A wheel of moment of inertia $5 imes 10^{- 3}kg - m^{2}$ is making 20 revolutions/sec. The torque required to stop it in $10\ sec$ is

Question

A wheel of moment of inertia $5 	imes 10^{- 3}kg - m^{2}$ is making 20 revolutions/sec. The torque required to stop it in $10\ sec$ is

Dr. Abubacker · Accepted Answer

(a) $\alpha = \frac{2\pi(n_{2} - n_{1})}{t} = \frac{2\pi(0 - 20)}{10} = - 4\pi\ rad/s^{2}$
Negative sign means retardation
Now $	au = I\alpha = 5 	imes 10^{- 3} 	imes 4\pi = 2\pi 	imes 10^{- 2}N - m$

Q. A wheel of moment of inertia $5 \times 10^{- 3}kg - m^{2}$ is making 20 revolutions/sec. The torque required to stop it in $10\ sec$ is

Explanation

Reviewed by Dr. Abubacker

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