Question

A body of mass $m$ slides down an incline and reaches the bottom with a velocity $v$. If the same mass was in the form of a ring which rolls down this incline, the velocity of the ring at the bottom would have been:

• A. $v$
• B. $\left(\sqrt{2}\right)v$
• C. $\frac{v}{\sqrt{2}}$
• D. $\left(\sqrt{\frac{2}{5}}\right)v$

Answer 1

The correct option is C $\frac{v}{\sqrt{2}}$

Let us that the $r$ is the radius of the ring and $m$ is its mass then,

Moment of inertial of the mass $I=m{r}^2.....\left(i\right)$

Now,

Total energy at the bottom of the inclined plane = Transnational KE + Rotational KE

Total energy at the top of the plane = Potential energy

Now, as the mass/ring slides/rolls down the inclined plane then potential energy gets converted into kinetic energy.

In the first case when it is not a ring. If $h$ is the height of the plane then,

$mgh=\frac{1}{2}m{v}^2$

$h=\frac{v^2}{2g}........\left(ii\right)$

Now,

According to the second case if $v^{\prime }$ is the velocity at the bottom of the plane then,

$mgh=\frac{1}{2}m{v}^{\prime 2}+\frac{1}{2}I{\omega }^2$

Or

$mgh=\frac{1}{2}m{v}^{\prime 2}+\frac{1}{2}m{r}^2\times {\left(\frac{v^{\prime }}{r}\right)}^2$ [Since $v^{\prime }=r\times \omega$]

Or,

$mgh=m{v}^{\prime 2}$

$h=\frac{v^{\prime 2}}{g}........\left(iii\right)$

Comparing (ii) and (iii)

$\frac{v^2}{2g}=\frac{v^{\prime 2}}{g}$

$v^{\prime 2}=\frac{v^2}{2}$

$v^{\prime }=\frac{v}{\sqrt{2}}$