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

Question

A solid ball rolls down a parabolic path ABC from a height h as shown in the figure. The portion AB of the path is rough while BC is smooth. How high will the ball climb in BC ?

TestHub · Accepted Answer

Using the conservation of mechanical energy between $A$ and $B$

At B, total kinetic energy = $m g h$

Here, $m$ = mass of the ball

The ratio of rotational to translational kinetic energy would be,

$\frac{K_{R}}{K_{T}} = \frac{2}{5}$

∴ $K_{R} = \frac{2}{7} mg h and K_{T} = \frac{5}{7} mgh$

In portion $BC$ , 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 climbs in $BC$ , then

$m g H = K_{T}$

or $mgH = \frac{5}{7} mg h or H = \frac{5}{7} h$

Physics - Rotation Question with Solution | TestHub

Options:

Doubts & Discussion