Question

A solid sphere of mass $m$ is released from rest from the rim of a hemispherical cup so that it rolls along the surface. Find the normal contact force between the solid sphere and the cup at the bottom most point.

• A. $\frac{17}{7}mg$
• B. $\frac{15}{7}mg$
• C. $\frac{12mg}{7}$
• D. $\frac{11mg}{7}$

Answer 1

The correct option is A $\frac{17}{7}mg$


Applying conservation of mechanical energy between $A$ and $B$, we get

$K_A+U_A=K_B+U_B$
$\Rightarrow O+mgR=\frac{1}{2}m{v}^2[1+\frac{k^2}{R^2}]+0\Rightarrow mgR=\frac{1}{2}m{v}^2[1+\frac{2}{5}](\because \frac{k^2}{R^2}=\frac{2}{5})\Rightarrow m{v}^2=\frac{10mgR}{7}$

At $B$, normal force $N=mg+\frac{m{v}^2}{R}=\frac{17mg}{7}$