## Kerala Plus One Physics Notes Chapter 7 Systems of Particles and Rotational Motion

Introdution:

When we consider the size of the body we have to consider it as a system of particles. If the body is rigid, the distance between the particles do not change. This cannot be followed strictly as bodies bend and stretch when forces are applied. But we shall consider all solids as I rigid bodies in this chapter and proceed.

Regid Body:

A body is said to be rigid if the distance between any two points in the body always remains the same even under the action of external forces.

What Kind of Motion can a Rigid Body Have?

- Pure translational motion eg., when a body is sliding on a smooth surface. Then all the particles of the body will have the same velocity.
- Pure rotational motion: In such a motion a rigid body rotates about a fixed axis. Every particle moves in a circle whose centre is called the axis of rotation, eg..ceiling fan, mixie blade etc.
- Combination of translational and rotational motion: When a cylinder rolls downs an inclined plane, its motion is both translational (as it moves down the plane) and rotational (as it is rotating about an axis).
- Centre Of mass :The centre of mass of a system of particles is the point where all the mass of the system may be assumed to be concentrated.

Centre of mass of a Two particle system:

Consider two particles of masses m_{1} and m_{2} with position vectors respectively with respect to the origin O. Now the position co-ordinate of the center of mass C is defined as

Special Care:

Thus the centre of mass of two equal masses lies exactly at the center of the line joining the two masses.

x, y and z coordinates of centre of mass of a two particle system are

Centre of mass of n-particle system

Consider a system of n particles of mass m_{1} ,m_{2},….. m_{n} with position vectors respectively. Then the centre of mass of the system of n particle is defined as

CAQ:

Question 1.

Three masses 3,4 and 5 kg are located at the comers of an equilateral triangle of side 1 m. Locate the centre of mass of the system.

Answer:

Suppose the equilateral triangle lies in the XY-plane with mass 3 kg at the origin. Let

(x, y) be the co-ordinates of CM.

Example of centre of mass motion:

Explosion of a bomb in air. When a bomb is projected upwards it will follow a parabolic path (remember in the 4th chapter). After it explodes, the fragments will travel in their own parabolic path. Since the force of explosion is internal force, the centre of mass of the system will continue to follow the same parabolic path of the shell before explosion.

Centre of mass of regular bodies:

For bodies having regular geometrical shape and uniform mass (or density)

Linear Momentum of a System of Particles:

Momentum conservation and centre of mass

The equation means that, the total 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.

Momentum Conservation and Centre of mass motion:

The total momentum of the centre of mass is conserved if no external force acts.

If the total external force acting on the system is zero; the centre of mass moves with a constant velocity.

CAQ

Question 2.

A shell is fired from a cannon with a speed of 100 m/s at an angle 600 with the horizontal (positive x-direction). At the highest point of its trajectory, the shell exposed into two equal fragments. One of the fragments moves along the negative x-direction with a speed of 50 m/s. What is the speed of the other fragment at the time of explosion.

Answer:

The velocity of the shell at the highest point of trajectory is

v_{m} = u cos θ = 100 cos 60° = 50 m/s

Let v_{1} be the speed of the fragment which moves along the negative x-direction and the other fragment has speed v_{2} which must be along the positive x-direction. Now from momentum conservation, we have

Angular velocity (ω) and its relationship with linear velocity (v):

Consider a rotating body. All the particles in it are moving in circles about an axis.

Torque and Angular momentum:

Torque (Moment of a force)

Consider opening a door. The door turns on its hinges when you apply force on its handle. The larger the force, the more it turns. Also you will notice that it is easier to open the door by applying force near the end than near the hinges. This means that the turning effect is more when the distance of force from axis of rotation is more. Torque or moment of force about a point is measured as the product of force and the perpendicular distance between the point and the line of action of force.

Torque = Force × perpendicular distance

Consider a rod which is free to rotate about an axis O. Let a force F act on the end P of the rod to rotate it. Then torque,

x =F× OQ = F× d

Note:

Torque is taken positive if turning tendency is anticlockwise and negative if clockwise.

CAQ

Question 3.

Find the torque of a force about the origin which acts on a particle whose position vector is

Answer:

Angular momentum:

The angular momentum is the moment of momentum of the particle about a point. It is the product of momentum of the particle and perpendicular distance between the point and direction of momentum.

Consider the same particle P, used in the above example rotating about an axis O.

