A ball of radius 11 cm and mass 8 kg rolls from rest down a ramp of length 2m. The ramp is inclined at 35° to the horizontal. When the ball reaches the bottom, its velocity is (sin 35° = 0.57)?

Options

A. 2 m/s
B. 5 m/s
C. 4 m/s
D. 6 m/s

Answer

To solve the problem of a ball rolling down an inclined ramp, we will use the principle of conservation of energy. The potential energy at the top of the ramp will be converted into both translational and rotational kinetic energy at the bottom.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the ball, \( m = 8 \, \text{kg} \)
- Radius of the ball, \( r = 11 \, \text{cm} = 0.11 \, \text{m} \)
- Length of the ramp, \( L = 2 \, \text{m} \)
- Angle of inclination, \( \theta = 35^\circ \)
- Gravitational acceleration, \( g = 9.8 \, \text{m/s}^2 \)
- \( \sin(35^\circ) = 0.57 \)

2. **Calculate the Height of the Ramp:**
The height \( H \) of the ramp can be calculated using the sine function:
\[ H = L \sin(\theta) = 2 \times \sin(35^\circ) = 2 \times 0.57 = 1.14 \, \text{m} \]

3. **Write the Conservation of Energy Equation:**
The potential energy at the top of the ramp is converted into kinetic energy at the bottom. The equation is:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
For a solid sphere, the moment of inertia \( I \) is given by:
\[ I = \frac{2}{5} m r^2 \]
The relationship between linear velocity \( v \) and angular velocity \( \omega \) is:
\[ \omega = \frac{v}{r} \]

4. **Substituting the Moment of Inertia:**
Substitute \( I \) into the energy equation:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v^2}{r^2}\right) \]
Simplifying this gives:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \left(\frac{1}{2} + \frac{1}{5}\right) mv^2 = \frac{7}{10} mv^2 \]

5. **Canceling Mass and Rearranging:**
Cancel \( m \) from both sides:
\[ gh = \frac{7}{10} v^2 \]
Rearranging gives:
\[ v^2 = \frac{10gh}{7} \]

6. **Substituting Values:**
Substitute \( g = 9.8 \, \text{m/s}^2 \) and \( h = 1.14 \, \text{m} \):
\[ v^2 = \frac{10 \times 9.8 \times 1.14}{7} \]
Calculate the right-hand side:
\[ v^2 = \frac{111.12}{7} \approx 15.87 \]

7. **Taking the Square Root:**
Finally, take the square root to find \( v \):
\[ v \approx \sqrt{15.87} \approx 4 \, \text{m/s} \]

### Final Answer:
The velocity of the ball when it reaches the bottom of the ramp is approximately \( 4 \, \text{m/s} \).