Question

A sphere of mass $M$ and radius $r$ shown in the figure, slips on rough horizontal plane. At some instant it has translational velocity $v_0$ and rotational velocity about centre $v_0/2r$. The percentage change of translational velocity after the sphere start pure rolling:

• A. $14.28$%
• B. $7.14$%
• C. $21.42$%
• D. None

Answer 1

The correct option is A $14.28$%

Initially the velocity of COM of sphere is $v_o$ and angular velocity is $w_i$.

When the pure rolling starts, let the velocity of COM of sphere be $v$ and angular velocity be $w_f$.

Now $f=Ma⟹a=\frac{f}{M}$

$v=v_0-at⟹v=v_o-\frac{f}{M}t$ ..............(1)

Also, $\tau =I\alpha$

$fr=\frac{2}{5}M{r}^2\alpha ⟹\alpha =\frac{5f}{2Mr}$

$w_f=w_i+\alpha t$

$w_f=\frac{v_0}{2r}+\frac{5ft}{2Mr}⟹\frac{2}{5}r{w}_f=\frac{v_o}{2}\times \frac{2}{5}+\frac{ft}{M}$ ..............(2)

Adding (1) and (2), $v+\frac{2}{5}r{w}_f=\frac{v_o}{5}+v_o$

Also $v=r{w}_f$ (pure rolling)

$⟹v=\frac{6}{7}{v}_o$

Now $\frac{v_o-v}{v_o}\times 100=\frac{1}{7}\times 100=14.28$%