The M.I. of a solid cylinder about its axis is I. It is allowed to rool down an incline plane without slipping. If its angular velocity at the bottom be `omega`, then kinetic energy of rolling cylinder will be

Options

A. `Iomega^2`
B. `3/2Iomega^2`
C. `2Iomega^2`
D. `1/2Iomega^2`

Answer

To find the kinetic energy of a solid cylinder rolling down an incline without slipping, we can follow these steps:

### Step 1: Understand the moment of inertia
The moment of inertia (I) of a solid cylinder about its axis is given by:
\[ I = \frac{1}{2} m r^2 \]
where \( m \) is the mass of the cylinder and \( r \) is its radius.

### Step 2: Write the expression for kinetic energy
The total kinetic energy (KE) of the rolling cylinder can be expressed as the sum of its translational kinetic energy and rotational kinetic energy:
\[ KE = KE_{\text{translational}} + KE_{\text{rotational}} \]
This can be written as:
\[ KE = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 \]

### Step 3: Relate linear and angular velocity
For a rolling object without slipping, the relationship between linear velocity (v) and angular velocity (ω) is given by:
\[ v = r \omega \]
We can substitute this into the kinetic energy equation.

### Step 4: Substitute v in the kinetic energy equation
Substituting \( v = r \omega \) into the translational kinetic energy term:
\[ KE_{\text{translational}} = \frac{1}{2} m (r \omega)^2 = \frac{1}{2} m r^2 \omega^2 \]

### Step 5: Substitute I in the kinetic energy equation
Now, substituting the moment of inertia \( I = \frac{1}{2} m r^2 \) into the rotational kinetic energy term:
\[ KE_{\text{rotational}} = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{1}{2} m r^2\right) \omega^2 = \frac{1}{4} m r^2 \omega^2 \]

### Step 6: Combine the kinetic energy terms
Now, we can combine both kinetic energy terms:
\[ KE = \frac{1}{2} m r^2 \omega^2 + \frac{1}{4} m r^2 \omega^2 \]
Factoring out \( m r^2 \omega^2 \):
\[ KE = \left(\frac{1}{2} + \frac{1}{4}\right) m r^2 \omega^2 = \frac{3}{4} m r^2 \omega^2 \]

### Step 7: Substitute \( m r^2 \) in terms of I
Since \( I = \frac{1}{2} m r^2 \), we can express \( m r^2 \) in terms of \( I \):
\[ m r^2 = 2I \]
Substituting this into the kinetic energy expression:
\[ KE = \frac{3}{4} \cdot 2I \cdot \omega^2 = \frac{3}{2} I \omega^2 \]

### Final Answer
Thus, the kinetic energy of the rolling cylinder is:
\[ KE = \frac{3}{2} I \omega^2 \]

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