A hollow cylinder of mass m and radius r is kept on a smooth horizontal surface. Horizontal force F is applied at a height above the centre. Find the value of h/R so that ring performs pure rolling on the surface.

Options

A. 1
B. `1/2`
C. `1/3`
D. `2/5`

Explanation

To solve the problem of finding the value of \( \frac{h}{r} \) such that a hollow cylinder performs pure rolling on a smooth horizontal surface when a horizontal force \( F \) is applied at a height \( h \) above the center, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a hollow cylinder (ring) of mass \( m \) and radius \( r \) on a smooth horizontal surface.
- A horizontal force \( F \) is applied at a height \( h \) above the center of the cylinder.

2. **Conditions for Pure Rolling**:
- For pure rolling to occur, the point of contact with the surface must have zero velocity relative to the surface. This means that the linear acceleration \( a \) of the center of mass and the angular acceleration \( \alpha \) must satisfy the relation:
\[ a = \alpha r \]

3. **Equations of Motion**:
- The net force acting on the cylinder in the horizontal direction is given by:
\[ F = m a \]
- The torque \( \tau \) about the center of mass due to the force \( F \) applied at height \( h \) is:
\[ \tau = F \cdot (h + r) \]

4. **Moment of Inertia**:
- The moment of inertia \( I \) of a hollow cylinder about its center is:
\[ I = m r^2 \]

5. **Torque and Angular Acceleration**:
- The torque can also be expressed in terms of angular acceleration:
\[ \tau = I \alpha \]
- Substituting the moment of inertia:
\[ F \cdot (h + r) = (m r^2) \alpha \]

6. **Substituting for Angular Acceleration**:
- From the condition for pure rolling, we have \( \alpha = \frac{a}{r} \). Substituting this into the torque equation gives:
\[ F \cdot (h + r) = (m r^2) \left(\frac{a}{r}\right) \]
- This simplifies to:
\[ F \cdot (h + r) = m r a \]

7. **Relating Force and Acceleration**:
- From the net force equation \( F = m a \), we can express \( a \) as:
\[ a = \frac{F}{m} \]
- Substituting this into the torque equation yields:
\[ F \cdot (h + r) = m r \left(\frac{F}{m}\right) \]
- This simplifies to:
\[ F \cdot (h + r) = F r \]

8. **Solving for \( h \)**:
- Dividing both sides by \( F \) (assuming \( F \neq 0 \)):
\[ h + r = r \]
- Rearranging gives:
\[ h = r - r = 0 \]
- This implies:
\[ h = r \]

9. **Finding the Ratio**:
- Therefore, the ratio \( \frac{h}{r} \) is:
\[ \frac{h}{r} = \frac{r}{r} = 1 \]

### Final Answer:
\[ \frac{h}{r} = 1 \]