In the above probllem, the normal force between the ball and the shell in position B is (m=mass of ball)

Options

A. `(12)/(7)mg`
B. `(7)/(9)mg`
C. `(17)/(7)mg`
D. `(10)/(7)mg`

Answer

To solve the problem regarding the normal force between the ball and the shell in position B, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Energy at Point A**:
At point A, the ball is at a height \( h = r \) (where \( r \) is the radius of the shell). The potential energy at this point is given by:
\[ PE_A = mgh = mg \cdot r \]

2. **Identify the Energy at Point B**:
At point B, the ball is at the bottom of the shell. The total mechanical energy at this point consists of both translational and rotational kinetic energy. The translational kinetic energy is given by \( \frac{1}{2} mv^2 \) and the rotational kinetic energy for a solid sphere is given by \( \frac{1}{2} I \omega^2 \). The moment of inertia \( I \) for a solid sphere is \( \frac{2}{5} m r^2 \), and since \( \omega = \frac{v}{r} \), we can express the rotational kinetic energy as:
\[ KE_{rot} = \frac{1}{2} \cdot \frac{2}{5} m r^2 \cdot \left(\frac{v}{r}\right)^2 = \frac{1}{5} mv^2 \]
Therefore, the total energy at point B is:
\[ KE_B = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \frac{7}{10} mv^2 \]

3. **Conservation of Energy**:
Since energy is conserved, we can equate the potential energy at point A to the total kinetic energy at point B:
\[ mg r = \frac{7}{10} mv^2 \]
Simplifying this equation (and canceling \( m \)):
\[ g r = \frac{7}{10} v^2 \]
Rearranging gives:
\[ v^2 = \frac{10}{7} g r \]

4. **Centripetal Force Calculation**:
At point B, the ball is moving in a circular path, and the net force towards the center (centripetal force) is provided by the normal force \( N \) and the weight of the ball \( mg \). Therefore, we can write:
\[ N - mg = \frac{mv^2}{r} \]
Substituting \( v^2 \) from the previous step:
\[ N - mg = \frac{m \cdot \frac{10}{7} g r}{r} \]
Simplifying gives:
\[ N - mg = \frac{10}{7} mg \]
Rearranging this equation to solve for \( N \):
\[ N = mg + \frac{10}{7} mg = \frac{7}{7} mg + \frac{10}{7} mg = \frac{17}{7} mg \]

5. **Final Result**:
The normal force between the ball and the shell in position B is:
\[ N = \frac{17}{7} mg \]