Question

A body rotating at 20 rad /s is acted upon by a constant torque providing it a deceleration of 2 rad/s², At what time will the body have kinetic energy same as the initial value if the torque continues to act . ?

Answer 1

Given in thew question :-

Initial velocity = 20 rad / second.

therefore $\alpha$ = 20 rad/second²

Due to the torque the effect of declaration on the body , the body will start accelerating in opposite direction.

Therefore the initial Kinetic energy = K.e at any instant.

Hence Init. velocity = velocity at any instant

Now Fin. velocity $\omega'$ =(-ω)= -20 rad/ second.

Init. velocity $\omega = 20 rad/second$

Therefore we know the formula ,

$\omega' = \omega - \alpha t$

since $\omega' = -\omega$

Hence,

$-\omega = \omega - \alpha t$

2.ω = α t

t = (2 × ω) / α

t = (2 × 20 )/2

t = 20 second.

Hope it Helps :-)

Answer 2

Torque in rotatory motion is similar to force in linear motion (though the units are different).

Now, let us assume that the body is rotating in clockwise direction(+ve sign),

As the torque continues to act, the deceleration will exist even after the body stops rotating clockwise; that is the body will now start rotating in anti-clockwise direction(-ve sign).

Now, the time taken by the body to attain 20 rad/s in the anti-clockwise direction is asked. (When the angular velocity becomes -20 rad/s, the kinetic energy will still be the same as K.E = ¹/₂ Iω²)

We know ω = ω₀ + αt (similar to v = u+at)

ω → final angular velocity

ω₀ → initial angular velocity

α → angular acceleration

→ -20 = 20 + (-2)t

→ -2t = -40

→ t = 20 seconds

Therefore, the time taken to reach the same velocity in anticlockwise direction (that is same kinetic energy) is 20 seconds.

Hope it helps!!