Question

A heavy ring of mass \[m\] is fitted onto the periphery of a light circular disc. A small particle of equal mass is clamped at the centre of the disc. The system is rotated in such a way that the centre of mass moves in a circle of radius \[r\] with uniform speed \[v\]. We conclude that an external force:
A. \[\dfrac{{m{v^2}}}{r}\] must be acting on the central particle
B. \[\dfrac{{2m{v^2}}}{r}\] must be acting on the central particle
C. \[\left( {\dfrac{{2m{v^2}}}{r}} \right)\] must be acting on the system
D. \[\left( {\dfrac{{2m{v^2}}}{r}} \right)\] must be acting on the ring

Answer 1

Hint: Use the formula for the centripetal force acting on an object performing circular motion. Determine the mass of the total system of the light circular ring, heavy ring on the periphery and small particle at the centre of the ring. Hence, determine the magnitude of the external force acting on the system to balance the centripetal force acting on the system in circular motion.

Formula used:
The centripetal force \[{F_C}\] acting on an object is the circular motion is
\[{F_C} = \dfrac{{m{v^2}}}{R}\] …… (1)
Here, \[m\] is the mass of the object, \[v\] is the velocity of the object and \[R\] is the radius of the circular path.

Complete step by step answer:
We have given that a heavy ring with mass \[m\] is placed on the circumference of a light circular disc. A small of the same mass \[m\] is placed at the centre of the light ring. The system of the heavy ring and the small mass rotates in such a way that the centre of mass of the system is rotated in a circle of radius \[r\] and uniform speed \[v\]. We have asked to determine the value of the external force and state that it acts on a system, ring or central particle.

Since the light circular ring on which the heavy ring and small mass are placed has negligible mass, we can neglect the mass of this light circular ring.The mass of the system of the heavy ring and a small mass is
\[M = m + m = 2m\]
When the system of the heavy ring and small mass is rotating in a circular path, there must be a centripetal force acting on the system in the inward direction along the radius.
\[{F_C} = \dfrac{{2m{v^2}}}{R}\]
To balance this centripetal force, there must be an external force \[F\] acting on the system and the magnitude of this external force must be the same as that of the centripetal force.
\[\therefore F = \dfrac{{2m{v^2}}}{R}\]
Therefore, an external force of \[\dfrac{{2m{v^2}}}{R}\] must be acting on the system.

Hence, the correct option is C.

Note:The students should keep in mind that the ring on which the heavy ring and small mass are placed is a light circular ring. So, we can neglect its mass from the total mass of the system. Also the external force acting on the system must equal the centripetal force on the system is stated according to Newton’s second law of motion.