Question

Two bodies, a ring and a solid cylinder of same material are rolling down without slipping an inclined plane. The radii of the bodies are same. The ratio of velocity of the centre of mass at the bottom of the inclined plane of the ring to that of the cylinder is $\frac{\sqrt{x}}{2}$. Then, the value of $x$ is _____.

Answer 1

Suppose both the bodies start rolling from vertical height $h$.

Let,
$m_c=$ mass of cylinder,
$m_r=$ mass of ring,
$R=$Radius of ring or cylinder,

Now, applying energy conservation between top & bottom of the inclined plane,

$m_{r}gh=\frac{1}{2}{I}_r{\omega }_r^2+\frac{1}{2}{m}_r{v}_r^2$

Here, $I_r=m_r{R}^2;v_r=R{\omega }_r$

$\Rightarrow m_{r}gh=\frac{1}{2}\left(m_r{R}^2\right)(\frac{v_r}{R}{)}^2+\frac{1}{2}{m}_r{v}_r^2$

$\Rightarrow v_r=\sqrt{gh}$

Similarly, for solid cylinder,

$m_{c}gh=\frac{1}{2}{I}_c{\omega }_c^2+\frac{1}{2}{m}_c{v}_c^2$

Here, $I_c=m_c{R}^2/2;v_c=R{\omega }_c$

$\Rightarrow m_{c}gh=\frac{1}{2}\left(m_c{R}^2/2\right)(\frac{v_c}{R}{)}^2+\frac{1}{2}{m}_c{v}_c^2$

$\Rightarrow v_c=\sqrt{\frac{4gh}{3}}$

$\therefore \frac{v_r}{v_c}=\frac{\sqrt{3}}{2}$

So, correct answer : $3$

Alternate solution: Let $I_{poc}$ be the MOI about point of contact. Now, applying energy conservation between top & bottom of the inclined plane, $m_{r}gh=\frac{1}{2}{I}_{r,poc}{\omega }_r^2$ Here, $I_{r,poc}=2m_r{R}^2;v_r=R{\omega }_r$ $\Rightarrow m_{r}gh=\frac{1}{2}\left(2m_r{R}^2\right)(\frac{v_r}{R}{)}^2$ $\Rightarrow v_r=\sqrt{gh}$ Similarly, for solid cylinder, $m_{c}gh=\frac{1}{2}{I}_{c,poc}{\omega }_c^2$ Here, $I_{c,poc}=3m_c{R}^2/2;v_c=R{\omega }_c$ $\Rightarrow m_{c}gh=\frac{1}{2}\left(3m_c{R}^2/2\right)(\frac{v_c}{R}{)}^2$ $\Rightarrow v_c=\sqrt{\frac{4gh}{3}}$ $\therefore \frac{v_r}{v_c}=\frac{\sqrt{3}}{2}$