A disc of mass 3 kg rolls down an inclined plane of height 5 m. The translational kinetic energy of the disc on reaching the bottom of the inclined plane is

Options

A. 50 J
B. 100 J
C. 150 J
D. 175 J

Answer

To solve the problem of finding the translational kinetic energy of a disc of mass 3 kg rolling down an inclined plane of height 5 m, we can follow these steps:

### Step 1: Understand the Energy Conservation Principle
When the disc rolls down the inclined plane, its potential energy at the top will convert into translational and rotational kinetic energy at the bottom. The total mechanical energy is conserved.

### Step 2: Calculate the Potential Energy at the Top
The potential energy (PE) at the height \( h \) is given by the formula:
\[ PE = mgh \]
where:
- \( m = 3 \, \text{kg} \) (mass of the disc),
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( h = 5 \, \text{m} \) (height of the inclined plane).

Substituting the values:
\[ PE = 3 \times 10 \times 5 = 150 \, \text{J} \]

### Step 3: Set Up the Energy Conservation Equation
At the bottom of the inclined plane, the potential energy will have converted into translational kinetic energy (TKE) and rotational kinetic energy (RKE):
\[ PE = TKE + RKE \]

### Step 4: Express the Kinetic Energies
The translational kinetic energy (TKE) is given by:
\[ TKE = \frac{1}{2} mv^2 \]
The rotational kinetic energy (RKE) for a disc is given by:
\[ RKE = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia of the disc about its center, which is \( \frac{1}{2} m r^2 \), and \( \omega \) is the angular velocity.

### Step 5: Relate Linear and Angular Velocity
For rolling without slipping:
\[ v = r \omega \implies \omega = \frac{v}{r} \]

### Step 6: Substitute for RKE
Substituting \( \omega \) into the RKE formula:
\[ RKE = \frac{1}{2} \left(\frac{1}{2} m r^2\right) \left(\frac{v}{r}\right)^2 = \frac{1}{4} mv^2 \]

### Step 7: Combine the Kinetic Energies
Now, substituting TKE and RKE into the energy conservation equation:
\[ 150 = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 \]
Combining the terms:
\[ 150 = \left(\frac{1}{2} + \frac{1}{4}\right) mv^2 = \frac{3}{4} mv^2 \]

### Step 8: Solve for \( v^2 \)
Rearranging gives:
\[ v^2 = \frac{150 \times 4}{3m} \]
Substituting \( m = 3 \):
\[ v^2 = \frac{150 \times 4}{3 \times 3} = \frac{600}{9} = \frac{200}{3} \]

### Step 9: Calculate TKE
Now, substituting \( v^2 \) back into the TKE formula:
\[ TKE = \frac{1}{2} mv^2 = \frac{1}{2} \times 3 \times \frac{200}{3} = 100 \, \text{J} \]

### Final Answer
The translational kinetic energy of the disc on reaching the bottom of the inclined plane is:
\[ \boxed{100 \, \text{J}} \]