Question

A sphere has only translational velocity $V_0=2r\omega$ when placed on rough horizontal surface with frictional coefficient $\mu$. If $r$ is radius then the time in which pure rolling starts is

• A. $\frac{2r\omega }{7\mu g}$
• B. $\frac{4r\omega }{7\mu g}$
• C. $\frac{r\omega }{7\mu g}$
• D. $\frac{2\sqrt{2}r\omega }{7\mu g}$

Answer 1

The correct option is B $\frac{4r\omega }{7\mu g}$

When sphere is positioned on the horizontal surface friction($\mu mg$) acts to retard translation and support rotation.
As a result pure rolling starts after some time when $V=r{\omega }^{\prime }$
Where $V$ is velocity of sphere after time $t$
${\omega }^{\prime }$ is angular velocity of sphere after time $t$
From Newtons Laws of Motion
$a=\frac{f}{m}=\mu g\&\alpha =\frac{\tau }{I}=\frac{fr}{2m{r}^2/5}=\frac{5\mu g}{2r}$

velocity after time t.

$V=V_0-\mu gt{\omega }^{\prime }=\alpha t=\frac{5\mu g}{2r}t$

Pure rolling start when

$V=r{\omega }^{\prime }\Rightarrow t=\frac{4r\omega }{7\mu g}$

OR

Conserving angular momentum about an axis passing through $AB$
Initial angular momentum at A
$L_i=m{V}_{0}r=2m{r}^2\omega$
When it starts rolling at B
Final angular momentum
$L_f=m{V}_{t}r+I_{cm}{\omega }_t=m{V}_{t}r+\left(\frac{2m{r}^2}{5}\right){\omega }_t$
As $V_t=r{\omega }_t$
$L_f=\frac{7m{V}_{t}r}{5}$
By conserving angular momentum
$L_i=L_f$
$2m{r}^2\omega =\frac{7m{V}_{t}r}{5}$
$\Rightarrow V_t=\frac{10r\omega }{7}$
Now From equations of motion

$V_t=V_0-\mu gt$
$\frac{10r\omega }{7}=2r\omega -\mu gt$
$\mu gt=2r\omega -\frac{10r\omega }{7}$
$t=\frac{4\omega r}{7\mu g}$