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If rotational inertia parameter of a body rolling down a rough inclined plane (of inclination $\theta$ 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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$t=\frac{1}{\sin \theta} \sqrt{\frac{(1+\beta) 2 h}{g}}$
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
$\begin{aligned} t & =\sqrt{\frac{2 h(1+\beta)}{g \sin ^2 \theta}} \\ & =\frac{1}{\sin \theta} \sqrt{\frac{2 h(1+\beta)}{g}}\end{aligned}$
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
$\begin{aligned} t & =\sqrt{\frac{2 h(1+\beta)}{g \sin ^2 \theta}} \\ & =\frac{1}{\sin \theta} \sqrt{\frac{2 h(1+\beta)}{g}}\end{aligned}$
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