Question

A solid sphere of mass m is lying at rest on a rough horizontal surface.the coefficient of friction between ground and sphere is ú.the maximum value of F so that sphere will not slip,is equal to

Answer 1

For pure translational motion, the net linear force acting on the sphere is given by,

$\longrightarrow\sf{F-f=ma}$

$\longrightarrow\sf{F-\mu\,mg=ma\quad\quad\dots(1)}$

where a is the net acceleration of the sphere, and f is the frictional force.

For pure rotational motion, the net torque acting on the sphere is given by,

$\longrightarrow\sf{\tau=I\alpha}$

And since the torque is due to friction,

$\longrightarrow\sf{fr=I\alpha}$

where I is the moment of inertia of the sphere and $\alpha$ is angular acceleration.

Moment of inertia of sphere is $\sf{\dfrac{2}{5}\,mr^2}$ and angular acceleration, $\sf{\alpha=\dfrac{a}{r}.}$ Then,

$\longrightarrow\sf{\mu\,mgr=\dfrac{2}{5}\,mr^2\cdot\dfrac{a}{r}}$

$\longrightarrow\sf{\mu\,mgr=\dfrac{2}{5}\,mar}$

$\longrightarrow\sf{\mu g=\dfrac{2}{5}\,a}$

$\longrightarrow\sf{a=\dfrac{5}{2}\,\mu g}$

Then (1) becomes,

$\longrightarrow\sf{F-\mu\,mg=\dfrac{5}{2}\,\mu\,mg}$

$\longrightarrow\sf{\underline{\underline{F=\dfrac{7}{2}\,\mu mg}}}$