UNIT – 5 CENTRE OF MASS & ROTATIONAL MOTION  

In previous units, we have studded object having single particle of finite dimensions, but in this unit, we will study the objects having system of particles, so we have to define a rigid body & their centre of mass

1.Rigid Body:-

A body which has definite shape & definite size is called rigid body.

The distance between the different pairs of particles should not change by applying any force, however no anybody is perfectly rigid body.

2. The concept of system, internal force and external force:-

• System:-

A collection of any number of particles which interact with each other is called a system, any object of finite size may be considered as a system.

• Internal Force:-

The force exerted by the particles of a system on, one another is called internal forces.

• External forces:-

The force exerted by any external agencies on a system is called external forces; the motion of the body is due to external force acting on the body.

3. Types of motion:- 

There may be of three types of motion of a body

• Pure Translatory motion:-

If all the particles of a body are moving with the same velocity, than the motion of body is called of body is called translator motion or straight line motion.

• Pure rotational motion:-

If all the particles of a rigid body are rotating about a fixed axis in circular path perpendicular to the axis, then the motion of body is called pure rotational motion.

In pure rotational motion, the velocity of the different particle is different – different

However their angular velocity is same irrespective of their distance from the axis of rotation

4. Concept of centre of mass:-^imp

Centre of mass of a rigid body may be defined as a point at which whole mass of the body may be supposed to exist. 

All the external force exerted on the system, are applied only on the centre of mass of the body.

Centre of mass is only a mathematical concept. 

Centre of mass may or may not be inside a body, it may also be at outside for system of particles.

5. Position of centre of mass of bodies of regular shape:-

     Uniform hollow sphere –    centre of sphere

     Uniform solid sphere – centre of sphere 

     Uniform circular ring –centre of ring

     Uniform circular disc –centre of disc

     Uniform rod –centre of rod 

     Cube, square, rectangle – point of intersection of diagonal 

     Triangle –point of intersection of medians 

     Solid, hollow cylinder –middle point of the axis of cylinder 

     Cone – on the axis of cone at a point distant 3h/4 from the vertex o where h= OA is height of cone

6. Centre of mass of a two particle system:- ^imp

Suppose a system of two particles having mass  &  placed at A & B, having position vectors  &. Now again suppose that  &  are the external & internal force on A &  &  are internal & internal force on B.

Now from Newton’s 2^nd law of motion 

 = +  ….1   ( is force on   due to )

&  = +…2   ( is force on  due to )

Adding eqⁿ  1 and 2 we get    

++

As from Newton’s 3^rd law of motion 

=   

So   = +

=  =   (total external force = +)

Or   

Or 

Or   

Dividing & multiplying  L.H.S. we get

 

So                

As = 

Comparing with   

We get position vector of centre of mass as       

 

Or  

Thus product of centre of mass with its position vector is equal to sum of product of the particles & their respective position vectors.

Centre of mass of two particle system lies at the line joining of the points. 

If centre of mass lies at origin 

Then 

If  >   then < 

i.e. centre of mass always lies near to heavy body.

7. Centre of mass of n-particle system:-

Suppose a system of n particles having masses , ,…... Whose position vector are  having velocities . 

Now again suppose that  are the external forces acting on the system & each particle experience internal forces due to rest (n-1) particles as 

Here  is force on i^th particle due to j^th particle 

Now form Newton’s 2^nd law of motion          

 

We can write n such eqⁿ for n particle, adding all these eqⁿ s, we get.

The internal force of all the particle cancel out in pairs i,e. 

So       ( = total external force)

   

Or   

Multiplying and dividing L.H.S. by M, we get     

Comparing with                        

We get     

 

Here  is the position vector of centre of mass of n particle system 

8. Motion of centre of mass:-

As we know position vector of centre of mass is given by 

 

OR  

OR  

Now differentiating above eqⁿ both sides 

We get  

Or  

Again differentiating above eqⁿ both sides, 

we get  

Thus product of centre of mass & its acceleration is equal to vector sum of all the forces acting on the system of particles.

The centre of mass changes its position under the Translatory motion & remains uncharged under rotatory motion

9. Momentum  conservation & centre of mass:- ^imp

As we know  

If  

Here m 0 so          

 

Thus velocity of centre of mass remains unchanged if no any external force is acting on the body, which is Newton’s first law of motion.

Also   

Means if no external force is acting on the system of particle then total momentum of system remains unchanged.

