Question

A small disc is set on rolling with speed $10\text{m/s}$ on the horizontal part of the track from right to left. There is curved part also as shown in the figure. To what height will the disc climb up on the curved path ? Assuming all track is rough and there is no slipping occur anywhere.

• A. $5\text{m}$
• B. $7.5\text{m}$
• C. $10\text{m}$
• D. $15\text{m}$

Answer 1

The correct option is B $7.5\text{m}$

Let the mass of the disc is $m\text{kg}$ and radius $R$.

Disc is having the rotational as well as translational kinetic energy at the initial position on the horizontal track.
$\Rightarrow$At final position, when it reaches height $h\text{m},\omega =0,v=0$
$v=R\omega$: condition for pure rolling of disc
$\Rightarrow$Applying the conservation of mechanical energy between point $A$ and $B$.
$\therefore K{E}_{Trans}+K{E}_{Rot}+PE=\text{constant}$
$\frac{1}{2}m{v}^2+\frac{1}{2}{I}_{\text{CM}}{\omega }^2+0=0+0+mgh...\left(i\right)$
where, $I_{\text{CM}}=$ moment of inertia of disc about centre.
$I_{\text{CM}}=\frac{m{R}^2}{2},\omega =\frac{v}{R}$
From Eq$...\left(i\right)$
$\Rightarrow \frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{m{R}^2}{2}\right)\times \left(\frac{v^2}{R^2}\right)=mgh$
$\frac{1}{2}m{v}^2+\frac{1}{4}m{v}^2=mgh$

$\frac{3}{4}m{v}^2=mgh$
$\Rightarrow h=\frac{3}{4}\frac{v^2}{g}$
$\therefore h=\frac{3}{4}\times \frac{(10{)}^2}{10}=7.5\text{m}$