You have an inclined plane. It is 2m long and 1m high. A cylinder of mass 300 grams and 3cm radius is
initially at rest at the top of the hill. What is its angular velocity when it reaches the bottom of the hill?
To find the angular velocity of the cylinder when it reaches the bottom of the inclined plane, we can use the principles of rotational motion and conservation of energy.
Here are the steps to find the angular velocity:
1. Convert the given values to SI units:
- Convert the mass of the cylinder from grams to kilograms: 300 grams = 0.3 kg
- Convert the radius of the cylinder from centimeters to meters: 3 cm = 0.03 m
2. Calculate the potential energy at the top of the hill:
- Potential energy (PE) = mass × gravitational acceleration × height
- PE = 0.3 kg × 9.8 m/s^2 × 1 m
- PE = 2.94 J
3. Calculate the final kinetic energy at the bottom of the hill:
- Since the cylinder is initially at rest, its initial kinetic energy is zero.
- At the bottom of the hill, the potential energy is fully converted to kinetic energy.
- Kinetic energy (KE) = Potential energy
- KE = 2.94 J
4. Calculate the moment of inertia of the cylinder:
- Moment of inertia (I) for a solid cylinder = (1/2) × mass × radius^2
- I = (1/2) × 0.3 kg × (0.03 m)^2
- I = 1.35 × 10^-4 kg·m^2
5. Apply the conservation of energy principle:
- The total mechanical energy of the system (potential energy + kinetic energy) is conserved.
- PE + KE = (1/2) × I × angular velocity^2
- 2.94 J + 2.94 J = (1/2) × (1.35 × 10^-4 kg·m^2) × angular velocity^2
- 5.88 J = (6.75 × 10^-5) × angular velocity^2
6. Solve for the angular velocity:
- angular velocity^2 = (5.88 J) / (6.75 × 10^-5 kg·m^2)
- angular velocity^2 ≈ 87074.07 rad^2/s^2
- angular velocity ≈ √(87074.07 rad^2/s^2)
- angular velocity ≈ 295.15 rad/s
Therefore, the angular velocity of the cylinder when it reaches the bottom of the hill is approximately 295.15 rad/s.