Question

A coin is placed at the edge of a horizontal disc rotating about a vertical axis through speed \[2\,{\text{rad/s}}\]. The radius of the disc is 50cm. Find the minimum coefficient of friction between disc and coin so that the coin does not slip (\[g = 10\,{\text{m}}{{\text{s}}^{ - 2}}\]).
A. 0.1
B. 0.2
C. 0.3
D. 0.4

Answer 1

Hint: The coin will not slip from the disc when the friction force between the coin and disc is equal to the centrifugal force on the coin. The magnitude of centrifugal force is equal to the magnitude of centripetal force. Recall the expression for the centripetal force and equate it with the frictional force to determine the value of coefficient of friction.

Formula used:
Centripetal force, \[{F_C} = \dfrac{{m{v^2}}}{R}\]
where, m is the mass, v is the velocity and R is the radius of circular motion.
Frictional force, \[f = \mu mg\]
where, \[\mu \] is the coefficient of friction and g is the acceleration due to gravity.

Complete step by step solution:
We have given that the angular velocity of the disc is \[\omega = 2\,{\text{rad/s}}\] and the radius of the circular motion of the coin is \[R = 50\,{\text{cm}}\].
The coin will not slip from the disc when the friction force between the coin and disc is equal to the centrifugal force on the coin. Therefore, we can write,
\[\mu mg = \dfrac{{m{v^2}}}{R}\]
\[ \Rightarrow \mu = \dfrac{{{v^2}}}{{Rg}}\]
Here, \[\mu \] is the coefficient of friction, v is the linear velocity of the coin, R is the distance of the coin from the centre of the disc and g is the acceleration due to gravity.

But, we have, \[v = R\omega \]. Therefore, the above equation becomes,
\[\mu = \dfrac{{{{\left( {R\omega } \right)}^2}}}{{Rg}}\]
\[ \Rightarrow \mu = \dfrac{{R{\omega ^2}}}{g}\]
Substituting \[R = 50\,{\text{cm}}\], \[\omega = 2\,{\text{rad/s}}\] and \[g = 10\,{\text{m}}{{\text{s}}^{ - 2}}\] in the above equation, we get,
\[\mu = \dfrac{{\left( {0.5} \right){{\left( 2 \right)}^2}}}{{10}}\]
\[ \Rightarrow \mu = \dfrac{2}{{10}}\]
\[ \therefore \mu = 0.2\]

Therefore, the correct answer is option (B).

Note: The direction of centrifugal force on the body performing circular motion is away from the centre of circular motion. The force that acts on the body in circular motion is the centrifugal force and not the centripetal force. The magnitude of the centripetal force is equal to the friction force on the coin since it is the force responsible for the circular motion of the coin.