10. Some example of motion of centre of mass:-

     Explosion of fire cracker ( rocket bomb):-

Suppose a fire cracker is fired from a point, now its explosion occurs at point P due to its chemicals, here no any external force is acting on the cracker, so that motion  of different fragments if cracker moves in such  a way that the motion of centre of mass remain unaffected by explosion, it will follow same parabolic path.

     Motion of earth moon system around the sun:-

Suppose earth and moon are having a common centre of mass lying near to earth, as both are revolving around the sun, in such a way that their centre of mass always remain in a fixed circular orbit. For this phenomenon, they always remains in opposite to one another during their motion, to have uncharged centre of mass.

Q.1 Two bodies of mass 1 kg and 2 kg are located at (1, 2) and (-1, 3) respectively, calculate the co-ordinate of their centre of mass.

Ans.     

Q.2 Three particles of mass 0.50kg, 1.0kg, 1.5kg are placed at the corners of a right angle triangle as shown in fig then find centre of mass?

 

 

  Q.3 Four particles A, B, C, D having mass m respectively, are placed at the corners of a square of side a. Locate centre of mass?

Ans.   

Q.4 Two uniform roads A and B of the same diameter, having lengths 2m and 3m and masses per unit length as 4kg/m and 6kg/m respectively are joined ends to ends. Find the position of centre of mass of combined rod from the free end of A?

Ans. Mass of rod  

Mass of rod B= 

Thus      

 Q.5 Find the centre of mass of uniform L shaped lamina of mass 3kg as shown in fig?

So                    

 

                                              :- ROTATION MOTION :-

11. Angular Displacement:-

Suppose a body moves in circular path from P to Q making angle  at the centre of a circle of radius r. Then angular displacement  of object in time t is given by 

 

Here S is linear displacement covered by the body.        

Or   

Angular displacement is a vector quantity and its direction may be given by right hand thumb rule.

Angular displacement is dimensionless having unit radian.

12. Angular velocity (ω):-

The time rate of change of angular displacement of a body is called angular velocity. It is denoted by omega (ω)

Suppose an object moves in circular path from P to Q in small time ∆t and covers a angular displacement. 

Then angular velocity  may be given as         

Angular velocity is a vector quantity, 

For anticlockwise rotation   is in upward direction and for clockwise direction ω is in downward direction, its direction can be given by right hand thumb rule.

Unit of is rad s^-1 

13. Relation between angular velocity and linear velocity:-

As we know  

Or          

here r is radius which is constant 

14. Angular Acceleration:-

The time rate of change of angular velocity of the body is called angular acceleration. It is denoted by α

As   

S.I. unit of angular acceleration is rad s^-2 

15. Relation between linear acceleration a and angular acceleration  :-

As   

Or   

Or    

Or   α

Or    

16. Uniform circular motion:-

If a point object moves on a circular path with a constant speed, then the motion of the body is called uniform circular motion.

Time period:-

The time taken by the object to complete one revolution on its circular path is called time period. 

It is denoted by T.

Frequency:-

Number of revolution completed by the body in one second is called its frequency. 

It is denoted by ν

i.e  

Relation between angular velocity, frequency and time period.

As 

17. Rotational kinetic equations:-

When a body is rotating with constant angular acceleration, simple relation between kinematic variable  can be given as

As we know   

Or  

When     t=0, than =  & when t= t   than  

So the above eqⁿ becomes   

 

Or   

Or  

As we know  

Or   ω

When t=0 than  

Now integrating both sides, we get  

 

                               

                                                                                                                                         

 

As we know that    

When 

Integrating eqⁿ 

We get    

Q.6 a flywheel rotating at 420 rpm slows down at a constant rate of 2 rads^-2. 

In how much time will it stop? 

Ans. Here  

s^-2   t=? 

As  

⇒ 

        

So   

Q.7 a grinding stone of radius 2m revolving at 120rpm accelerates to 660 rpm in 9s. 

18. Find angular acceleration & linear acceleration.

Ans. s^-2, 

a =r 

Q.8 Moon is at 3.824  10⁵ km from the centre of earth & require 27.3 days for a revolution. What is angular velocity and centripetal acceleration of the moon?

Ans.

a_c= ms^-2

19. Centripetal Acceleration:-

The acceleration acting on the body undergoing uniform circular motion is called centripetal acceleration. It acts on the object along the radius toward the centre of circular path as shown in fig. 

