A uniform solid sphere of radius $R$ is placed on a rough horizontal surface. A horizontal force $F$

Question

A uniform solid sphere of radius $R$ is placed on a rough horizontal surface. A horizontal force $F$ is applied directly at its topmost point (height $h=R$ from the center). Assuming the sphere rolls without slipping, what is the magnitude and direction of the required frictional force $f$ at the contact point?

Professor PC Thomas Classes · Answer

(b) : For translation: $F + f = Ma_{cm}$ (assuming friction $f$ acts forward). For rotation about the CM, torque $	au = I_{cm} \alpha$. The applied force $F$ creates a clockwise torque $FR$, while forward friction $f$ creates a counter-clockwise torque $-fR$. Thus, $FR - fR = (\frac{2}{5}MR^2) \alpha$. For pure rolling, $a_{cm} = \alpha R$, so $FR - fR = \frac{2}{5}M R a_{cm}$. Substitute $Ma_{cm} = F+f$: $F - f = \frac{2}{5}(F+f)$. This gives $5F - 5f = 2F + 2f \Rightarrow 3F = 7f \Rightarrow f = \frac{3}{7}F$. The positive sign confirms friction acts in the forward direction.