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

Answer:T=65.33 N

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

$T\times r=I\times \alpha$

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

$T\times \frac{d}{2}=Mr^2\times \alpha$

$T=\frac{Mr\alpha }{2}$

Substitute the value of T in $(i)$

$mg-\frac{Mr\alpha }{2}=ma\quad \text{[Pure rolling}\ a=\alpha r]$

$mg-\frac{Ma}{2}=ma$

$mg=(\frac{M}{2}+m)a$

$a=\frac{20}{10+20}\times g$

$a=\frac{2}{3}\times g$

Substitute the value of a in Tension equation

$T=\frac{Ma}{2}$

$T=10\times \frac{2}{3}\times g$

$T=65.33\ N$