Question

Find the total work performed by the sliding friction force acting on the cylinder.

• A. $A=-\frac{m{\omega }_0^2{R}^2}{6}$
• B. $A=\frac{2m{\omega }_0^2{R}^2}{6}$
• C. $A=\frac{m{\omega }_0^2{R}^2}{6}$
• D. $A=-\frac{2m{\omega }_0^2{R}^2}{6}$

Answer 1

The correct option is A $A=-\frac{m{\omega }_0^2{R}^2}{6}$

When the cylinder is placed on the horizontal plane, the frictional force acts in the forward direction such that its linear velocity increases, and angular velocity decreases till it rolls without slipping.

Let the time after which it starts to roll without slipping be $t$.

Frictional force is $\mu mg$

So after time $t$, linear velocity is $v_t=0+\left(\mu mg/m\right)t=\mu gt$

angular velocity after time $t$ is, ${\omega }_t={\omega }_o-\frac{\mu mgR}{m{R}^2/2}t={\omega }_o-2\mu gt/R$

for pure rolling,

$v_t={\omega }_{t}R⟹\mu gt={\omega }_o-2\mu gt/R⟹t={\omega }_{o}R/3\mu g$

So $v_t={\omega }_{o}R/3$ and ${\omega }_t={\omega }_o/3$

Intial Kinetic energy of the cylinder is

$K_i=\frac{1}{2}I{\omega }_o^2=\frac{1}{2}\times \left(m{R}^2/2\right)\times {\omega }_o^2=\frac{m{\omega }_o^2{R}^2}{4}$

Final Finetic Energy,

$K_f=\frac{1}{2}m{v}_t^2+\frac{1}{2}I{\omega }_t^2=\frac{m{\omega }_o^2{R}^2}{12}$

So total work done by the sliding friction is ,

$K_f-K_i=\frac{m{\omega }_o^2{R}^2}{12}-\frac{m{\omega }_o^2{R}^2}{4}=-\frac{m{\omega }_o^2{R}^2}{6}$