Question

A torque of 30 N-m is applied on a 5 kg wheel
whose moment of inertia is 2 kg-m² for 10 sec.
The angle covered by the wheel in 10 sec will be
(a) 750 rad (b) 1500 rad
(c) 3000 rad (d) 6000 rad

Answer 1

$\Large\boxed{\sf{(a)\quad\!\!750\ rad}}$

Given,

$\sf{\tau=30\ N\ m}$

$\sf{m=5\ kg}$

$\sf{I=2\ kg\ m^2}$

$\sf{t=10\ s}$

The angular acceleration,

$\longrightarrow\sf{\alpha=\dfrac{\tau}{I}}$

$\longrightarrow\sf{\alpha=\dfrac{30}{2}\ rad\ s^{-2}}$

$\longrightarrow\sf{\alpha=15\ rad\ s^{-2}}$

Let the initial angular velocity be 0. Then angular displacement is given by,

$\longrightarrow\sf{\theta=\omega_0t+\dfrac{1}{2}\alpha t^2}$

$\longrightarrow\sf{\theta=\dfrac{1}{2}\times15\times10^2\ rad}$

$\longrightarrow\sf{\underline{\underline{\theta=750\ rad}}}$