A small ball of radius r rolls without sliding in a big hemispherical bowl. of radius R. what would be the ratio of the translational and rotaional kinetic energies at the bottom of the bowl.

Options

A. `2:1`
B. `3:2 `
C. `4:3`
D. `5:2 `

Answer

To solve the problem of finding the ratio of translational and rotational kinetic energies of a small ball rolling without slipping in a big hemispherical bowl, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Kinetic Energies**:
- The translational kinetic energy (TKE) of the ball is given by:
\[ TKE = \frac{1}{2} mv^2 \]
- The rotational kinetic energy (RKE) of the ball is given by:
\[ RKE = \frac{1}{2} I \omega^2 \]
where \(I\) is the moment of inertia of the ball and \(\omega\) is the angular velocity.

2. **Moment of Inertia**:
- For a solid sphere, the moment of inertia \(I\) is:
\[ I = \frac{2}{5} m r^2 \]

3. **Relation between Linear and Angular Velocity**:
- Since the ball rolls without slipping, the relationship between the linear velocity \(v\) and the angular velocity \(\omega\) is:
\[ v = r \omega \quad \Rightarrow \quad \omega = \frac{v}{r} \]

4. **Substituting \(\omega\) in RKE**:
- Substitute \(\omega\) in the expression for RKE:
\[ RKE = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v}{r}\right)^2 \]
- Simplifying this gives:
\[ RKE = \frac{1}{2} \cdot \frac{2}{5} m r^2 \cdot \frac{v^2}{r^2} = \frac{1}{5} mv^2 \]

5. **Finding the Ratio of TKE to RKE**:
- Now, we can find the ratio of translational kinetic energy to rotational kinetic energy:
\[ \text{Ratio} = \frac{TKE}{RKE} = \frac{\frac{1}{2} mv^2}{\frac{1}{5} mv^2} \]
- The \(mv^2\) terms cancel out:
\[ \text{Ratio} = \frac{\frac{1}{2}}{\frac{1}{5}} = \frac{1}{2} \cdot \frac{5}{1} = \frac{5}{2} \]

6. **Final Result**:
- Therefore, the ratio of translational kinetic energy to rotational kinetic energy at the bottom of the bowl is:
\[ \frac{TKE}{RKE} = \frac{5}{2} \]