Question

A solid ball rolls down a parabolic path $\text{ABC}$ from a height $h$ as shown in figure. Portion $\text{AB}$ of the path is rough while $\text{BC}$ is smooth. How high will the ball climb in $\text{BC}$ ?

• A.
  $\frac{5}{7}h$
• B.
  $\frac{7}{5}h$
• C.
  $\frac{2}{5}h$
• D.
  $\frac{h}{5}$

Answer 1

The correct option is A

$\frac{5}{7}h$
Since in the portion $\text{AB}$, the surface is rough, so the ball will have both rotational as well as translational motion.

Applying the conservation of mechanical energy between the points $\text{A}$ and $\text{B}$,

Potential energy at point $\text{A}$ = Rotational kinetic energy at point $\text{B}+$Translational kinetic energy at point $\text{B}$

$mgh=\frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}^2...\left(1\right)$

where,
moment of inertia about centre, $I=\frac{2}{5}m{r}^2$
$v$ is the velocity at point $\text{B}$.

Since, the ball have pure rolling motion on the rough surface,
so $v=\omega r$

Substituting the values in the above equation, we get

$mgh=\frac{1}{2}\times \frac{2}{5}m{r}^2\times {\left(\frac{v}{r}\right)}^2+\frac{1}{2}m{v}^2$

$\Rightarrow gh=\frac{7}{10}{v}^2$

$\Rightarrow v^2=\frac{10gh}{7}....\left(1\right)$

In portion $\text{BC}$, friction is absent. Therefore, rotational kinetic energy will remain constant and translational kinetic energy will convert into potential energy. Hence, if $H$ be the height to which ball climb in $\text{BC}$, then

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

Substituting the value of $v^2$,

$\Rightarrow \frac{1}{2}m\frac{10gh}{7}=mgH$

$\therefore H=\frac{5}{7}h$

Hence, option (a) is correct answer.