A small sphere of mass m and radius r rolls without slipping on the insides of a large hemisphere of radius R whose axis of symmetry is vertical. Its starts at the top from rest. What is the kinetic energy of the small sphere at the bottom? What fraction is rotational and what transiational? What normal force does the small spheres exert on the hemisphere at the bottom?
[Hint: consider circular motion of center of mass at the bottom to calcualte normal reaction. For others consider conservation of energy.?

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the potential energy at the top
The small sphere starts from rest at the top of the hemisphere. The potential energy (PE) at the top is given by:
\[ PE = mgh \]
where \( h = R - r \) is the height from which the sphere falls. Thus,
\[ PE = mg(R - r) \]

### Step 2: Apply conservation of energy
As the sphere rolls down without slipping, the potential energy at the top will convert into kinetic energy (KE) at the bottom. The total kinetic energy at the bottom consists of translational kinetic energy and rotational kinetic energy:
\[ KE = KE_{translational} + KE_{rotational} \]
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]

### Step 3: Substitute the moment of inertia and relate \( \omega \) to \( v \)
For a solid sphere, the moment of inertia \( I \) is:
\[ I = \frac{2}{5} m r^2 \]
Since the sphere rolls without slipping, we have:
\[ \omega = \frac{v}{r} \]
Thus,
\[ \omega^2 = \frac{v^2}{r^2} \]
Substituting this into the kinetic energy equation gives:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} \left( \frac{2}{5} m r^2 \right) \left( \frac{v^2}{r^2} \right) \]
\[ KE = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \frac{7}{10} mv^2 \]

### Step 4: Set potential energy equal to kinetic energy
Using conservation of energy:
\[ mg(R - r) = \frac{7}{10} mv^2 \]
Cancelling \( m \) from both sides:
\[ g(R - r) = \frac{7}{10} v^2 \]
Solving for \( v^2 \):
\[ v^2 = \frac{10g(R - r)}{7} \]

### Step 5: Calculate the total kinetic energy at the bottom
Now, substituting \( v^2 \) back into the kinetic energy equation:
\[ KE = \frac{7}{10} mv^2 = \frac{7}{10} m \left( \frac{10g(R - r)}{7} \right) = mg(R - r) \]

### Step 6: Determine the fraction of translational and rotational kinetic energy
1. **Rotational Kinetic Energy**:
\[ KE_{rotational} = \frac{1}{2} I \omega^2 = \frac{1}{2} \left( \frac{2}{5} m r^2 \right) \left( \frac{v^2}{r^2} \right) = \frac{1}{5} mv^2 \]
Substituting \( v^2 \):
\[ KE_{rotational} = \frac{1}{5} m \left( \frac{10g(R - r)}{7} \right) = \frac{2mg(R - r)}{7} \]

2. **Translational Kinetic Energy**:
\[ KE_{translational} = \frac{1}{2} mv^2 = \frac{1}{2} m \left( \frac{10g(R - r)}{7} \right) = \frac{5mg(R - r)}{7} \]

### Step 7: Calculate the normal force at the bottom
At the bottom, the forces acting on the sphere are the normal force \( N \) and the weight \( mg \). The centripetal force required for circular motion is provided by the net force:
\[ N - mg = \frac{mv^2}{R - r} \]
Substituting \( v^2 \):
\[ N - mg = \frac{m \left( \frac{10g(R - r)}{7} \right)}{R - r} \]
\[ N - mg = \frac{10mg}{7} \]
Thus,
\[ N = mg + \frac{10mg}{7} = \frac{17mg}{7} \]

### Summary of Results
- **Kinetic Energy at the Bottom**: \( KE = mg(R - r) \)
- **Fraction of Rotational Kinetic Energy**: \( \frac{2}{7} \)
- **Fraction of Translational Kinetic Energy**: \( \frac{5}{7} \)
- **Normal Force at the Bottom**: \( N = \frac{17mg}{7} \)