Question 65: Work, Energy and Power & Circular Motion

Question

A coin placed on a rotating table just slips when it is placed at a distance of $$1 \mathrm{~cm}$$ from the center. If the angular velocity of the table in halved, it will just slip when placed at a distance of _________ from the centre :

Accepted Answer

When a coin is placed on a rotating table and is just about to slip, the centrifugal force acting on the coin equals the maximum static friction force. Let's denote the mass of the coin as $$m$$, the initial angular velocity as $$\omega_1$$, and the final angular velocity as $$\omega_2$$.
Initially, when the coin is placed at a distance of 1 cm from the center, the centrifugal force acting on the coin is:
$$F_1 = m r_1 \omega_1^2$$
where $$r_1 = 1 \mathrm{~cm}$$.
When the angular velocity is halved ($$\omega_2 = \frac{1}{2}\omega_1$$), the centrifugal force acting on the coin when it just slips is:
$$F_2 = m r_2 \omega_2^2 = m r_2 \left(\frac{1}{2}\omega_1\right)^2$$
Since the coin is just about to slip in both cases, the maximum static friction force remains the same. Therefore, we can equate the centrifugal forces:
$$m r_1 \omega_1^2 = m r_2 \left(\frac{1}{2}\omega_1\right)^2$$
Canceling the mass $$m$$ and the initial angular velocity $$\omega_1^2$$ from both sides, we get:
$$r_1 = r_2 \left(\frac{1}{2}\right)^2$$
Now, we can solve for $$r_2$$:
$$r_2 = \frac{r_1}{\left(\frac{1}{2}\right)^2} = \frac{1 \mathrm{~cm}}{\frac{1}{4}} = 4 \mathrm{~cm}$$
So, the coin will just slip when placed at a distance of 4 cm from the center when the angular velocity is halved.

Answer

1 cm

Answer

8 cm

Answer

4 cm

Answer

2 cm

Work, Energy and Power & Circular Motion

A coin placed on a rotating table just slips when it is placed at a distance of $1 \mathrm{~cm}$ from the center. If the angular velocity of the table in halved, it will just slip when placed at a distance of _________ from the centre :

Choose the correct answer: