Question

A small object placed on a rotating horizontal table, just slips when it is placed at a distance $4\text{cm}$ from axis of rotation. If the angular velocity of the table is doubled, the object will slip when its distance from axis of rotation is:

• A. $1\text{cm}$
• B. $2\text{cm}$
• C. $4\text{cm}$
• D. $8\text{cm}$

Answer 1

The correct option is A $1\text{cm}$


Applying vertical equilibrium condition on particle, we get

$N=mg$

Here centripetal force is provided by the frictional force acting between object and table, so equation of circular dynamics,

$f=F_{centripetal}$

At just slipping condition,

$f=f_{max}=\mu N$

$\therefore \mu N=m{\omega }^{2}r$

$\mu mg=m{\omega }^{2}r$

${\omega }^{2}r=\mu g=\text{constant}$

Thus applying the above relation for both cases,

${\omega }_1^2{r}_1={\omega }_2^2{r}_2$

Here, ${\omega }_1=\omega ;{\omega }_2=2\omega ;r_1=4\text{cm}$

${\omega }^2\times 4=(2\omega {)}^2{r}_2$

$\therefore r_2=1\text{cm}$

Hence, option $\left(a\right)$ is correct.