Question

A solid sphere and a solid cylinder having the same mass and radius, roll down the same incline. Find the ratio of their acceleration.

Answer 1

Given: A solid sphere and a solid cylinder having the same mass and radius, roll down the same incline.

To find the ratio of their acceleration.

Solution:

From the above fig:

$mg\sin \theta -f=ma$,

where $f$ is the frictional force

Now for solid cylinder,

$f_{1}R=I\alpha ⟹f_{1}R=\frac{m{R}^2}{2}\times \frac{a_1}{R}⟹f_1=\frac{m{a}_1}{2}\therefore mg\sin \theta =m{a}_1+f_1⟹mg\sin \theta =m{a}_{+}\frac{m{a}_1}{2}=\frac{3}{2}m{a}_1⟹a_1=\frac{2}{3}\sin \theta ......\left(i\right)$

Now for solid sphere,

$f_{2}R=I\alpha ⟹f_{2}R=\frac{2m{R}^2}{5}\times \frac{a_2}{R}⟹f_2=\frac{2m{a}_2}{5}\therefore mg\sin \theta =m{a}_2+f_2⟹mg\sin \theta =m{a}_2+\frac{2m{a}_2}{5}=\frac{7}{2}m{a}_2⟹a_2=\frac{2}{7}\sin \theta ......\left(i\right)$

Hence the ratio of their acceleration is:

$a_2:a_1=\frac{2\sin \theta }{7}:\frac{2\sin \theta }{3}=3:7$