Question

A solid sphere (mass 2 M) and a thin spherical shell (mass M) both of the same size, roll down an inclined plane, then

• A. Solid sphere will reach the bottom first
• B. Hollow spherical shell will reach the bottom first
• C. Both will reach at the same time
• D. Cannot be predicted as the data is insufficient
• E. None of the above

Answer 1

The correct option is A Solid sphere will reach the bottom first

Acceleration for a body rolling down an inclined plane is given by $a=\frac{gsin\theta }{1+\frac{K^2}{R^2}}=\frac{Mgsin\theta }{[M+\frac{I}{R^2}]}[\because K={\left[\frac{I}{M}\right]}^{1/2}]$
$\Rightarrow a_{shell}=\frac{M^{\prime }}{M^{\prime }+\frac{I_{sphere}}{R^2}gsin\theta }$
$[\because Given{M}^{\prime }=2M]$
Substituting, $I_{sphere=\frac{2}{3}}{M}^{\prime }{R}^2=\frac{2}{3}\left(2M\right)R^2$
$a_{shell}=\frac{2M}{2M+\frac{\frac{2}{3}\left(2M\right)R^2}{R^2}}gsin\theta$
$=\frac{3}{5}gsin\theta .$
Similarly, $a_{sphere}=\frac{M}{M+\frac{I_{shell}}{R^2}}gsin\theta$
Substituting, $I_{shell}=\left(2/5\right)M{R}^2,weget{a}_{sphere}=\frac{M}{M+\frac{\left(2/5\right)M{R}^2}{R^2}}$
$=\frac{5}{7}gsin\theta$ ...(2)
Hence, from (1) and (2), it is clear that $a_{sphere}>a_{shell}$
So, sphere will reach the bottom first.