Question

A small object placed on a rotating horizontal turntable just slips when it is placed at a distance of 4 cm from the axis of rotation. If the angular velocity on the turn-table is doubled, the object slips when its distance from the axis of rotation is –
A) 1 cm
B) 2 cm
C) 4 cm
D) 8 cm

Answer 1

We have to spend our knowledge on the rotational torque and the angular velocity involved in the situation given above to understand and analyze the dependence of the angular velocity in stabilizing the system without any slipping of the object.

Complete step-by-step solution
We need to understand the forces which hold the object in position during the rotational motion. We know that the rotational motion involves the centripetal force acting on the object which is given as –
${{F}_{C}}=m{{\omega }^{2}}r$
Where, m is the mass of the object, $\omega$ is the angular velocity of rotation and r is the distance from the axis of rotation.
Now, the force which opposes this centripetal force is the force of friction that acts on the body, which will be constant as long as the object is at rest. The force of static friction is given by –
${{F}_{f}}=\mu mg=\text{constant}$
Where, $\mu$ is the coefficient of static friction, m is the mass of the object and g is the acceleration due to gravity.
When the object is at rest, the two forces must be balancing, or the centripetal force will be less than or equal to the force of friction. At the verge of slipping, the both forces can be equated as –

& {{F}_{C}}={{F}_{f}} \\\ & \Rightarrow m{{\omega }^{2}}r=\mu mg \\\ & \Rightarrow {{\omega }^{2}}r=\mu g=\text{constant} \\\ \end{aligned}$$ From this, we understand that the distance from the axis of rotation and the angular velocity of the rotation are related as – $$\text{r}\propto \dfrac{1}{{{\omega }^{2}}}$$ Now, we can find the ratio of two situations to find the required distance of the object from the axis of rotation. In the first situation, the distance from axis is given as 4 cm and in the second situation, the angular velocity is doubled. We can conclude as – $$\begin{aligned} & \text{r}\propto \dfrac{1}{{{\omega }^{2}}} \\\ & \Rightarrow \dfrac{{{r}_{1}}}{{{r}_{2}}}=\dfrac{{{\omega }_{2}}^{2}}{{{\omega }_{1}}^{2}} \\\ & \Rightarrow \dfrac{4cm}{{{r}_{2}}}=\dfrac{{{(2\omega )}^{2}}}{{{\omega }^{2}}} \\\ & \therefore {{r}_{2}}=1cm \\\ \end{aligned}$$ So, when the angular velocity is doubled, the object should be 1 cm from the axis of rotation to avoid slipping. This is the required solution. **Note:** When we say that the frictional force is constant throughout the body, we mean that the frictional force always balances the opposing force up to its maximum ability. The centripetal force of lesser magnitude is opposed by the same magnitude of the friction to avoid any movement in any direction.