A cylinder of mass 2 kg is released from rest from the top of an inclined plane of inclination `30^(@)` and length 1m. If the cylinder rolls without slipping, then its speed when it reaches the bottom, is [Take, g = 10 m/`s^(2)`]

Options

A. `(20)/(3)` m/s
B. `sqrt((20)/(3))` m/s
C. `(10)/(3) ` m/s
D. `sqrt((10)/(3))` m/s

Explanation

To solve the problem, we will use the principle of conservation of energy. The potential energy lost by the cylinder as it rolls down the incline will be converted into kinetic energy (both translational and rotational).

### Step-by-Step Solution:

1. **Identify the parameters:**
- Mass of the cylinder, \( m = 2 \, \text{kg} \)
- Length of the incline, \( L = 1 \, \text{m} \)
- Angle of inclination, \( \theta = 30^\circ \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the height (h):**
The height \( h \) can be calculated using the sine of the angle:
\[ h = L \sin(\theta) = 1 \cdot \sin(30^\circ) = 1 \cdot \frac{1}{2} = 0.5 \, \text{m} \]

3. **Calculate the potential energy (PE) at the top:**
The potential energy at the top of the incline is given by:
\[ PE = mgh = 2 \cdot 10 \cdot 0.5 = 10 \, \text{J} \]

4. **Set up the kinetic energy (KE) at the bottom:**
When the cylinder reaches the bottom, its energy will be in the form of translational kinetic energy and rotational kinetic energy:
\[ KE = KE_{\text{translational}} + KE_{\text{rotational}} = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
For a solid cylinder, the moment of inertia \( I \) is:
\[ I = \frac{1}{2} m r^2 \]
The relationship between linear velocity \( v \) and angular velocity \( \omega \) is:
\[ \omega = \frac{v}{r} \]

5. **Substituting the moment of inertia:**
Substituting \( I \) and \( \omega \) into the kinetic energy equation:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{1}{2} m r^2\right) \left(\frac{v}{r}\right)^2 \]
This simplifies to:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2 \]

6. **Using conservation of energy:**
Setting the potential energy equal to the total kinetic energy:
\[ mgh = \frac{3}{4} mv^2 \]
Canceling \( m \) from both sides (since \( m \neq 0 \)):
\[ gh = \frac{3}{4} v^2 \]

7. **Substituting values:**
Substituting \( g = 10 \, \text{m/s}^2 \) and \( h = 0.5 \, \text{m} \):
\[ 10 \cdot 0.5 = \frac{3}{4} v^2 \]
Simplifying gives:
\[ 5 = \frac{3}{4} v^2 \]

8. **Solving for \( v^2 \):**
Multiplying both sides by \( \frac{4}{3} \):
\[ v^2 = \frac{4 \cdot 5}{3} = \frac{20}{3} \]

9. **Finding \( v \):**
Taking the square root:
\[ v = \sqrt{\frac{20}{3}} \approx 2.58 \, \text{m/s} \]

### Final Answer:
The speed of the cylinder when it reaches the bottom of the incline is approximately \( 2.58 \, \text{m/s} \).

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