A hollow cylinder of mass 5 kg rolls down with a speed of 20 m/s without slipping. What will be its kinetic energy?

Answer

To calculate the kinetic energy of a hollow cylinder rolling down without slipping, we need to consider both its translational and rotational kinetic energy.

### Step-by-Step Solution:

1. **Identify the mass and speed of the cylinder:**
- Mass (m) = 5 kg
- Speed (v) = 20 m/s

2. **Calculate the translational kinetic energy (TKE):**
- The formula for translational kinetic energy is given by:
\[ \text{TKE} = \frac{1}{2} mv^2 \]
- Substitute the values:
\[ \text{TKE} = \frac{1}{2} \times 5 \, \text{kg} \times (20 \, \text{m/s})^2 \]
- Calculate \( (20 \, \text{m/s})^2 = 400 \, \text{m}^2/\text{s}^2 \):
\[ \text{TKE} = \frac{1}{2} \times 5 \times 400 = \frac{2000}{2} = 1000 \, \text{J} \]

3. **Calculate the rotational kinetic energy (RKE):**
- For a hollow cylinder, the moment of inertia (I) about its axis is given by:
\[ I = m r^2 \]
- The relationship between linear velocity (v) and angular velocity (ω) is:
\[ v = \omega r \quad \Rightarrow \quad \omega = \frac{v}{r} \]
- The formula for rotational kinetic energy is:
\[ \text{RKE} = \frac{1}{2} I \omega^2 \]
- Substitute \( I \) and \( \omega \):
\[ \text{RKE} = \frac{1}{2} (m r^2) \left(\frac{v}{r}\right)^2 = \frac{1}{2} m r^2 \frac{v^2}{r^2} = \frac{1}{2} mv^2 \]
- Thus, the rotational kinetic energy is the same as the translational kinetic energy for a hollow cylinder:
\[ \text{RKE} = 1000 \, \text{J} \]

4. **Calculate the total kinetic energy (TKE + RKE):**
- Total Kinetic Energy (KE) is the sum of translational and rotational kinetic energies:
\[ \text{KE} = \text{TKE} + \text{RKE} = 1000 \, \text{J} + 1000 \, \text{J} = 2000 \, \text{J} \]

### Final Answer:
The total kinetic energy of the hollow cylinder is **2000 J**.