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.

Answer 2

The tension force is directed over the length of the rope and pulls energy equally on the bucket and cylinder at the ends.

The tension force in the rope is 49.66 N.

What is tension force?

The tension force is defined as the force generated when a load is applied at one or more ends of a rope or wire.

Given that the mass m of the bucket is 23.0 kg. The diameter d of the cylinder is 0.300 m and mass m' is 13 kg. The height h from where the bucket falls is 10.5 m.

If the acceleration is a and gravitational acceleration is g on the bucket then the tension force T in the rope of the bucket is given below.

$mg - T = ma$.....................equation 1.

The tension in the rope due to cylinder is given as,

$T = \dfrac {Ia}{r^2}$

Where I is the moment of the inertia which is given as,

$I = \dfrac {m'r^2}{2}$

The tension force is,

$T = \dfrac {m'r^2\times a}{r^2 \times2}$

$T = \dfrac{m'a}{2}$

Substituting the value of T in equation 1,

$mg - \dfrac {m'a}{2} = ma$

$mg = ma + \dfrac {m'a}{2}$

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

$a = \dfrac {23 \times 9.8}{23 + \dfrac {13}{2}}$

$a = 7.64 \;\rm m/s^2$

The tension in the rope is given below.

$T = \dfrac {13\times 7.64}{2}$

$T = 49.66 \;\rm N$

Hence we can conclude that the tension force in the rope is 49.66 N.

To know more about the tension force, follow the link given below.

https://brainly.com/question/2287912.