Question

A circular disc reaches from top to bottom of an inclined plane of length ' $L$ '. When it slips down the plane, it takes time ' $t_{1}{ }^{\prime}$. When it rolls down the plane, it takes time $t_{2}$. The value of $\frac{t_{2}}{t_{1}}$ is $\sqrt{\frac{3}{x}}$. The value of $x$ will be_______.

Answer 1

If disk slips on inclined plane, then it's acceleration
$a_{1}=g \sin \theta $
$L=\frac{1}{2} a_{1} t_{1}{ }^{2} $
$\Rightarrow t_{1}=\sqrt{\frac{2 L}{a_{1}}} \ldots$(i)
If disk rolls on inclined plane, its acceleration,
$a_{2}=\frac{g \sin \theta}{1+\frac{I}{m R^{2}}}$
$a_{2}=\frac{g \sin \theta}{1+\frac{m R^{2}}{2 m R^{2}}}$
$a_{2}=\frac{2}{3} g \sin \theta$
Now $ L=\frac{1}{2} a_{2} \cdot t_{2}{ }^{2} $
$\Rightarrow t_{2}=\sqrt{\frac{2 L}{a_{2}}} \ldots$ (ii)
Now $\frac{t_{2}}{t_{1}}=\sqrt{\frac{a_{1}}{a_{2}}}=\sqrt{\frac{3}{2}}$
$\Rightarrow x=2$