Question

A uniform solid circular cylinder of radius 'r' is placed on a rough horizontal surface. It is given a linear velocity v = 2$\omega$r and angular velocity $\omega$ as shown in the figure. The speed of cylinder when it starts rolling is

• A. $\omega R$
• B. 3/2$\omega$R
• C. 5/3$\omega$R
• D. 2/3 $\omega$R

Answer 1

The correct option is C 5/3$\omega$R

$\Rightarrow v=$ Velocity of center of mass

$f=$ Frictional force $=\mu mg$

$\omega =$ angular velocity

Now, $u=25\omega$

$\Rightarrow v=u+at$

$u=v-at$

$2r\omega =v-at$

Also, $\tau =I\alpha$

$\Rightarrow f.\frac{r}{I}=\alpha$

$\frac{\mu mg}{\frac{m{r}^2}{2}}=\alpha$

$\Rightarrow \alpha =\frac{2\mu g}{r}$

$\therefore 2r\omega =v=\mu gt$

$\Rightarrow t=v-\frac{2r\omega }{\mu g}$

$\because {\omega }_f=\omega -\alpha t$

$\Rightarrow 0=\omega -\left(\frac{2\mu g}{r}\right)\left(\frac{v-2r\omega }{\mu g}\right)$

or, solving for $v$ we get,

$\Rightarrow v=\frac{5}{3}\omega R.$

Hence, the answer is $\frac{5}{3}\omega R.$