Question

A horizontal smooth groove is formed inside of a solid sphere of radius R = 1m and a charge Q = 1µC is uniformly distributed in the sphere in whole volume. The sphere rotates with constant angular velocity ω = 1 rad/s as shown in the figure. A charge particle of mass m = 1 gm and charge q= $(\frac{1}{3})$ µC is released at rest just right of the centre along the groove. The speed with which the ball leaves the groove is

• A. 2 m/s
• B. $\sqrt{3}$ m/s
• C. $\sqrt{5}$ m/s
• D. 1 m/s

Answer 1

Answer: (A)

2 m/s

Answer 2

The problem is solved by considering the motion of the charged particle in the rotating frame of reference. The forces acting on the particle along the groove are the electrostatic force due to the uniformly charged sphere and the centrifugal force. The electrostatic force is repulsive and proportional to the distance from the center. The centrifugal force is also directed outwards and proportional to the distance from the axis of rotation. Since the groove is along a radius in the horizontal plane of rotation, the distance from the center and the distance from the axis of rotation are the same, denoted by x. The total force along the groove is the sum of the electrostatic and centrifugal forces. We use the work-energy theorem in the rotating frame, which states that the total work done by the forces along the displacement in the rotating frame is equal to the change in kinetic energy in the rotating frame. The particle starts from rest at the center (approximately) and moves to the edge of the sphere at $x=R$. We calculate the work done by the net force along the groove and equate it to the final kinetic energy since the initial kinetic energy is zero. This gives an equation for the square of the speed at $x=R$. Substituting the given values, we find the speed. Alternatively, we can use the conservation of effective energy in the rotating frame, which includes the potential energy from conservative forces (electrostatic force) and a potential-like term from the centrifugal force. By equating the initial effective energy at the center (at rest) to the final effective energy at the edge, we can find the speed at the edge. Both methods yield the same result.