A cylinder rolls down two different inclined planes of the same height but of different inclinations

Options

A. In both cases the speed and time of descent will be different
B. In both cases the speed and time of descent will be same
C. The speed will be different but time of descent will be same
D. The time of descent will be different but speed will be same

Answer

To solve the problem of a cylinder rolling down two different inclined planes of the same height but with different inclinations, we will analyze the energy transformations involved and derive the equations governing the motion.

### Step-by-Step Solution:

1. **Understanding the Energy Conservation Principle**:
- When the cylinder rolls down the incline, it loses potential energy and gains kinetic energy. The potential energy lost is given by:
\[ \text{Potential Energy} = mgh \]
- Here, \( m \) is the mass of the cylinder, \( g \) is the acceleration due to gravity, and \( h \) is the height of the incline.

**Hint**: Remember that potential energy is related to height and mass.

2. **Kinetic Energy of the Rolling Cylinder**:
- The kinetic energy of the rolling cylinder consists of two parts: translational kinetic energy and rotational kinetic energy.
- The translational kinetic energy is:
\[ \text{Translational KE} = \frac{1}{2} mv^2 \]
- The rotational kinetic energy is:
\[ \text{Rotational KE} = \frac{1}{2} I \omega^2 \]
- For a solid cylinder, the moment of inertia \( I \) is given by \( I = \frac{1}{2} mk^2 \), where \( k \) is the radius of the cylinder. The relationship between angular velocity \( \omega \) and linear velocity \( v \) is \( \omega = \frac{v}{r} \).

**Hint**: Recall the formulas for translational and rotational kinetic energy.

3. **Setting Up the Energy Equation**:
- The total kinetic energy when the cylinder reaches the bottom of the incline is:
\[ \text{Total KE} = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
- Substituting \( I \) and \( \omega \):
\[ \text{Total KE} = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{1}{2} mk^2\right) \left(\frac{v^2}{r^2}\right) \]
- This simplifies to:
\[ \text{Total KE} = \frac{1}{2} mv^2 \left(1 + \frac{k^2}{r^2}\right) \]

**Hint**: Ensure you understand how to substitute and simplify expressions involving kinetic energy.

4. **Equating Potential Energy to Kinetic Energy**:
- By conservation of energy, we set the potential energy equal to the total kinetic energy:
\[ mgh = \frac{1}{2} mv^2 \left(1 + \frac{k^2}{r^2}\right) \]
- Canceling \( m \) from both sides gives:
\[ gh = \frac{1}{2} v^2 \left(1 + \frac{k^2}{r^2}\right) \]

**Hint**: Remember that mass cancels out when using conservation of energy.

5. **Solving for Velocity**:
- Rearranging the equation to solve for \( v^2 \):
\[ v^2 = \frac{2gh}{1 + \frac{k^2}{r^2}} \]
- This shows that the final speed \( v \) depends on the height \( h \) and the geometry of the cylinder (through \( k \) and \( r \)).

**Hint**: Focus on how the geometry of the cylinder affects its speed.

6. **Analyzing Time of Descent**:
- The time of descent will depend on the angle of inclination. A steeper incline will generally result in a shorter time of descent.
- Since the inclines have the same height but different angles, the acceleration of the cylinder will be different, leading to different times of descent.

**Hint**: Consider how the angle of inclination affects acceleration and time.

### Conclusion:
From the analysis, we conclude that the speed of the cylinder will be different for the two inclines due to their different inclinations, while the time of descent will also differ. Therefore, the correct answer is that the speed and time of descent will be different for the two inclined planes.