A wheel of moment of inertia 2.5 Kg-$m^{2}$ has an initial angular velocity of 40 ${rads\ }^{- 1}$. A constant torque of 10 Nm acts on the wheel. The time during which the wheel is accelerated to 60 ${rads}^{- 1}$ is

Question

Dr. Abubacker · Accepted Answer

(c) Given, $MI = 2.5\ kgm^{- 2}$
$W = 40\ rad\ s^{- 1}$
$T = 10\ Nm$
As $T = I\alpha$
$10 = 2.5\alpha$
$\alpha = 4\ rad\ s^{- 2}$
Now, $\omega = \omega_{0} + \alpha t$
$60 = 40 + 4 	imes t$
$20 = 4t$
$t = 5\ s$

Q. A wheel of moment of inertia 2.5 Kg-$m^{2}$ has an initial angular velocity of 40 ${rads\ }^{- 1}$. A constant torque of 10 Nm acts on the wheel. The time during which the wheel is accelerated to 60 ${rads}^{- 1}$ is

Explanation

Reviewed by Dr. Abubacker

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