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Laws for Rotary Motion

First law. It states that a body continues in its state of rest or of rotation about an axis with constant angular velocity unless it is compelled by an external torque to change the state.

In actual practice, we see that when a body is rotating about an axis with constant angular velocity, the body does not continue in its state of rotation, but comes to rest after some time. This is due to air resistance and friction between the body and its bearing or axis. If these forces had been absent, the body would have gone on rotating indefinitely.

Second law. It states that the rate of change of angular momentum of a rotating body
is proportional to the external torque applied on the body and takes place in the direction of the torque.

Consider a body of moment of inertia I rotating with an angular velocity

t = Time in seconds.

Initial angular momentum of the body

= Moment of Inertia × Initial angular velocity = I × ω0 =Iω0

Final angular momentum =I × ω = Iω

Change of angular momentum

= Final angular momentum-Initial angular moments

Laws for Rotary Motion

Laws for Rotary Motion

Problem 5.75. Two blocks weighing 100 N and 40 N are supported at the ends of a rope of negligible weight which is passing over the rough surface of a pulley mounted on a horizontal axle. The pulley may be assumed as a solid disc with a weight of 50 N. Friction in the bearings of the pulley may be neglected. Find the tension on the two parts of the two rope and the linear acceleration of the blocks.

(AMIE Summer, 1984)

Sol. Given:

Bigger weight, W1 = 100 N

Smaller weight, W2 = 40 N

Weight of pulley, W0 = 50 N

Pulley is given as a solid disc.

Let a = Linear acceleration of the system

T1 = Tension in the rope to which bigger weight is attached

T2 = Tension in the rope to which smaller weight is attached.

(i) For acceleration, using equation (5.48),

Problem 5.76.A cage weighing 2940N is raised by a rope, one end of which is wrapped round a drum 120 cm in diameter and weighing 735 N and having a radius of gyration of 55 cm. The drum is rotated by an electric motor, which exerts constant torque of 4000 N m. If the rope is tight when the drum begins to rotate, determine:

(i)                           the acceleration of the cage,

(ii)                        the tension in the rope, and

(iii)                       the time required to raise the cage 20 m from ground. Neglect the loss of energy due to friction and weight of the rope. Take g = 9.8 m/s2.

(AMIE Winter, 1982)

Sol. Given:

Weight of cage, W= 2940N

Diameter of drum, D = 120 cm = 1.20 m

5.10.6. D’ AIembert’s Principle Applicable to Rotary Motion. It states that when
external torques (also called active torques) acts on a system having rotating motion, then the algebraic sum of all the torques acting on the system due to external forces and reversed active forces including the inertia torques (taken in the opposite direction of the angular acceleration) is zero.

To illustrate the above principle, consider the two following cases:

1. Rotation due to a weight W attached to one end of a string, passing over a pulley of weight W0 (Refer to Art. 5.10.3 on page 421 where this problem is done by another method).

A weight W is attached to one end of an inextensible string, which passes over a pulley of weight W0 the other end of the string is attached to the periphery of the pulley as shown in Fig. 5.47. When the weight W moves downwards, the rotation to the pulley is caused in clockwise direction.

Let a = Linear acceleration of the weight W

α = Angular acceleration of the pulley

R = Radius, of the pulley

1= Moment of inertia of the pulley about the axis of rotation.

From equation (5.13), we know

Linear acceleration = Angular acceleration × Radius.

a = α × R

Reversed effective force on weight W

= – (Mass of W) × Acceleration

Laws for Rotary Motion