A disc can roll wihtout slippingg, without applying any external force on a

Options

A. (a)rough inclined plane
B. smooth inclined plane
C. (c)rough horizontal surface
D. (d)smooth horizontal surface

Answer

It seems that the provided transcript is not related to the question about a disc rolling without slipping. However, I will provide a step-by-step solution to the question regarding the rolling disc.

### Question: A disc can roll without slipping.

### Solution Steps:

1. **Understanding the Concept of Rolling Without Slipping**:
- A disc rolls without slipping when the point of contact with the ground does not slide. This means that the linear velocity of the center of mass (v) is related to the angular velocity (ω) by the equation:
\[ v = R \cdot \omega \]
where \( R \) is the radius of the disc.

2. **Identifying Forces and Motion**:
- When a disc rolls down an incline or on a flat surface without any external force, it experiences gravitational force acting downwards and the normal force acting perpendicular to the surface. The frictional force at the point of contact prevents slipping.

3. **Using Energy Conservation**:
- The total mechanical energy of the disc is conserved when rolling without slipping. The potential energy (PE) at the height is converted into kinetic energy (KE) as it rolls down:
\[ PE = KE_{translational} + KE_{rotational} \]
- The translational kinetic energy is given by:
\[ KE_{translational} = \frac{1}{2} mv^2 \]
- The rotational kinetic energy is given by:
\[ KE_{rotational} = \frac{1}{2} I \omega^2 \]
- For a solid disc, the moment of inertia \( I \) is:
\[ I = \frac{1}{2} m R^2 \]

4. **Relating Linear and Angular Velocities**:
- Substitute \( \omega \) in terms of \( v \):
\[ \omega = \frac{v}{R} \]
- Therefore, the rotational kinetic energy becomes:
\[ KE_{rotational} = \frac{1}{2} \left(\frac{1}{2} m R^2\right) \left(\frac{v}{R}\right)^2 = \frac{1}{4} mv^2 \]

5. **Setting Up the Energy Equation**:
- The total kinetic energy when rolling without slipping is:
\[ KE_{total} = KE_{translational} + KE_{rotational} = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2 \]
- Setting the potential energy equal to the total kinetic energy:
\[ mgh = \frac{3}{4} mv^2 \]
- The mass \( m \) cancels out:
\[ gh = \frac{3}{4} v^2 \]

6. **Solving for Linear Velocity**:
- Rearranging the equation to solve for \( v \):
\[ v^2 = \frac{4gh}{3} \]
\[ v = \sqrt{\frac{4gh}{3}} \]

### Final Result:
The linear velocity \( v \) of the disc rolling without slipping is given by:
\[ v = \sqrt{\frac{4gh}{3}} \]