Question

A circular disc reaches from top to bottom of an inclined plane of length $l$. When it slips down the plane, it takes $t$ s. When it rolls down the plane, then it takes $\left( \frac{\alpha}{2} \right)^{1/2} t \, \text{s},$ where $\alpha$ is __________ .

Answer 1

For slipping, acceleration $a_s = g \sin \theta$. For rolling, acceleration $a_r = \frac{g \sin \theta}{1 + k^2/r^2}$, where $k = r/\sqrt{2}$ (for a disc):
$a_r = \frac{g \sin \theta}{1 + 0.5} = \frac{2g \sin \theta}{3}.$
Time ratio:
$\frac{t_r}{t_s} = \sqrt{\frac{a_s}{a_r}} = \sqrt{\frac{g \sin \theta}{\frac{2g \sin \theta}{3}}} = \sqrt{\frac{3}{2}}.$
Given $t_r = \sqrt{\frac{\alpha}{2}} t_s$, equating:
$\sqrt{\frac{\alpha}{2}} = \sqrt{\frac{3}{2}}, \quad \alpha = 3.$

Answer 2

For slipping, acceleration $a_s = g \sin \theta$. For rolling, acceleration $a_r = \frac{g \sin \theta}{1 + k^2/r^2}$, where $k = r/\sqrt{2}$ (for a disc):
$a_r = \frac{g \sin \theta}{1 + 0.5} = \frac{2g \sin \theta}{3}.$
Time ratio:
$\frac{t_r}{t_s} = \sqrt{\frac{a_s}{a_r}} = \sqrt{\frac{g \sin \theta}{\frac{2g \sin \theta}{3}}} = \sqrt{\frac{3}{2}}.$
Given $t_r = \sqrt{\frac{\alpha}{2}} t_s$, equating:
$\sqrt{\frac{\alpha}{2}} = \sqrt{\frac{3}{2}}, \quad \alpha = 3.$