A wheel is at rest in horizontal position. Its M.I. about vertical axis passing through its centre is 'I'. A…
A wheel is at rest in horizontal position. Its M.I. about vertical axis passing through
its centre is 'I'. A constant torque ' $\tau$ 'acts on it for ' $\mathrm{t}$ ' second. The change in rotational
kinetic energy is
$\frac{\tau^{2} t^{2}}{2 I}$
$\left[\frac{\tau t}{2 I}\right]$
$\left[\frac{\tau t}{2 I}\right]^{\frac{1}{2}}$
$\left[\frac{\tau t}{2 I}\right]^{2}$
Solution
If $\alpha$ is the angular acceleration, then the angular velocity after time $t$ is given by $\omega=\alpha t$ but $\alpha=\frac{\tau}{I} \quad \therefore \omega=\frac{\tau}{I} \cdot \mathrm{t}$
kinetic energy $\mathrm{K}=\frac{1}{2} \mathrm{I} \omega^{2}=\frac{1}{2} \mathrm{I} \cdot \frac{\tau^{2}}{\mathrm{I}^{2}} \cdot \mathrm{t}^{2}=\frac{\tau^{2} \mathrm{t}^{2}}{2 \mathrm{I}}$