always at the geometrical centre (b) lies always inside the body (c) lies always outside the body (d) may lie within or outside the body 17. Find the position of centre of mass of a system consisting of two particles of masses m 1 and m2 separated by a distance L apart. [mL/(m1+m2)] 18. Four identical spheres each of radius 10 cm and mass 1 kg are placed on a horizontal surface touching one another so that their centre are located at the corners of square of side 20 cm. What is the distance of their centre of mass from centre of either sphere? 19. From a square sheet of uniform density, a portion is removed as shown in the figure. Find the centre of mass of the remaining portion if the side of the square is a.

a

[Ans: 7a/12 , a/2] 20. A solid sphere and a ring both of radius 7 cm are placed on a horizontal table simultaneously with initial angular speed of 10 rad/s. If coefficient of kinetic friction is 0.18, which of the two will start to roll earlier? [Ans: sphere] 21. Four particles of mass m, m 2m and 2m are placed at the four corners of a square of side a. Find the centre of mass of the system. [Ans: a/2 , 2a/3] 22. From a uniform circular disc of diameter d, a circular hole of diameter d/6 and having its centre at a distance of d/4 from the centre of the disc is scooped out. Find the centre of mass of the remaining portion. [Ans: The cm of the remaining portion of the disc is at a distance of d/140 towards left of the origin O.] 23. A 2kg body and 3kg body are moving along x-axis. At a particular instant, the 2 kg body is 1 m from the origin and has a velocity of 3 m/s and the 3 kg body is 2 m from the origin and has a velocity of -1 m/s. find the position and velocity of the centre of mass and also find the total momentum. [Ans:1.6 m, and 0.6 m/s] 24. A disc of mass m and radius R rolls on a horizontal surface. If the velocity of the disc is v, find the height h to which the disc will rise on the incline of angle θ. [Ans: 3v2/4g] 25. Calculate the angular momentum of the electron of the hydrogen atom about its own nucleus. Given that Mass of the electron = 9.1×10-31 kg. Radius of the electron orbit = 0.58×10-10m. Time for one revolution of the electron in its orbit = 1.51×10-16s [Ans: 1.27×10-34kgm2s-1] CONCEPTUALS 26. What is conserved when a planet revolves around a star? 27. Why spokes are provided in a bicycle wheel? 28. A cat is able is land on her feet after a fall. Which principle of physics is being used by her? 29. A ballet dancer, an acrobat and an ice skater make use of an im0portant principle in Physics. Name that principle. 30. Would the equations v =rω and a = rα hold good even in the case of rolling with slipping? 31. What is the moment of inertia of a rod of mass M, length l about an axis perpendicular to it through one end? 32. Given two s[spheres of equal masses. One rolls down a smooth inclined plane of height h. The other falls freely through height h. In which case the work done is more? 33. Two satellites of equal masses are orbiting at different heights. Will their moments of inertia be the same or different? 15

PHYSICS

ROTATIONAL MOTION

CLASS: XI

PHYSICS

ROTATIONAL MOTION

CLASS: XI

What is the magnitude of angular momentum of the cylinder about its axis? [Ans: 3125 J, 62.51 s] 58. (a) A child stands at the centre of a turn table with his two arms outstretched. The turn table is set rotating with an angular speed of 40 rpm. How much is angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction. (b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy? [Ans: 100 rpm] 59. A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30N? What is the linear acceleration of the rope? Assume that there is no slipping. [Ans: 25 rad/s2 ; 10m/s2] 60. A cylinder of mass 5 kg and radius 30 cm and free to rotate about its axis, receives an angular impulse of 3kgm2s-1 initially followed by a similar impulse after every 4s. What is the angular speed of the cylinder 30 sec after the initial impulse? The cylinder is at rest initially. MOMENT OF INERTIA

PHYSICS

ROTATIONAL MOTION

CLASS: XI

PHYSICS

ROTATIONAL MOTION

CLASS: XI

87. A wheel and its axle have a total mass of 50 kg. The axle has a diameter of 10 cm. A 2500g mass in fastened to a cord wrapped round the axle. When started from rest the mass moves a distance of 280 cm in 7 s. find the radius of gyration of the rotating system. Neglect friction. [10.3 cm]

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