Question

A solid cylinder of mass 3kg is rolling on a horizontal plane with velocity of 4 ms^-1. It strikes with a spring of force constant 200 Nm^-1, what will be maximum compression in the spring?

(a) 0.5 m

(b) 0.6 m

(c) 0.7 m

(d) 0.2 m

Answer 1

The rolling cylinder strike a spring,

∴ Loss in kinetic energy = Increase in potential energy of spring

or \(\frac{1}{2}mv^2(1 + \frac{K^2}{R^2}) = \frac{1}{2}Kx^2_{max}\) 

or x_max = 0.6 m

Answer 2

To solve this problem, we'll use the principle of conservation of energy. The kinetic energy of the rolling cylinder will be completely converted into the potential energy stored in the spring when it compresses to its maximum extent.

### Step-by-Step Solution:

1. **Kinetic Energy of the Rolling Cylinder:**
   - A rolling solid cylinder has both translational and rotational kinetic energy.
   - Translational kinetic energy: \( KE_{trans} = \frac{1}{2} m v^2 \)
   - Rotational kinetic energy: \( KE_{rot} = \frac{1}{2} I \omega^2 \)
   - For a solid cylinder, the moment of inertia \( I = \frac{1}{2} m r^2 \).
   - The angular velocity \( \omega \) is related to the linear velocity \( v \) by \( \omega = \frac{v}{r} \).
   - Therefore, \( KE_{rot} = \frac{1}{2} \times \frac{1}{2} m r^2 \times \left(\frac{v}{r}\right)^2 = \frac{1}{4} m v^2 \).
   - Total kinetic energy: \( KE = KE_{trans} + KE_{rot} = \frac{1}{2} m v^2 + \frac{1}{4} m v^2 = \frac{3}{4} m v^2 \).

2. **Potential Energy in the Spring:**
   - When the cylinder compresses the spring, the potential energy stored in the spring is given by \( PE = \frac{1}{2} k x^2 \), where \( x \) is the maximum compression.

3. **Equating Kinetic and Potential Energy:**
   - At maximum compression, all the kinetic energy will have been converted into potential energy.
   - Therefore, \( \frac{3}{4} m v^2 = \frac{1}{2} k x^2 \).

4. **Solving for Maximum Compression \( x \):**
   - Rearrange the equation to solve for \( x \):
     \[
     x = \sqrt{\frac{3 m v^2}{2 k}}
     \]

5. **Substitute the Given Values:**
   - Mass \( m = 3 \) kg
   - Velocity \( v = 4 \) m/s
   - Spring constant \( k = 200 \) N/m

   \[
   x = \sqrt{\frac{3 \times 3 \times 4^2}{2 \times 200}} = \sqrt{\frac{144}{400}} = \sqrt{0.36} \approx 0.6 \text{ meters}
   \]

### Final Answer:
The maximum compression in the spring is approximately **0.6 meters**.