Question

A 23.0 kg bucket of water is suspended by a very light rope wrapped around a solid uniform cylinder 0.300 m in diameter with mass 13.0 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 10.5 m to the water.

Answer 1

Answer:

The tension in the rope is 49.66 N.

Explanation:

Given that,

Mass of bucket m= 23.0 kg

Diameter = 0.300 m

Mass of rope M= 13.0 kg

Height = 10.5 m

Suppose we need to find the tension in the rope while the bucket is falling.

We need to calculate the acceleration

Using balance equation

$mg-T=ma$..(I)

We need to calculate the tension in the rope

Using formula of tension

$Tr=I\alpha$

$Tr=\dfrac{Mr^2}{2}\times\dfrac{a}{r}$

$T=\dfrac{Ma}{2}$....(II)

Put the value of T in the equation (I)

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

$a=\dfrac{mg}{m+\dfrac{M}{2}}$

Put the value into the formula

$a=\dfrac{23.0\times9.8}{23.0+\dfrac{13.0}{2}}$

$a=7.64\ m/s^2$

Now, put the value of a in equation (II)

$T=\dfrac{13.0\times7.64}{2}$

$T=49.66\ N$

Hence, The tension in the rope is 49.66 N.