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 principle of conservation of energy. The initial kinetic energy of each object will be converted into gravitational potential energy as they climb the incline.

### Step-by-Step Solution:

1. **Identify the Initial Kinetic Energy of the Cylindrical Shell:**
- The initial speed of the cylindrical shell is \( v \).
- The total kinetic energy (KE) of the cylindrical shell consists of translational and rotational components:
\[ KE_{\text{shell}} = \frac{1}{2} m v^2 + \frac{1}{2} I_{\text{shell}} \omega^2 \]
- For a thin cylindrical shell, the moment of inertia \( I_{\text{shell}} = m r^2 \) and \( \omega = \frac{v}{r} \). Thus:
\[ KE_{\text{shell}} = \frac{1}{2} m v^2 + \frac{1}{2} (m r^2) \left(\frac{v}{r}\right)^2 = \frac{1}{2} m v^2 + \frac{1}{2} m v^2 = mv^2 \]

2. **Set Up the Energy Conservation Equation for the Cylindrical Shell:**
- As the shell climbs to height \( h \), its kinetic energy is converted into potential energy (PE):
\[ mv^2 = mgh \]
- This simplifies to:
\[ v^2 = gh \quad \text{(Equation 1)} \]

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

4. **Set Up the Energy Conservation Equation for the Solid Cylinder:**
- As the solid cylinder climbs to height \( h \), its kinetic energy is converted into potential energy:
\[ \frac{3}{4} m' (v')^2 = m'gh \]
- This simplifies to:
\[ \frac{3}{4} (v')^2 = gh \quad \text{(Equation 2)} \]

5. **Equate the Two Height Equations:**
- From Equation 1 and Equation 2, we have:
\[ gh = v^2 \quad \text{and} \quad gh = \frac{3}{4} (v')^2 \]
- Thus:
\[ v^2 = \frac{3}{4} (v')^2 \]

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

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