QJEEAdv07Phfe6c: JEE Advanced 2007 Physics - Class 11 JEEAdv07Phfe6c Solution: A small object of uniform density rolls up a curved surface…

By applying conservation of Energy, we have,

$\frac{1}{2} m v^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=m g h_{\max }$ ..... (i)

For pure rolling,

$v=R\omega$

$\Rightarrow \quad \omega=v / \mathrm{R}$ ..... (ii)

Putting eq. (ii) in eq. (i), we get

$\Rightarrow \frac{1}{2} m v^{2}+\frac{1}{2} \mathrm{I}\left(\frac{v^{2}}{R^{2}}\right)=m g \times \frac{3 v^{2}}{4 g}$ $\left(\because h_{\max } \times \frac{3 v^{2}}{4 g}\right)$

$\Rightarrow \quad \mathrm{I}=\frac{1}{2} \mathrm{MR}^{2}$

Hence, this is a required formula of moment of Inertia of Disc. Therefore, the object a disc.

$2^{\text {nd }}$ Method : For pure rolling, the velocity of point of contact is zero i.e., $v=0$ , So, $\Delta x$ (displacement) is zero. Then work done by the frictional force should be

$W=f_{0} \Delta x=0$

If work is done by dissipative force, work done should be zero. It means that work done of system is zero. So energy should be conserved. Initially, the object has both translation as well as rotational motion. If it attains ' $h_{\max }$ ' kinetic energy changes into potential energy.

From energy conservation principle,

$\mathrm{K} . \mathrm{E}_{\text {translational }}+\mathrm{K} . \mathrm{E}_{\text {rotational }}=$ Potential Energy

$\Rightarrow \quad \frac{1}{2} m v^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=m g \times \frac{3 v^{2}}{4 g}$

$\Rightarrow \quad m v^{2}+\mathrm{I} \omega_{2}=\frac{3 v^{2}}{2} m$

$\Rightarrow \quad \mathrm{I} \omega^{2}=\frac{1}{2} m v^{2}$

$\Rightarrow \quad \mathrm{I}=\frac{1}{2} \frac{m v^{2}}{\omega^{2}}$ ..... (i)

For No slipping,

$\Rightarrow$ $\begin{aligned} & v=\omega R \\& \omega=\frac{v}{R}\end{aligned}$ ..... (ii)

Put eq (ii) in eq (i)

Hence, $\quad \mathrm{I}=\frac{1}{2} \mathrm{MR}^{2}$

$\mathrm{I}=\frac{1}{2} \mathrm{M} \frac{v^{2}}{v^{2}} \times \mathrm{R}^{2}$

this is moment of inertia of disc.

So, object is a Disc.