A bucket hangs from a rope that is attached to a cylindrical, massive pulley wit
ID: 1342001 • Letter: A
Question
A bucket hangs from a rope that is attached to a cylindrical, massive pulley with a radius of m and a mass of 2.22 kg. The pulley spins as the bucket falls, undergoing an angular The bucket starts at rest and falls a distance 5.51 m before hitting the water. During this time the pulley undergoes an angular acceleration of 3.19 rad/s2, which means the bucket is falling with a linear acceleration of 1.72 m/s2 (a=r?).
1.) What is the moment of inertia of the pulley? (remember, this is a solid cylinder.)
2.)What is the magnitude of the torque that the rope exerts on the pulley? (Think about the rotational version of Newton's Second Law.)
3.)Since you know the torque, what is the force that the rope is exerting on the pulley?
4.) How much time elapses between when the bucket is released and when it hits the water?
Explanation / Answer
Assuming that the by "frictionless" the question means no friction only on the axle of the cylinder and present elsewhere.
We will right the linear and angular motions.
For the bucket,
Mg - T = Ma
(1.5)(9.8) - T = 1.5(a) ...(1)
For pulley,
T x R = I = (MR²/2) ...(2)
a = R ...(3)
This relation comes from the fact that as the pulley rotates, it loosens rope and rope is used by the bucket to go down. Its called string constraint.
Substituting R in (2)
TR = (MR/2)a
T = Ma/2
Substitue T in (1)
Mg - Ma/2 = Ma
Mg = 3Ma/2
a = 2g/3 = 6.533 m/s²
= a/R = 6.533 / 0.2 = 32.66 rad / sec²
T = Ma/2 = (1.5)(6.533)/2 = 4.9 N
Torque = T x R = 4.9 x 0.2 = 0.98 N-m
I = Torque/ = 0.98 / 32.66 = 0.03 kg m²
M = 2I/r² = 2 x 0.03 / 0.04 = 1.5 kg
Change in potential energy = Mgh where h is descended height
= (1.5)(9.8)(8.19) = 120.4 J
v = u + at = 0 + (6.533)3 = 19.6 m/sec
Just as a = R, v = R.
So we can either find the velocity of the bucket when it hits the water and use this relation or directly calculate using angular acceleration.
I am gonna use the former one.
v = 19.6
= v/r = 19.6 / 0.2 = 98 rad/sec
Angular momentum = I = 0.03 x 98 = 2.93 kg m^2 / sec
Kinetic energy = (1/2)mv² = (1/2)(1.5)(19.6)² = 288.12 J
Rotationl Kinetic energy = (1/2)I² = 144.06 J
Total = 288.12 + 144.06 = 432.18 J
The total kinetic energy is coming out to be greater than potential energy. (I have no idea why this is coming. It should have been equal)
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