Q7.30     A solid disc and a ring, both of radius $10cm$  are placed on a horizontal table               simultaneously, with initial angular speed equal to $10\Pi rad\: s^{-1}$ . Which of the two will start to roll              earlier ? The co-efficient of kinetic friction is $\mu _{k}=0.2$.

Friction is the cause for motion here.

So using Newton's law of motion we can write :

$f\ =\ ma$

$\mu _k mg\ =\ ma$

or                                           $a\ =\ \mu _k g$

Now by the equation of motion, we can write :

$v\ =\ u\ +\ at$

or                                            $v\ =\ \mu _k gt$

The torque is given by :

$\tau\ =\ I \alpha$

or                                  $r\times f \ =\ -I \alpha$

or                                   $\mu _k mgr \ =\ -I \alpha$

or                                             $\alpha\ =\ -\frac{\mu _k mgr}{I}$

Now using the equation of rotational motion we can write :

$\omega\ =\ \omega _o\ +\ \alpha t$

or                                      $\omega\ =\ \omega _o \ +\ -\frac{\mu _k mgr}{I}t$

Condition for rolling is         $v\ =\ \omega r$

So we can write :

$v =\ r\left ( \omega _o \ +\ -\frac{\mu _k mgr}{I}t \right )$

For ring the moment of inertia is :   $mr^2$

So we have :

$t\ =\ \frac{r \omega _o}{2 \mu _k g}$

or                                            $=\ \frac{0.1\times 10\times 3.14}{2\times 0.2 \times 10}\ =\ 0.80\ s$

Now in case of the disc, the moment of inertia is :

$I\ =\ \frac{1}{2}mr^2$

Thus                                $t\ =\ \frac{r \omega _o}{3 \mu _k g}$

or                                          $=\ \frac{0.1\times 10\times 3.14}{3 \times 0.2 \times 9.8}\ =\ 0.53\ s$

Hence disc will start rolling first.

Exams
Articles
Questions