Question

A disc rotates about its axis of symmetry in a horizontal plane at a steady rate of $3.5$ revolutions per second. A coin placed at a distance of $1.25\, cm$ from the axis of rotation remains at rest on the disc. The coefficient of friction between the coin and the disc is : $(g = 10 \, m/s^2)$

• A. 0.5
• B. 0.3
• C. 0.7
• D. 0.6

Answer 1

The Correct Option is D

To solve this problem, we need to calculate the coefficient of friction that is necessary to keep the coin at rest on the rotating disc.

The centripetal force required to keep the coin at rest is provided by the frictional force between the coin and the disc. The frictional force can be calculated using the formula for centripetal force:

F_c = \frac{mv^2}{r}

where m is the mass of the coin, v is the linear velocity of the coin, and r is the radius (distance from the axis of rotation to the coin), which is 1.25 \, cm = 0.0125 \, m.

The frictional force F_f is given by:

F_f = \mu mg

where \mu is the coefficient of friction and g = 10 \, m/s^2.

Since the frictional force provides the necessary centripetal force, we have:

\mu mg = \frac{mv^2}{r}

Cancelling m from both sides:

\mu g = \frac{v^2}{r}

The linear velocity v is related to the angular velocity \omega by:

v = \omega r

Given that the disc rotates at 3.5 revolutions per second, the angular velocity \omega in radians per second is:

\omega = 2\pi \times 3.5 = 7\pi \, \text{rad/s}

Substitute v = \omega r into the equation:

\mu g = \frac{(\omega r)^2}{r} = \omega^2 r

Substitute the known values:

\mu \times 10 = (7\pi)^2 \times 0.0125

Simplifying further:

\mu = \frac{49\pi^2 \times 0.0125}{10}

Calculate the value:

\mu \approx \frac{49 \times 9.87 \times 0.0125}{10} \approx 0.6

Thus, the coefficient of friction is 0.6.