Question

A solid sphere and a spherical shell roll down on an inclined plane from rest from the same height. The ratio of the times taken by them is:
A. √21/25
B. 21/25
C. √25/21
D. 25/21

Answer 1

The net force acting on them is the resultant of weight and torque and the acceleration is calculated from this net force. From the kinematic equation, value of time is calculated and ratio is obtained on dividing time taken by first body to second body.

Complete step by step answer:
When a solid sphere and a spherical shell roll down on inclined plane, the body performs both translational and rotational motion i.e., the translational motion is due to the bodies going straight down the inclined plane and rotational motion as they are also rotating about their axis while going down. Therefore, the force acting on the bodies are due to resultant of weight and torque (turning effect of force about its axis) i.e.,
F=F1(due to the weight)-τ(torque)=$F1 - \dfrac{{F2}}{r}$
$ma = mg - \dfrac{{I\alpha }}{r}$ where I is moment of inertia, m is mass, g is acceleration due to gravity and α is angular acceleration and r is the radius of the body.
$mar = mgr - I\alpha \\\ mar = mgr - I\left( {\dfrac{a}{r}} \right) \\\ ma{r^2} = mg{r^2} - Ia \\\ mg{r^2} = ma{r^2} + Ia \\\ mg{r^2} = \left( {m{r^2} + I} \right)a \\\ a = \dfrac{{mg{r^2}}}{{m{r^2} + I}} \\\$
Now, $S = ut + \dfrac{1}{2}a{t^2}$ where S is the distance travelled by both of them, t is time taken by them.
$S = 0 \times t + \dfrac{1}{2}a{t^2}$ [ as the bodies are in a rest position. So initial velocity i.e. u =0]
S = 0 + \dfrac{1}{2}a{t^2} = \dfrac{1}{2}a{t^2} \\\ t = \sqrt {\dfrac{{2S}}{a}} \\\ \
Time taken by solid sphere
{t_1} = \sqrt {\dfrac{{2S}}{{\dfrac{{mg{r^2}}}{{m{r^2} + I}}}}} \\\
{t_1} = \sqrt {\dfrac{{2S\left( {m{r^2} + I} \right)}}{{mg{r^2}}}} \\\
{t_1} = \sqrt {\dfrac{{2S\left( {m{r^2} + \dfrac{2}{5}m{r^2}} \right)}}{{mg{r^2}}}} = \sqrt {\dfrac{{2Sm{r^2}\left( {1 + \frac{2}{5}} \right)}}{{mg{r^2}}}} \\\\\sqrt {\dfrac{{2S\left( {\dfrac{7}{5}} \right)}}{g}} = \sqrt {\dfrac{{14S}}{{5g}}} \\\ \

Time taken by spherical shell,
${t_2} = \sqrt {\dfrac{{2S}}{{{a_1}}}} \\\ {t_2} = \sqrt {\dfrac{{2S}}{{\dfrac{{mg{r^2}}}{{m{r^2} + I}}}}} = \sqrt {\dfrac{{2S\left( {m{r^2} + I} \right)}}{{mg{r^2}}}} \\\ \sqrt {\dfrac{{2S\left( {m{r^2} + \dfrac{2}{3}m{r^2}} \right)}}{{mg{r^2}}}} = \sqrt {\dfrac{{2S\left( {\dfrac{{3m{r^2} + 2m{r^2}}}{3}} \right)}}{{mg{r^2}}}} \\\ \sqrt {\dfrac{{2S\left( {\dfrac{{5m{r^2}}}{3}} \right)}}{{mg{r^2}}}} = \sqrt {\dfrac{{10S}}{{3g}}} \\\ \$
Ratio of time taken by solid sphere to spherical shell i.e. $\dfrac{{{t_1}}}{{{t_2}}}$
= \dfrac{{\sqrt {\dfrac{{14S}}{{5g}}} }}{{\sqrt {\dfrac{{10S}}{{3g}}} }} = \sqrt {\dfrac{{14S \times 3g}}{{5g \times 10S}}} \\\ \sqrt {\dfrac{{21}}{{25}}} \\\ \

So, the correct answer is “Option A”.

Note:
We have to remember that the mass, value of g and r in both the cases i.e., solid sphere and spherical shell are the same as both are of the same mass and radius. The value of S is also the same as both are rolling down the same length of inclination.