Sol. (i) The change in the potential energy is caused by the change in the height of the mass M,

Δ u = u (2) – u (1) = –Mgh – 0 = –MgR (1 – cos θ ) …...(i)

Where the indices (1) and (2) refer to the initial and final states, respectively.

(ii) Assuming that the mass is initially at rest, and that there is no friction, we may obtain from the law of conservation of energy:

E k (2) = E k (1) – (u (2) – u (1) ) …..(ii)

Hence, (E k (1) = 0)

E k (2) = – Δ u = MgR (1 – cos θ ) …..(iii)

(iii) The radial acceleration is given by:

| | = = = = 2g (1 – cos θ ) …...(iv)

Since the acceleration is directed towards the center of the sphere,

= –2g (1 – cos θ ) …...(v)

The tangential acceleration is

a tangent = …...(vi)

The tangent force is Mg sin θ , Therefore,

a tangent = g sin θ …...(vii)

(iv) We will first calculate the forces acting on M. In this section, we are only interested in the components. The forces are = N , the normal force and the force of gravity. The projection of the force of gravity on is –Mgcos θ . The mass acceleration in the direction was already determined in the previous section,

a r = –2g (1 – cos θ )

Therefore, according to Newton's second law of motion, while the mass is still on the surface,

–2Mg (1 – cos θ ) = N – Mg cos θ …...(viii)

The condition mass M must satisfy in order to leave the surface is N = 0. Hence,

2(1 – cos θ 0 ) = cos θ 0 ~ cos θ 0 = …...(ix)

(v) With friction, the kinetic energy accumulated by the mass is smaller and therefore |a r | is smaller. Hence, θ > θ 0 .