Question

A solid sphere of radius R is set into motion on a rough horizontal surface with a linear speed $v_0$ in forward direction ad an angular velocity ${\omega }_0=\frac{v_0}{2R}$ in counter clockwise direction as shown in figure. If co-efficient of friction is $\mu$, then find

(i)The time after which sphere starts pure rolling,

(ii)The work done by friction over a long time

• A. $t=\frac{2}{17}\frac{v_0}{mug},W=\frac{-5}{13}m{v}_0^2$
• B. $t=\frac{3v_0}{7mug},W=\frac{-9}{28}m{v}_0^2$
• C. $t=\frac{7}{19}\frac{v_0}{mug},W=-3m{v}_0^2$
• D. None of these

Answer 1

The correct option is B

$t=\frac{3v_0}{7mug},W=\frac{-9}{28}m{v}_0^2$

As the sphere is slipping, kinetic friction will act,

$f_k=\mu mg,a_{cm}=\frac{f_k}{m}=\mu g,$

$\alpha =\frac{f_{k}R}{I_{CM}}=\frac{\mu mgR}{\frac{2}{5}m{R}^2}=\frac{5\mu g}{2R}$

when the sphere starts pure rolling, velocity v of com will be $\omega R$.

$v=v_0-\mu gt$

$\omega =-{\omega }_0+\alpha t$

for pure rolling,

$v=\omega R$

$\Rightarrow v_0-\mu gt=-(-{\omega }_0+\alpha t)R$

$\Rightarrow v_0-\mu gt=(\frac{-v_0}{2R}+\frac{5\mu gt}{2R})R$

$\Rightarrow \frac{3v_0}{2}=\frac{7}{2}\mu gt\Rightarrow t_0=\frac{3v_0}{7\mu g}$

(b) Friction will not do any work when pure rolling starts.

∴friction does work only when

$0<t<\frac{3v_0}{7\mu g}$

$W_f=$ Change in KE

$K{E}_f=\frac{1}{2}m{v}^2+\frac{1}{2}{I}_{CM}{\omega }_2$

$v=v_0-\mu g{t}_0=v_0-\mu g\frac{3v_0}{7\mu g}=\frac{4v_0}{7}$

$\omega =\frac{4}{7}\frac{v_0}{R}$

$K{E}_f=\frac{1}{2}\times m\times (\frac{4v_0}{7}{)}^2+\frac{1}{2}\times \frac{2}{5}m{R}^2(\frac{4}{7}\frac{v_0}{R}{)}^2$

$K{E}_i=\frac{1}{2}m{v}_0^2+\frac{1}{2}\left(\frac{2}{5}m{R}^2\right)\left(\frac{v_0}{2R}{)}^2\right)$

$\Rightarrow W_f=\frac{-9}{28}m{v}_0^2$