Let p = mv be the momentum of the particle, which makes an angle 0 with r.

Angular momentum, L=p×d

L=p×r sin θ

Note:

It can be proved that the torque on a particle is equal to its rate of change of angular momentum, just like force is the rate of change of linear momentum (Newton’s 2 law).

So,

Couple:

A pair of equal and opposite force acting on a body along two different lines of action makes a couple, eg., turning a pipe.

Moment of couple:

It can be found by taking the moments of the two forces about any point and then adding them. Consider the figure below,

Moment of couple about O is

Z = F×OA + F×OB

Z = F ×(OA + OB) = F × AB

z=Fd

So, moment of a couple = force× perpendicular distance between the two forces.

Centre of gravity :

Any irregular body can be balanced on a vertical support if we balance it carefully. The point where it balances is called the centre of, gravity. The body is now in both translational and rotational equilibrium.

The torques produced on the left side is cancelled by the torques produced on the i right side. That is, the total gravitational torque I about the centre of gravity is zero. Thus we define the centre of gravity of a body as that point where the total gravitational torque is zero.

Moment of Inertia:

We know that interia in straight line motion is the inability of a body to change its state of rest or of unifrom motion, without the help of an external force. The same property in rotational motion is called moment of inertia.

“Moment of inertia of a body about a given axis i is defined as the property of the body due to which it is unable to change its position of rest or of uniform rotational motion without the help of an external torque.”

It depends on two factors:

1. Mass of the body.

2. Distribution of mass about the axis of rotation.

Thus moment of inertia of a particle about an axis is the product of the mass of the particle and the square of the distance of the particle from the axis.

SI unit is kg m^{2}.

Moment of inertia of a rigid body:

Consider a rigid body capable of rotation about an axis AB. Let the body consist of particles of masses. m_{1} m_{2}, m_{3 }……. m_{n} at distance r_{1} r_{2}, r_{3 }……. r_{n}

Radius of Gyration (K):

It is the defined as the distance from axis of rotation at which, if the whole mass of the body was concentrated, then its moment of inertia about that point would be the same as the moment of inertia of actual distribution of mass.

I=Mk^{2}

Radius of gyration (k) of a body is the square root of ratio of moment of inertia and total mass of the body.

ie., radius of gyration,

Note : We can determine the radius of gyration (k) of the ring about any axis by equating its M.l. about that axis to M k2. For example, the radius of gyration of a thin ring about any diameter is given by

Theorems of Perpendicular and Parallel axes:

1. Theorem of perpendicular axes:

This theorem is good for flat bodies which means, thickness much less than length and breadth. The theorem states that “the moment of inertia of a planar body about an axis perpendicular to its plane is equal to the sum of moments of inertia about two mutually perpendicular axes passing through the same point, lying on the plane”.

If I_{x} and I_{y }are the moments of inertia of the planer body about the perpendicular axes OX and OY ,then moment of inertia about z axis (oz) I_{z }= I_{x} +I_{y }

2. Theorem of parallel axes:

This theorem is good for any shape. The moment of inertia of the body about any axis is equal to the sum of moment of inertia of a parallel axis passing through the centre of mass and product of its mass of the body and square of the distance between the two parallel axes.

I=I_{cm}+Mh^{2 }where l is the moment of inertia of the body about the axis PQ.

Work done by a torque:

Consider a force F acting at the point P of a body which undergoes an angular

displacement Δθ.

Work done

So, work done by torque = torque × angular displacement.

Power delivered by a torque:

So, Power = torque × angular velocity

Note: The kinetic energy of a rotating body is given by K.E = 1/2|ωa^{2} and its angular momentum can also be expressed as L =|ω.

Relation between torque and M.l:

Consider a rigid body rotating about a fixed axis with angular velocity ω. We know,

Law of conservation of angular momentum:

When no external torque acts on a system of particles, the angular momentum of the system remains constant.

Example.

An ice skater can increase his angular velocity by bringing his arms close to his body and bringing his stretched legs close together. When the body is close together / = Mt^{2 }decreases so co increases to keep /ω constant.

Rotational and Translational analogues:

Equations of Rotational Motion:

CAQ

Question 4.

A flywheel of mass 25 kg has a radius of 0.2 m. What force should be applied tangentially to the rim of the flywheel so that it acquires an angular acceleration of 2

rad s^{2}?

Answer:

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