In the previous question the smallest kinetic energy at the bottom of the incline will be achieved by

A. the solid sphere
B. the hollow sphere
C. the disc
D. all will achieve same kinetic energy

Options

A. the solid sphere
B. the hollow sphere
C. the disc
D. all will achieve same kinetic energy

Explanation

To solve the problem of determining which object achieves the smallest kinetic energy at the bottom of an incline, we will analyze the kinetic energy of three different objects: a solid sphere, a hollow sphere, and a disc. We will consider their moments of inertia and how they affect their kinetic energy.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have three objects: a solid sphere, a hollow sphere, and a disc, all with the same mass (m) and radius (r).
- They are rolled down an incline, and we need to determine which one has the smallest kinetic energy at the bottom.

2. **Kinetic Energy Components**:
- The total kinetic energy (KE) at the bottom of the incline consists of translational kinetic energy and rotational kinetic energy:
\[ KE = KE_{\text{trans}} + KE_{\text{rot}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \]
- Where \(I\) is the moment of inertia and \(\omega\) is the angular velocity.

3. **Relating Linear and Angular Velocity**:
- For rolling without slipping, the relationship between linear velocity (v) and angular velocity (\(\omega\)) is given by:
\[ v = r\omega \quad \Rightarrow \quad \omega = \frac{v}{r} \]

4. **Substituting for \(\omega\)**:
- Substitute \(\omega\) in the kinetic energy equation:
\[ KE = \frac{1}{2}mv^2 + \frac{1}{2}I\left(\frac{v}{r}\right)^2 \]
- This simplifies to:
\[ KE = \frac{1}{2}mv^2 + \frac{1}{2}\frac{I}{r^2}v^2 = \left(\frac{1}{2}m + \frac{1}{2}\frac{I}{r^2}\right)v^2 \]

5. **Moments of Inertia**:
- The moments of inertia for the three objects are:
- Solid Sphere: \(I = \frac{2}{5}mr^2\)
- Hollow Sphere: \(I = \frac{2}{3}mr^2\)
- Disc: \(I = \frac{1}{2}mr^2\)

6. **Calculating Effective Mass for Kinetic Energy**:
- For each object, we can express the effective mass in the kinetic energy equation:
\[ KE = \frac{1}{2}\left(m + \frac{I}{r^2}\right)v^2 \]
- Substitute the moments of inertia:
- Solid Sphere: \(m + \frac{2}{5}m = \frac{7}{5}m\)
- Hollow Sphere: \(m + \frac{2}{3}m = \frac{5}{3}m\)
- Disc: \(m + \frac{1}{2}m = \frac{3}{2}m\)

7. **Comparing Effective Masses**:
- The effective mass for each object is:
- Solid Sphere: \(\frac{7}{5}m\)
- Hollow Sphere: \(\frac{5}{3}m\)
- Disc: \(\frac{3}{2}m\)

8. **Determining the Smallest Kinetic Energy**:
- The object with the highest effective mass will have the smallest kinetic energy at the bottom of the incline because they all start with the same potential energy and convert it into kinetic energy.
- Comparing the effective masses:
- \(\frac{7}{5}m > \frac{5}{3}m > \frac{3}{2}m\)
- Thus, the hollow sphere has the highest effective mass and will have the smallest kinetic energy.

### Conclusion:
The smallest kinetic energy at the bottom of the incline will be achieved by the **hollow sphere**.