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.

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Answer:

The tension in the rope is 41.38 N.

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

[tex]\tau=F\times r[/tex]

[tex]I\times\alpha=F\times r[/tex]

Where, I = moment of inertia

[tex]\alpha[/tex] = angular acceleration

[tex]\dfrac{Mr^2}{2}\times\dfrac{a}{r}=F\times r[/tex]

[tex]F=\dfrac{M}{2}a[/tex]...(I)

Here, F = tension

The force is

[tex]F=m(g-a)[/tex]...(II)

Where, F = tension

a = acceleration

From equation (I) and (II)

[tex]\dfrac{M}{2}a=m(g-a)[/tex]

[tex]a=\dfrac{g}{1+\dfrac{M}{2m}}[/tex]

Put the value into the formula

[tex]a=\dfrac{9.8}{1+\dfrac{12.1}{2\times14.0}}[/tex]

[tex]a=6.84\ m/s^2[/tex]

We need to calculate the tension in the rope

Using equation (I)

[tex]F=\dfrac{M}{2}a[/tex]

Put the value into the formula

[tex]F=\dfrac{12.1}{2}\times6.84[/tex]

[tex]F=41.38\ N[/tex]

Hence, The tension in the rope is 41.38 N.