Question

A solid disc and a ring, both of radius 10 cm 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 ${\mu }_k=0.2$.

Answer 1

Radii of the ring and the disc, r = 10 cm = 0.1 m
Initial angular speed, ${\omega }_0=10\pi rad{s}^{-1}$
Coefficient of kinetic friction, ${\mu }_k=0.2$
Initial velocity of both the objects, u = 0
Motion of the two objects is caused by frictional force. As per Newton’s second law of motion, we have frictional force, f = ma
$m{u}_k$ mg= ma where, a = Acceleration produced in the objects m = Mass $a={\mu }_k$ g ... (i)
As per the first equation of motion, the final velocity of the objects can be obtained as:
v = u + at
$=0+{\mu }_k$ gt
$={\mu }_{k}gt...\left(ii\right)$
The torque applied by the frictional force will act in perpendicularly outward direction and cause reduction in the initial angular speed.
Torque, $\tau =–I\alpha$ where $\alpha$ = Angular acceleration
${\mu }_k$ mgr = –$I\alpha$
$\therefore \alpha =\frac{-{\mu }_{k}mgr}{I}......\left(iii\right)$
Using the first equation of rotational motion to obtain the final angular speed:
$\omega ={\omega }_0+\alpha t={\omega }_0+\frac{-{\mu }_{k}mgr}{I}t.......\left(iv\right)$
Rolling starts when linear velocity, $v=r\omega$
$\therefore v=r({\omega }_0-\frac{{\mu }_{k}gmrt}{I})=r{\omega }_0-\frac{{\mu }_{k}gm{r}^{2}t}{I}.....\left(iv\right)$
$Forthering:I=m{r}^2\therefore {\mu }_{k}g{t}_r=r{\omega }_0-\frac{{\mu }_{k}gm{r}^2{t}_r}{m{r}^2}=r{\omega }_0-{\mu }_{k}g{t}_r\Rightarrow 2{\mu }_{k}g{t}_r=r{\omega }_0\therefore t_r=\frac{r{\omega }_0}{2{\mu }_{k}g}=\frac{0.1\times 10\times 3.14}{2\times 0.2\times 9.8}=0.80s.....\left(vii\right)Forthedisc:I=\frac{1}{2}m{r}^2\therefore {\mu }_{k}g{t}_d=r{\omega }_0-\frac{{\mu }_{k}gm{r}^2{t}_d}{\frac{1}{2}m{r}^2}=r{\omega }_0-2{\mu }_{k}g{t}_d\Rightarrow 3{\mu }_{k}g{t}_d=r{\omega }_0\therefore t_d=\frac{r{\omega }_0}{3{\mu }_{k}g}=\frac{0.1\times 10\times 3.14}{3\times 0.2\times 9.8}=0.53s....\left(viii\right)$
Since $t_d>t_r$, the disc will start rolling before the ring.