Appling BAC we get change in velocity as

 

POQ &  ABC are similar  so   

  

Dividing both sides by , 

We get    

    

0r   

This is the required expression for centripetal acceleration

Direction of 

As  act along the centre of circle, so that centripetal acceleration act toward centre along radius of circle, its direction changes according to velocity at every point remains  to velocity & toward centre. The magnitude of  remains constant at all point.

Q.9 Calculate the angular speed of flywheel making 420 revolutions per minutes?

Here revolution minute  

So    

Q.10 a body of mass 10kg revolves in a circle of diameter 0.40m making 1000 revolutions per minute. Calculate its linear velocity & centripetal acceleration?

Here m=10kg r = 0.20m, 

Angular speed 

So

m
Q.11 An insect trapped in a circular groove of radius 12cm moves along the groove steadily & completes revolution in 100sec.

 (i) 

 is acceleration vector a constant vector, what is its linear displacement.

Here r=12cm 

Is not constant vector

 

Linear displacement is zero, as insect completing its revolution, comes on same point.

20. Centripetal Force:-

A force required to move a body in a circular path with uniform speed is called centripetal force.

It always acts along the radius & toward the centre of circular path.

     Expression for centripetal force:-‘

As we know if a body is moving in circular path with  velocity, than its centripetal acceleration may be given as

If m is the mass of the body than centripetal force may be given as                                             

21. Application of centripetal force

The tension provides a necessary centripetal force to a stone rotating in circular path.

The centripetal force is provided by the gravitational force of sun to the planets for their circular motion around sun.

The frictional force provides the necessary centripetal force to car taking circular turn on a road.

To e^– the centripetal force is provided by the electrostatic force between e^– and P.

22. Centrifugal force:-

While moving in a circular path the body has a constant tendency to regain its natural straight line path, this nature gives rise to a force called centrifugal force.

Hence a force that arises when a body is moving actually along a circular path, by virtue of which it regain its natural straight line path.

This force acts along the radius & away from the centre of the circle.

Now we will discuss some application of centripetal & centrifugal force.

23. Rounding a level curved road:-

Suppose a car is going on a curved road of radius r, than necessary centripetal force is provided to car to take a circular turn by the frictional force between tyres and the road.

Now there are three forces acting on car as 

1. The weight of the car is acting in vertically downward direction.  

2. Normal rxⁿ R is acting vertically upward direction.

3. Frictional force facing toward centre.

As there is no any acceleration in vertical direction so 

R-mg=0

R=mg

Car will move without slipping if  

 

Hence maximum velocity of car without skidding is

 

If h is height of car &  is distance between tyres then moment of force at car must be equal & opposite as 

24.Banking of a road:-

The raising of outer edge of a curved road about the inner edge is called banking of road. The angle between outer edge & inner edge is called angle of banking.

Suppose a car of weight mg is moving on a banked curved road at  angle.

If r is the radius of road, than various forces acting on the car are as shown in fig

Now equating the forces along the horizontal & vertical directions respectively we get 

  = 

Also 

2

Dividing eqⁿ 1 and 2 we get    

 

Or    

(Dividing by cosθ both numerator and denominator)

Or   

Or   

 If no friction then

 

If h is height AB of outer edge of the road b=OA  

Then OB= 

From this eqⁿ, we can calculate h usually h<< b. 

therefore h² is negligibly, small compared to b² 

Or  

25. Bending of a cyclist:-

When a cyclist moves on a curved path, leans somewhat inward because horizontal component of the normal reaction provides the necessary centripetal force to take a turn.

Suppose the cyclist bends cycle inward through an angle θ from the vertical.

Then forces along horizontal & vertical directions are as  

 = —–1

 = mg—-2

Dividing eqⁿ 1 by 2 

we get   

Or    

If Speed & curve is sharp angle will be large.

Q.12 A band in a level road has radius 100m. Find the maximum speed with which car turn this road without skidding, if the coefficient of friction is 0.8?

Ans. = 

Q.13 a car of mass 1500kg is moving with 12.5ms^-1 speed on a curved road of radius 20m. What is frictional force? What is.

Ans.  

Q.14 a train has to move a curve of 400 m, by how much should the outer rail be raised with respect to inner rail for a speed of 48km/h. the distance between rails is 1m?

Ans. v= 48km/h  

=