A uniform, thin cylindrical shell and solid cylinder roll horizontally without slipping. The speed of the cylindrical shell is v. The solid cylinder and the hollow cylinder encounter an incline that they climb without slipping. If the maximum height they reach is the same, find the intial speed v' of the solid cylinder.

Options

A. `sqrt((4)/(3))v`
B. `sqrt((3)/(4))v`
C. `sqrt((3)/(5))v`
D. `sqrt((5)/(3))v`

Answer

To solve the problem, we will use the principles of conservation of energy and the moment of inertia for both the hollow cylindrical shell and the solid cylinder.

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- Let the initial speed of the hollow cylindrical shell be \( v \).
- Let the initial speed of the solid cylinder be \( v' \).
- Both objects roll without slipping and reach the same maximum height \( h \) on the incline.

2. **Write the Energy Conservation Equation**:
- The initial kinetic energy (KE) of the hollow cylindrical shell is given by:
\[ KE_{\text{hollow}} = \frac{1}{2} m v^2 + \frac{1}{2} I_{\text{hollow}} \omega^2 \]
- The moment of inertia \( I \) for a hollow cylindrical shell is \( I_{\text{hollow}} = m r^2 \).
- The angular velocity \( \omega \) can be expressed as \( \omega = \frac{v}{r} \).
- Thus, the kinetic energy becomes:
\[ KE_{\text{hollow}} = \frac{1}{2} m v^2 + \frac{1}{2} (m r^2) \left(\frac{v}{r}\right)^2 = \frac{1}{2} m v^2 + \frac{1}{2} m v^2 = m v^2 \]

3. **Kinetic Energy of the Solid Cylinder**:
- The initial kinetic energy of the solid cylinder is:
\[ KE_{\text{solid}} = \frac{1}{2} m v'^2 + \frac{1}{2} I_{\text{solid}} \omega^2 \]
- The moment of inertia for a solid cylinder is \( I_{\text{solid}} = \frac{1}{2} m r^2 \).
- Thus, the kinetic energy becomes:
\[ KE_{\text{solid}} = \frac{1}{2} m v'^2 + \frac{1}{2} \left(\frac{1}{2} m r^2\right) \left(\frac{v'}{r}\right)^2 = \frac{1}{2} m v'^2 + \frac{1}{4} m v'^2 = \frac{3}{4} m v'^2 \]

4. **Potential Energy at Maximum Height**:
- At the maximum height \( h \), the potential energy (PE) for both objects is:
\[ PE = mgh \]

5. **Set Up the Energy Conservation Equations**:
- For the hollow cylindrical shell:
\[ mgh = mv^2 \quad \text{(1)} \]
- For the solid cylinder:
\[ mgh = \frac{3}{4} m v'^2 \quad \text{(2)} \]

6. **Equate the Two Expressions**:
- From (1):
\[ h = \frac{v^2}{g} \]
- From (2):
\[ h = \frac{4}{3g} v'^2 \]
- Setting the two expressions for \( h \) equal gives:
\[ \frac{v^2}{g} = \frac{4}{3g} v'^2 \]

7. **Solve for \( v' \)**:
- Cancel \( g \) from both sides:
\[ v^2 = \frac{4}{3} v'^2 \]
- Rearranging gives:
\[ v'^2 = \frac{3}{4} v^2 \]
- Taking the square root:
\[ v' = \frac{\sqrt{3}}{2} v \]

### Final Answer:
The initial speed \( v' \) of the solid cylinder is:
\[ v' = \frac{\sqrt{3}}{2} v \]