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docker41 [41]
3 years ago
12

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.
Physics
1 answer:
ruslelena [56]3 years ago
6 0

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

\tau=F\times r

I\times\alpha=F\times r

Where, I = moment of inertia

\alpha = angular acceleration

\dfrac{Mr^2}{2}\times\dfrac{a}{r}=F\times r

F=\dfrac{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)

\dfrac{M}{2}a=m(g-a)

a=\dfrac{g}{1+\dfrac{M}{2m}}

Put the value into the formula

a=\dfrac{9.8}{1+\dfrac{12.1}{2\times14.0}}

a=6.84\ m/s^2

We need to calculate the tension in the rope

Using equation (I)

F=\dfrac{M}{2}a

Put the value into the formula

F=\dfrac{12.1}{2}\times6.84

F=41.38\ N

Hence, The tension in the rope is 41.38 N.

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