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. 41 m/sec
C. 51 m/sec
D. 51 cm/sec

Answer

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

### Step 1: Identify the given information
- Diameter of the cylinder = 30 cm, hence radius \( r = \frac{30}{2} = 15 \) cm = 0.15 m
- Height of the inclined plane \( h = 2.0 \) m
- Initial velocity \( u = 0 \) (the cylinder is released from rest)

### Step 2: Write the conservation of mechanical energy equation
According to the conservation of mechanical energy, the total mechanical energy at the top of the incline (potential energy) equals the total mechanical energy at the bottom (kinetic energy).

At the top:
- Potential Energy (PE) = \( mgh \)
- Kinetic Energy (KE) = 0 (since it is at rest)

At the bottom:
- Potential Energy (PE) = 0 (since it is at the reference level)
- Kinetic Energy (KE) = Translational KE + Rotational KE
- Translational KE = \( \frac{1}{2} mv^2 \)
- Rotational KE = \( \frac{1}{2} I \omega^2 \)

### Step 3: Determine the moment of inertia and relate linear and angular velocity
For a solid cylinder, the moment of inertia \( I \) about its center of mass is given by:
\[ I = \frac{1}{2} m r^2 \]

Using the rolling condition, the relationship between linear speed \( v \) and angular speed \( \omega \) is:
\[ v = r \omega \]
Thus,
\[ \omega = \frac{v}{r} \]

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

### Step 5: Set up the energy conservation equation
Setting the total mechanical energy at the top equal to that at the bottom:
\[ mgh = \frac{3}{4} mv^2 \]

### Step 6: Cancel mass and solve for \( v^2 \)
Cancel \( m \) from both sides:
\[ gh = \frac{3}{4} v^2 \]
Rearranging gives:
\[ v^2 = \frac{4gh}{3} \]

### Step 7: Substitute the values of \( g \) and \( h \)
Using \( g = 9.8 \, \text{m/s}^2 \) and \( h = 2.0 \, \text{m} \):
\[ v^2 = \frac{4 \times 9.8 \times 2}{3} \]
\[ v^2 = \frac{78.4}{3} \]
\[ v^2 = 26.1333 \]

### Step 8: Calculate \( v \)
Taking the square root:
\[ v = \sqrt{26.1333} \approx 5.11 \, \text{m/s} \]

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

---