Water is drawn from a well in a bucket tied to the end of a rope whose other end wraps around a solid cylinder of mass 50 kg and diameter 25 cm. As this cylinder is turned with a crank, the rope raises the bucket. The mass of a bucket of water is 20 kg. Someone cranks the bucket up, lets go of the crank, and the bucket of water falls down to the bottom of the well. Without friction or air resistance, what is the angular acceleration of the 50-kg cylinder? A) 1.1  10² rad/s² B) 3.6 rad/s² C) 35 rad/s² D) 63 rad/s² E) 17 rad/s²

Question

Quizplus · Accepted Answer

The force causing the rotation of the cylinder is the tension in the rope. In the initial state, the tension in the rope is equal to the weight of the bucket and water:
T = (20 kg)(9.8 m/s^2) = 196 N

When the bucket is released and falls, the tension in the rope drops suddenly to zero. This causes an impulse in the rope that is transmitted to the cylinder. The impulse is equal to the change in momentum of the bucket and water:

J = Δp = (20 kg)(-9.8 m/s^2)(2 m) = -392 Ns

(The negative sign indicates that the change in momentum is in the opposite direction to the initial momentum.)

The impulse is also equal to the integral of the torque over the time it takes for the bucket to fall. Assuming the cylinder starts from rest, we can use the kinematic equation for rotational motion:

θ = 1/2 α t^2

where θ is the angular displacement, α is the angular acceleration, and t is the time. The angular displacement of the cylinder is equal to the length of rope that unwinds from the cylinder, which is equal to the distance the bucket falls:

θ = 2 m/R

where R is the radius of the cylinder (0.125 m). Substituting this into the kinematic equation and taking the derivative with respect to time gives:

dθ/dt = α t

Solving for α gives:

α = 2(dθ/dt)/t^2 = 4(dθ/dt)/(2θ/R)^2 = 16R^2(dθ/dt)/θ^2

We need to find dθ/dt, which is the angular velocity of the cylinder at the moment the bucket hits the bottom of the well. From conservation of energy, we can equate the initial gravitational potential energy of the bucket to the final kinetic energy of the bucket and cylinder:

mgh = (1/2)mb^2ω^2 + (1/2)Iω^2

where m is the mass of the bucket and water, h is the height of the well, b is the distance from the center of the cylinder to the end of the rope, ω is the final angular velocity of the cylinder and bucket, and I is the moment of inertia of the cylinder. Substituting in the given values and solving for ω gives:

ω = sqrt(2gh/(b^2 + I/m)) = sqrt(2(9.8 m/s^2)(20 m)/(0.125 m)^2 + 50 kg) = 7.8 rad/s

Plugging this into the equation for α gives:

α = 16(0.125 m)^2(7.8 rad/s)/(2 m)^2 = 35 rad/s^2

Therefore, the angular acceleration of the cylinder is 35 rad/s^2.

Water Is Drawn from a Well in a Bucket Tied