A bucket of water with total mass 23 kg is attached to a rope,which in turn is wound around a 0.050-m radius cylinder at the top of a well.A crank with a turning radius of 0.25 m is attached to the end of the cylinder and the moment of inertia of cylinder and crank is 0.12 kg*m².If the bucket is raised to the top of the well and released,what is the acceleration of the bucket as it falls toward the bottom of the well? (Assume rope's mass is negligible,that cylinder turns on frictionless bearings and that g = 9.8 m/s². ) A)3.2 m/s² B)6.3 m/s² C)7.4 m/s² D)9.8 m/s²

Question

Quizplus · Accepted Answer

The gravitational force acting on the bucket is F = mg, where m is the mass of the bucket and g is the acceleration due to gravity. At the beginning, the bucket is at rest, so the net force on it is zero. As the bucket falls, the tension in the rope decreases, causing the force on the bucket to decrease. At the bottom of the well, the tension in the rope is zero, and the net force on the bucket is F = mg.

To find the acceleration of the bucket, we need to use Newton's second law, which says that the net force on an object is equal to its mass times its acceleration: F = ma. In this case, the net force on the bucket is mg, and its mass is 23 kg (the mass of the water plus the bucket). The moment of inertia of the cylinder and crank is given as 0.12 kgm^2.

As the bucket falls, the cylinder and crank rotate, which means that work is being done on them. The work done is equal to the change in kinetic energy of the system. The potential energy of the bucket at the top of the well is mgh, where h is the height of the well. When the bucket reaches the bottom of the well, its potential energy is zero and its kinetic energy is (1/2)mv^2, where v is its velocity. The work done on the cylinder and crank is equal to the change in rotational kinetic energy, which is (1/2)Iω^2, where I is the moment of inertia and ω is the angular velocity.

Since the system is frictionless, the total work done is conserved, which means that the work done on the bucket is equal to the work done on the cylinder and crank. Therefore, we have:

mgh = (1/2)mv^2 + (1/2)Iω^2

Substituting the values given in the problem, we get:

(23 kg)(9.8 m/s^2)(0.050 m) = (1/2)(23 kg)v^2 + (1/2)(0.12 kg m^2)ω^2

Solving for v, we get:

v = sqrt[(23 kg)(9.8 m/s^2)(0.050 m)/(1/2)(23 kg)] = 3.2 m/s

Therefore, the acceleration of the bucket as it falls toward the bottom of the well is equal to the acceleration due to gravity, which is 9.8 m/s^2, and the correct answer is A.

A Bucket of Water with Total Mass 23 Kg Is