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

Correct Answer: 3

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: 
\[ \sqrt{\frac{\alpha}{2}} = \sqrt{\frac{3}{2}}, \quad \alpha = 3. \]