Question

A disc rotates with constant frequency of $60rpm$. The maximum distance from the center of the disc, where a coin can be placed before it filled off, is (The coefficient of frictions is $0.4$and$g = 10m/{s^2}$)
A. $1/{\pi ^2}$metre
B. $1/\pi $metre
C. ${\pi ^2}$metre
D. $\pi $metre

Answer 1

Hint: Rotation of the disc will apply centrifugal force on the coin in the direction away from the center of the disc. Because of which, the coin will be thrown off. The magnitude of the force applied on the coin as the distance between the center of the disk and the coin increases.

Complete step by step answer:
It is given in the question that the disc rotates with constant frequency, $W = 60rpm$
The coefficient of friction is given as $\mu = 0.4$
And the acceleration due to gravity, $g = 10m/{s^2}$.
Now, we know that the relation between frequency of a body and the time period is given by
$W = \dfrac{{2\pi }}{T}$ . . . (1)
And the relation between the distance and the time period is given by
Distance$ = \dfrac{{2g\mu }}{T}$ . . . (2)
Thus, Distance$ = \dfrac{1}{T} \times \dfrac{4}{{100}} \times \dfrac{{20}}{\pi }$
From equation (1), we get
$T = \dfrac{{2\pi }}{{60}} = \dfrac{\pi }{{30,}}$
From equation (1) and equation (2) we have,
Distance$ = \dfrac{{30}}{\pi } \times \dfrac{4}{{100}} \times \dfrac{{20}}{\pi } = \dfrac{1}{{{\pi ^2}}}m$.

Therefore, the correct option is option (A) $1/{\pi ^2}$metre.

So, the correct answer is “Option A”.

Note:
To solve this question, you need to know that the friction is applied on any body to prevent the motion of the body. That means, as long as the force applied on the coin due to the rotation of the disk is less than the force of friction that the coin can generate, the coin will not move. So the maximum distance will be the distance at which the centrifugal force on the coin will be equal to the frictional force of the coin. This is a simple question that you can solve if you know the required formulae. So it is important to know all the formulae.