If rotational inertia parameter of a body rolling down a rough inclined plane (of inclination $ heta$ and height $h$ ) without slipping is given by $\beta=I_{C M} / M R^2$, then time taken by the body to reach the bottom of the inclined plane is given by

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If rotational inertia parameter of a body rolling down a rough inclined plane (of inclination $	heta$ and height $h$ ) without slipping is given by $\beta=I_{C M} / M R^2$, then time taken by the body to reach the bottom of the inclined plane is given by

Aicharya · Accepted Answer

Given, mass of body $=M$ Radius $=R$ Angle of inclination $=\theta$ Height of inclined surface $=h$ Also, $\quad \beta=\frac{I_{\mathrm{CM}}}{M R^2}$

Using, the acceleration produced in body rolling down the inclined plane $a=\frac{g \sin \theta}{1+\frac{I_{\mathrm{CM}}}{M R^2}}=\frac{g \sin \theta}{1+\beta}$ ...(i) By using equation of motion, $s=u t+\frac{1}{2} a t^2$ where $u=0$ (initial velocity) $t=$ time taken to reach at bottom. $s=0+\frac{1}{2} \frac{g \sin \theta}{(1+\beta)} t^2$ $\Rightarrow \quad t^2=\frac{2 s(1+\beta)}{g \sin \theta}$ But, from the diagram, $\frac{h}{s}=\sin \theta$ $\Rightarrow \quad s=\frac{h}{\sin \theta}$ Substituting the value of $s$, we get $\begin{aligned} t & =\sqrt{\frac{2 h(1+\beta)}{g \sin ^2 \theta}} \\ & =\frac{1}{\sin \theta} \sqrt{\frac{2 h(1+\beta)}{g}}\end{aligned}$

If rotational inertia parameter of a body rolling down a rough inclined plane (of inclination $\theta$ and…

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