A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously,

Question:

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

Solution:

Disc

Radii of the ring and the disc, r = 10 cm = 0.1 m

Initial angular speed, $\omega_{0}=10 \pi \mathrm{rad} \mathrm{s}^{-1}$

Coefficient of kinetic friction, $\mu_{\mathrm{k}}=0.2$

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, $f=m a$

$\mu_{k} m g=m a$

Where,

$a=$ Acceleration produced in the objects

$m=$ Mass

$\therefore a=\mu_{\mathrm{k}} \mathrm{g} \ldots(i)$

As per the first equation of motion, the final velocity of the objects can be obtained as:

$v=u+a t$

$=0+\mu_{k} g t$

$=\mu_{\mathrm{k}} \mathrm{g} t \ldots$ (ii)

The torque applied by the frictional force will act in perpendicularly outward direction and cause reduction in the initial angular speed.

Torque, $\mathrm{T}=-l \alpha$

$\alpha=$ Angular acceleration

$\mu_{x} m g r=-l \alpha$

$\therefore \alpha=\frac{-\mu_{k} m g r}{I}$  $\ldots(i i i)$

Using the first equation of rotational motion to obtain the final angular speed:

$\omega=\omega_{0}+\alpha t$

$=\omega_{0}+\frac{-\mu_{\mathrm{k}} m g r}{I} t$  $\ldots(i v)$

Rolling starts when linear velocity, $v=r \omega$

$\mu_{\mathrm{k}} \mathrm{g} t=r\left(\omega_{0}-\frac{\mu_{\mathrm{k}} \mathrm{g} m r t}{I}\right)$

$=r \omega_{0}-\frac{\mu_{\mathrm{k}} \mathrm{g} m r^{2} t}{I}$  $\ldots(v i)$

For the ring: $I=m r^{2}$

$\therefore \mu_{\mathrm{k}} \mathrm{g} t=r \omega_{0}-\frac{\mu_{\mathrm{k}} \mathrm{g} m r^{2} t}{m r^{2}}$

$=r \omega_{0}-\mu_{\mathrm{k}} \mathrm{g} m t_{\mathrm{r}}$

$2 \mu_{\mathrm{k}} \mathrm{g} t=r \omega_{0}$

$=\frac{0.1 \times 10 \times 3.14}{2 \times 0.2 \times 9.8}=0.80 \mathrm{~s}$  $\ldots($ vii $)$

For the disc: $I=\frac{1}{2} m r^{2}$

$\therefore \mu_{\mathrm{k}} \mathrm{g} t_{\mathrm{d}}=r \omega_{0} \frac{\mu_{\mathrm{k}} \mathrm{g} m r^{2} t}{\frac{1}{2} m r^{2}}$

$=r \omega_{0}-2 \mu_{\mathrm{k}} \mathrm{g}$

$3 \mu_{\mathrm{k}} \mathrm{g} t_{\mathrm{d}}=r \omega_{0}$

$=\frac{0.1 \times 10 \times 3.14}{3 \times 0.2 \times 9.8}=0.53 \mathrm{~s}$  $\ldots($ viii $)$

Since $t_{d}>t_{r}$, the disc will start rolling before the ring.

 

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