A 4 kg block resting on a frictionless horizontal ledge is attached to a string that passes over a pulley and is attached to a hanging 2 kg block. The pulley is a uniform disk of radius 9 cm and mass 0.2 kg.
a. Find the speed of the 2 kg block after it falls from rest a distance of 2.5 m.
b. What is the angular velocity of the pulley at this time?
To solve these problems, we can apply principles of rotational and translational motion, as well as conservation of energy.
a. To find the speed of the 2 kg block after falling a distance of 2.5 m, we can use energy conservation. The potential energy of the block is converted into kinetic energy as it falls.
1. First, calculate the potential energy of the block:
Potential energy = mass × gravity × height
Potential energy = 2 kg × 9.8 m/s^2 × 2.5 m = 49 J
2. Next, find the total mechanical energy of the system. Since there is no friction and the pulley is considered an ideal object, the total mechanical energy is conserved.
Total mechanical energy = Potential energy
Total mechanical energy = 49 J
3. Determine the mechanical energy of the 2 kg block. As the block falls, it loses potential energy and gains kinetic energy.
Kinetic energy = 1/2 × mass × velocity^2
Initial kinetic energy = 0 J (since it starts from rest)
Final kinetic energy = 49 J
4. Set the final kinetic energy equal to the initial kinetic energy:
1/2 × 2 kg × velocity^2 = 49 J
5. Solve for the velocity:
velocity^2 = 49 J × (2/2 kg)^-1
velocity^2 = 49 J
velocity = sqrt(49 J) = 7 m/s
Therefore, the speed of the 2 kg block after falling a distance of 2.5 m is 7 m/s.
b. To find the angular velocity of the pulley at this time, we can use the principle of conservation of angular momentum.
1. The angular momentum of the system is conserved. Initially, the pulley is at rest, and when the two blocks start moving, the pulley acquires angular velocity.
2. The formula for angular momentum is:
Angular momentum = Inertia × angular velocity
Inertia of a uniform disk = (1/2) × mass × radius^2
Inertia = (1/2) × 0.2 kg × (0.09 m)^2 = 0.00081 kg·m^2 (rounded to three decimal places)
3. Initially, the angular momentum is zero, as the pulley is at rest:
Initial angular momentum = 0 kg·m^2/s
4. Finally, the angular momentum after the blocks start moving is equal to the sum of the individual angular momenta of the two objects (assuming the string is wrapped around the pulley with no slippage):
Angular momentum = (Inertia of the pulley) × (angular velocity of the pulley) + (mass of the block) × (velocity of the block) × (radius of the pulley)
We know the angular velocity of the pulley is what we are trying to find, and we can substitute the values we know:
7 m/s = (0.00081 kg·m^2) × (angular velocity) + (2 kg) × (7 m/s) × (0.09 m)
5. Rearranging the equation and solving for the angular velocity:
(0.00081 kg·m^2) × (angular velocity) = 7 m/s - (2 kg) × (7 m/s) × (0.09 m)
(0.00081 kg·m^2) × (angular velocity) = 7 m/s - 0.882 kg·m^2/s
(angular velocity) = (7 m/s - 0.882 kg·m^2/s) / (0.00081 kg·m^2)
angular velocity ≈ 8135.8 rad/s
Therefore, the approximate angular velocity of the pulley when the 2 kg block falls a distance of 2.5 m is 8135.8 rad/s.