Question

A bucket of water of mass 20.0 kg is suspended by a rope wrapped around a windlass in the form of a solid cylinder 0.20 m in diameter, also of mass 20.0 kg. The cylinder is pivoted on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 20.0 m to the water. Neglect the weight of the rope.

What is the tension in the rope while the bucket is falling?

Answer 1

The tension in the rope while the bucket is falling is 200 N.

Step-by-step explanation:

The tension in the rope while the bucket is falling can be found using the principle of conservation of energy. At the top of the well, the bucket has gravitational potential energy. As it falls, this potential energy is converted into kinetic energy. The work done by the tension in the rope is equal to the change in potential energy:

Work done by tension in the rope = Change in potential energy

Tension * Distance = m * g * h

where Tension is the tension in the rope, Distance is the distance the bucket falls, m is the mass of the bucket, g is the acceleration due to gravity, and h is the height of the well. Plugging in the given values:

Tension * 20.0 m = 20.0 kg * 9.8 m/s^2 * 20.0 m

Tension = 200 N

Answer 2

Answer:T=65.33 N

Step-by-step explanation:

Given

mass of bucket with water is
`m=20\ kg`

Diameter of cylinder
`d=0.2\ m`

mass of cylinder
`M=20\ kg`

bucket has fall a distance of
`h=20\ m`

Net force on bucket

`\sum F_(net)=mg-T=ma\quad \ldots (i)`

Consider downward direction to be positive

Tension(T) will provide torque to the cylinder

`T* r=I* \alpha`

where
`\alpha =\text{angular acceleration}`

`T* (d)/(2)=Mr^2* \alpha`

`T=(Mr\alpha )/(2)`

Substitute the value of T in
`(i)`

`mg-(Mr\alpha )/(2)=ma\quad \text{[Pure rolling}\ a=\alpha r]`

`mg-(Ma)/(2)=ma`

`mg=((M)/(2)+m)a`

`a=(20)/(10+20)* g`

`a=(2)/(3)* g`

Substitute the value of a in Tension equation

`T=(Ma)/(2)`

`T=10* (2)/(3)* g`

`T=65.33\ N`