A cylinder of mass M and radius R rolls on an inclined plane. The gain in kinetic energy is

Options

A. `1/2 Mv^2`
B. `1/2 I omega^2`
C. `3/4 Mv^2`
D. `3/4 I omega^2`

Answer

To find the gain in kinetic energy of a cylinder of mass \( M \) and radius \( R \) rolling down an inclined plane, we can follow these steps:

### Step 1: Understand the Motion
The cylinder is rolling down the inclined plane, which means it has both translational and rotational motion.

### Step 2: Write the Expressions for Kinetic Energy
1. **Translational Kinetic Energy (TKE)**:
\[ \text{TKE} = \frac{1}{2} M v^2 \]
where \( v \) is the linear velocity of the center of mass of the cylinder.

2. **Rotational Kinetic Energy (RKE)**:
The moment of inertia \( I \) for a solid cylinder about its axis is given by:
\[ I = \frac{1}{2} M R^2 \]
The angular velocity \( \omega \) is related to the linear velocity \( v \) by:
\[ \omega = \frac{v}{R} \]
Therefore, the rotational kinetic energy can be expressed as:
\[ \text{RKE} = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{1}{2} M R^2\right) \left(\frac{v^2}{R^2}\right) = \frac{1}{4} M v^2 \]

### Step 3: Total Kinetic Energy
Now, we can combine both forms of kinetic energy to find the total kinetic energy \( KE \):
\[ KE = \text{TKE} + \text{RKE} = \frac{1}{2} M v^2 + \frac{1}{4} M v^2 \]
To combine these, we find a common denominator:
\[ KE = \frac{2}{4} M v^2 + \frac{1}{4} M v^2 = \frac{3}{4} M v^2 \]

### Step 4: Conclusion
Thus, the total gain in kinetic energy of the cylinder as it rolls down the inclined plane is:
\[ \text{Gain in Kinetic Energy} = \frac{3}{4} M v^2 \]