Question

A stone is suspended from the free end of a wire that is wrapped around the outer rim of a pulley, as shown in the figure (see the figure (Figure 1) ). The pulley is a uniform disk with mass 10.0kg and radius 47.0cm and turns on frictionless bearings. You measure that the stone travels a distance 12.3m during a time interval of 3.20s starting from rest.

A.

Find the mass of the stone.

Take the free fall acceleration to be $9.80\;{\rm m/s^{2}}$ .

B.

Find the tension in the wire.

Take the free fall acceleration to be $9.80\;{\rm m/s^{2}}$ .

Answer 1

Concepts and reason

The concepts used to solve this problem are kinematic equation of motion, torque, andtension in the wire.

Initially, use the kinematic equation which relates the distance and acceleration to find the acceleration of the stone.

Then, use the torque on the pulley to find the tension in the wire.

Finally, use the net forces on the stone to find the mass of the stone.

Fundamentals

The relationship between the distance and accelerationis as follows:

$s=ut+at $

Here, the initial velocity is , the displacement is s, the time is , and the acceleration is .

Newton’s second law states that the external applied force on an object is equal to the rate of change of momentum of the object.

The expression for the force from Newton’s second lawis as follows:

$F=ma $

Here, the force is and the mass is .

The expression for the torque on the pulley in terms of tension and distance is as follows:

$T=TR $

Here, the torque is , the tension is , and the distance is .

The expression for the torque on the pulley in terms of moment of inertia is as follows:

$t=la $

Here, the moment of inertia is and the angular acceleration is .

The expression for the moment of inertia of the disk is as follows:

Here, the mass of the disk is and the radius is .

The expression for the angular acceleration is as follows:

(A)

The relationship between the distance and acceleration is as follows:

$s=ut+at $

Here, the initial velocity is zero. Then, the above equation will become as follows:

Rearrange the above expression for .

Substitute for and for .

The expression for the torque on the pulley in terms of tension and distance is as follows:

$T=TR $ …… (1)

The expression for the torque on the pulley in terms of moment of inertia is as follows:

$t=la $ …… (2)

CompareEquations(1) and (2).

$la=TR $ …… (3)

The expression for the moment of inertia of the disk is as follows:

The expression for the angular acceleration is as follows:

Substitute for and for in Equation (3).

Rearrange the above equation for .

Substitute for and for .

The expression for net force on the stone is as follows:

$F=mg-T $ …… (4)

Here, the acceleration due to gravity is .

The expression for the force from Newton’s second law is as follows:

$F=ma $ …… (5)

Compare expressions (4) and (5).

Rearrange the above expression for .

Substitute for , for , and for .

(B)

The expression for the tension in the wire is as follows:

Substitute for and for .

Therefore, the tension in the wire is .

Ans: Part A

Thus, the mass of the stone is .

Part B

Thus, the tension in the wire is .