A bucket of water with total mass 26 kg is attached to a rope,which in turn is wound around a 0.04-m radius cylinder at the top of a well.A crank with a turning radius of 0.20 m is attached to the end of the cylinder and the moment of inertia of cylinder and crank is 0.11 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)1.9 m/s² B)2.7 m/s² C)9.8 m/s² D)7.9 m/s² E)7.1 m/s²

Question

Quizplus · Accepted Answer

The gravitational force acting on the bucket is given by Fg = mg, where m is the mass of the bucket and g is the acceleration due to gravity. Since the rope is wound around the cylinder, the tension in the rope also exerts a force on the bucket, given by T = Iα/r, where I is the moment of inertia of the cylinder and crank, α is the angular acceleration, and r is the radius of the cylinder.

When the bucket is released, it begins to fall and the tension in the rope decreases. The net force on the bucket is then given by Fnet = mg - T. Using Newton's second law, Fnet = ma, we can find the acceleration a of the bucket as it falls.

To find T, we need to find α, which we can do using the conservation of energy. Initially, the bucket has no kinetic energy and is at a height h above the bottom of the well. When the bucket is released, it falls and gains kinetic energy, but loses potential energy as it descends. The total energy of the system (the bucket, rope, cylinder, and crank) is conserved, so we can write:

(mgh + Iω^2/2)initial = (mgh/2 + mv^2/2 + Iω^2/2)final

where ω is the angular velocity of the cylinder and crank, v is the velocity of the bucket, and the subscripts "initial" and "final" refer to the initial and final states of the system. Since the cylinder and crank start at rest and have a large moment of inertia compared to the bucket, we can assume that their initial angular velocity is zero. We can also assume that the bucket falls straight down and does not swing back and forth, so its velocity is purely vertical.

Solving for v, we get:

v = sqrt[(2gh - Iω^2/m)]

To find ω, we can use the fact that the linear velocity of the bucket is v = ωr, where r is the radius of the cylinder. This gives us:

ω = v/r = sqrt[(2gh - Iω^2/m)]/r

Solving for ω, we get:

ω = sqrt[(2gh)/(r^2 + I/m)]

Now we can find T using T = Iα/r, where α = a/r is the angular acceleration:

T = I(a/r^2) = I(r^2 + I/m)(a/2gh)

Substituting the values given in the problem, we get:

T = (0.11 kg·m2)(0.04 m + 0.11 kg·m2/26 kg)(a/2(9.8 m/s^2)(26 kg)) = 0.0209a N

The net force on the bucket is then given by:

Fnet = mg - T = (26 kg)(9.8 m/s^2) - 0.0209a N

Setting Fnet equal to ma and solving for a, we get:

a = (26 kg)(9.8 m/s^2)/(26 kg + 0.0209(26 kg)(9.8 m/s^2)) = 2.70 m/s^2

Therefore, the acceleration of the bucket as it falls toward the bottom of the well is 2.7 m/s^2, which corresponds to choice (B).

A Bucket of Water with Total Mass 26 Kg Is