A solid cylinder 30 cm in diameter at the top of an inclined plane 2.0 m high is released and rolls down the incline without loss of energy due to friction. Its linear speed at the bottom is

Options

A. 5.29 m/sec
B. `4.1xx10^3m/s`
C. 51 m/s
D. 51 cm/sec

Answer

To find the linear speed of a solid cylinder rolling down an inclined plane, we can use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Identify the given data
- Diameter of the cylinder = 30 cm, hence the radius \( r = \frac{30}{2} = 15 \) cm = 0.15 m
- Height of the incline \( h = 2.0 \) m
- Acceleration due to gravity \( g = 9.81 \) m/s²

### Step 2: Write the conservation of energy equation
At the top of the incline, the cylinder has potential energy (PE) and no kinetic energy (KE) since it is at rest. At the bottom, all potential energy will have converted into kinetic energy (both translational and rotational).

The potential energy at the top is given by:
\[ PE = mgh \]

The kinetic energy at the bottom is the sum of translational kinetic energy and rotational kinetic energy:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]

For a solid cylinder, the moment of inertia \( I \) is given by:
\[ I = \frac{1}{2} m r^2 \]

And the relationship between linear speed \( v \) and angular speed \( \omega \) is:
\[ \omega = \frac{v}{r} \]

### Step 3: Substitute the moment of inertia into the kinetic energy equation
Substituting \( I \) into the kinetic energy equation:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{1}{2} m r^2\right) \left(\frac{v}{r}\right)^2 \]
\[ = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 \]
\[ = \frac{3}{4} mv^2 \]

### Step 4: Set potential energy equal to kinetic energy
Setting the potential energy equal to the kinetic energy:
\[ mgh = \frac{3}{4} mv^2 \]

### Step 5: Simplify the equation
Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh = \frac{3}{4} v^2 \]

### Step 6: Solve for \( v^2 \)
Rearranging gives:
\[ v^2 = \frac{4gh}{3} \]

### Step 7: Substitute the values
Substituting \( g = 9.81 \) m/s² and \( h = 2.0 \) m:
\[ v^2 = \frac{4 \times 9.81 \times 2.0}{3} \]
\[ = \frac{78.48}{3} = 26.16 \]

### Step 8: Calculate \( v \)
Taking the square root:
\[ v = \sqrt{26.16} \approx 5.11 \text{ m/s} \]

### Final Answer
The linear speed of the solid cylinder at the bottom of the incline is approximately **5.11 m/s**.

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