A small mass m starts from rest and slides down the smooth spherical surface of F. Assume zero potential energy at the top. Find
(a) the change in potential energy,
(b) the kinetic energy,
(c) the speed ot the mass as a function of the angle `theta` made by the radius through the mass with the vertical.

Explanation

To solve the problem step by step, we will analyze the situation of a mass sliding down a smooth spherical surface.

### Step 1: Understanding the Setup
- We have a small mass \( m \) that starts from rest at the top of a smooth spherical surface of radius \( R \).
- The potential energy is considered zero at the top of the sphere.
- As the mass slides down, it descends a height \( h \) which can be expressed in terms of the angle \( \theta \) made with the vertical.

### Step 2: Finding the Change in Potential Energy
- At the top of the sphere, the height \( h \) is \( R \) (the radius).
- When the mass descends to an angle \( \theta \), the height \( h \) can be expressed as:
\[ h = R - R \cos \theta = R(1 - \cos \theta) \]
- The change in potential energy \( \Delta U \) as the mass descends is given by:
\[ \Delta U = -mgh \]
Substituting for \( h \):
\[ \Delta U = -mg(R(1 - \cos \theta)) = -mgR(1 - \cos \theta) \]

### Step 3: Finding the Kinetic Energy
- The change in kinetic energy \( \Delta K \) is equal to the change in potential energy (due to conservation of mechanical energy):
\[ \Delta K = -\Delta U = mgR(1 - \cos \theta) \]

### Step 4: Relating Kinetic Energy to Speed
- The kinetic energy can also be expressed as:
\[ K = \frac{1}{2} mv^2 \]
- Setting the two expressions for kinetic energy equal gives:
\[ \frac{1}{2} mv^2 = mgR(1 - \cos \theta) \]
- We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v^2 = gR(1 - \cos \theta) \]
- Solving for \( v^2 \):
\[ v^2 = 2gR(1 - \cos \theta) \]
- Taking the square root to find \( v \):
\[ v = \sqrt{2gR(1 - \cos \theta)} \]

### Final Answers
(a) The change in potential energy:
\[ \Delta U = -mgR(1 - \cos \theta) \]

(b) The kinetic energy:
\[ K = mgR(1 - \cos \theta) \]

(c) The speed of the mass as a function of the angle \( \theta \):
\[ v = \sqrt{2gR(1 - \cos \theta)} \]