Question

A solid sphere of mass 0⋅50 kg is kept on a horizontal surface. The coefficient of static friction between the surfaces in contact is 2/7. What maximum force can be applied at the highest point in the horizontal direction so that the sphere does not slip on the surface?

Answer 1

Let α be the angular acceleration produced in the sphere.



Rotational equation of motion,
$F\times R-f_r\times R=I\alpha$
$\Rightarrow F=\frac{2}{5}mR\alpha +\mu mg...\left(1\right)$

Translational equation of motion,
$$\begin{gathered} F=ma-\mu mg \\ \Rightarrow a=\frac{\left(F+\mu mg\right)}{m} \end{gathered}$$
For pure rolling, we have:
$\alpha =\frac{a}{R}$
$\Rightarrow \alpha =\frac{\left(F+\mu mg\right)}{mR}$

Putting the value of $\alpha$ in equation (1), we get:

$$\begin{gathered} F=\frac{2}{5}\frac{mR\left(F+\mu mg\right)}{mR}+\mu mg \\ \Rightarrow F=\frac{2}{5}\left(F+\mu mg\right)\mu mg \\ \Rightarrow F=\frac{2}{5}F+\left(\frac{2}{5}\times \frac{2}{7}\times 0.5\times 10\right)+\left(\frac{2}{7}\times 0.5\times 10\right) \\ \Rightarrow \frac{3F}{5}=\frac{4}{7}+\frac{10}{7}=2 \\ \Rightarrow F=\frac{5\times 2}{3}=\frac{10}{3}=3.3\mathrm{N} \end{gathered}$$