Question

A bucket of water of mass 14.0 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.260 m with mass 12.1 kg . The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls a distance 10.7 m to the water. You can ignore the weight of the rope.

Answer 1

The tension in the rope is 41.38 N.

Step-by-step explanation:

Given that,

Mass of bucket of water = 14.0 kg

Diameter of cylinder = 0.260 m

Mass of cylinder = 12.1 kg

Distance = 10.7 m

Suppose we need to find that,

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

We need to calculate the acceleration

Using relation of torque

`\tau=F* r`

`I*\alpha=F* r`

Where, I = moment of inertia

`\alpha` = angular acceleration

`(Mr^2)/(2)*(a)/(r)=F* r`

`F=(M)/(2)a`...(I)

Here, F = tension

The force is

`F=m(g-a)`...(II)

Where, F = tension

a = acceleration

From equation (I) and (II)

`(M)/(2)a=m(g-a)`

`a=(g)/(1+(M)/(2m))`

Put the value into the formula

`a=(9.8)/(1+(12.1)/(2*14.0))`

`a=6.84\ m/s^2`

We need to calculate the tension in the rope

Using equation (I)

`F=(M)/(2)a`

Put the value into the formula

`F=(12.1)/(2)*6.84`

`F=41.38\ N`

Hence, The tension in the rope is 41.38 N.