A bucket of water with total mass 21.0 kg is attached to a rope,which in turn is wound around a 0.050-m radius cylinder,with crank,at the top of a well.The moment of inertia of the cylinder and crank is 0.21 kg
M².The bucket is raised to the top of the well and released to fall back into the well.What is the kinetic energy of the cylinder and crank at the instant the bucket is moving with a speed of 5.4 m/s?

Question

A bucket of water with total mass 21.0 kg is attached to a rope,which in turn is wound around a 0.050-m radius cylinder,with crank,at the top of a well.The moment of inertia of the cylinder and crank is 0.21 kg

M².The bucket is raised to the top of the well and released to fall back into the well.What is the kinetic energy of the cylinder and crank at the instant the bucket is moving with a speed of 5.4 m/s?

Quizplus · Accepted Answer

The potential energy of the system (bucket + cylinder + crank) when the bucket is at the top of the well is given by:
PE = mgh
where m is the mass of the bucket, cylinder, and crank (21.0 kg), g is the gravitational acceleration (9.81 m/s^2), and h is the height of the well (the length of the rope), which we assume to be 10.0 m (although the value doesn't matter in this problem since it's canceled out). Thus:
PE = (21.0 kg)(9.81 m/s^2)(10.0 m) = 2058.6 J

At the instant the bucket is moving with a speed of 5.4 m/s, all of the potential energy has been converted into kinetic energy of the bucket, cylinder, and crank. Additionally, the rotational kinetic energy of the cylinder and crank should be taken into account, which is given by:
KE_rot = (1/2)Iω^2
where I is the moment of inertia of the cylinder and crank (0.21 kg m^2), and ω is the angular velocity of the cylinder and crank. Since the rope is wound around the cylinder, the angular velocity of the cylinder is related to the linear velocity of the bucket by:
v = rω
where v is the linear velocity of the bucket (5.4 m/s) and r is the radius of the cylinder (0.050 m). Thus:
ω = v/r = (5.4 m/s)/(0.050 m) = 108 rad/s

Substituting the values into the equation for KE_rot:
KE_rot = (1/2)(0.21 kg m^2)(108 rad/s)^2 = 1219.8 J

The total kinetic energy of the system is then:
KE_total = KE_linear + KE_rot
where KE_linear is the kinetic energy of the bucket due to its linear motion, and is given by:
KE_linear = (1/2)mv^2
where m is the mass of the bucket (21.0 kg), and v is the linear velocity of the bucket (5.4 m/s). Thus:
KE_linear = (1/2)(21.0 kg)(5.4 m/s)^2 = 307.4 J

Adding the values for KE_linear and KE_rot:
KE_total = 307.4 J + 1219.8 J = 1527.2 J ≈ 1.2E+3 J. Therefore, the answer is B.

A Bucket of Water with Total Mass 21