Torque in Circular Motion and Rotational Systems

Expert reviewed • 22 November 2024 • 10 minute read

What is Torque?

Torque is a fundamental concept in physics that plays a crucial role in understanding rotational motion and equilibrium. It is particularly important in the study of circular motion and mechanical systems.

Torque, often described as the rotational equivalent of force, is a measure of the tendency of a force to rotate an object around an axis or pivot point. It is a vector quantity, meaning it has both magnitude and direction.

The magnitude of torque is given by the equation:

\tau = rFsin θ

Where:

$\tau$ is the torque
$r$ is the distance from the axis of rotation to the point where the force is applied (lever arm)
$F$ is the magnitude of the applied force
$\theta$ is the angle between the force vector and the lever arm

When the force is perpendicular to the lever arm $(\theta=90\degree)$ , the equation simplifies to:

τ = r⊥F

Where:

r⊥ is the perpendicular distance from the axis of rotation to the line of action of the force.

The direction of torque follows the right-hand rule: If you curl the fingers of your right hand in the direction of rotation, your thumb points in the direction of the torque vector.

How is Torque Related to Circular Motion?

In the context of circular motion, torque is closely related to the centripetal force that keeps an object moving in a circular path. However, it’s important to note that torque and centripetal force serve different purposes:

Centripetal force causes circular motion by constantly changing the direction of velocity.
Torque, when applied tangentially to the circular path, can change the angular velocity of the object.

Practice Question 1

Consider a bicycle wheel with a radius of 0.3 m. A tangential force of 20 N is applied to the rim of the wheel. Calculate the torque applied to the wheel.

Since the force is applied tangentially, sin θ = 1. Therefore:

τ = rF \\= 0.3 × 20\\ = 6 \:\text{Nm}

The torque applied to the wheel is 6 Nm

What is the Relationship Between Work and Energy in Circular Motion?

The relationship between torque and work in circular motion is crucial for understanding energy transfer in rotational systems.

Work Done by Torque

The work done by a torque in rotating an object through an angle θ is given by:

W = τθ

Where:

$W$ is the work done
$τ$ is the torque
$\theta$ is the angle of rotation in radians

This equation is analogous to the linear work equation $W = Fd$ , where force and displacement are replaced by torque and angular displacement, respectively.

Uniform Circular Motion and Work

In uniform circular motion, where the speed remains constant, the net work done on the object is zero. This is because:

The centripetal force acts perpendicular to the displacement at all times.
Work is calculated as $W = Fdcos θ$ , where $\theta$ is the angle between the force and displacement vectors.
In uniform circular motion, $\theta=90\degree$ , so $cos90\degree=0$ , resulting in zero work done by the centripetal force.

This explains why the kinetic energy of an object in uniform circular motion remains constant:

KE = \frac{1}{2}mv^2

Where:

$KE$ is kinetic energy
$m$ is the mass of the object
$v$ is the velocity

Practice Question 2

A merry-go-round with a radius of 2 m is initially at rest. A constant tangential force of 50 N is applied to its edge, causing it to complete one full rotation. Calculate the work done by this force.

First, we calculate the torque:

τ = rF \\= 2 × 50 \\= 100 \:\text{Nm}

The angle for one full rotation is $2\pi$ radians. Therefore, the work done is:

W = τθ \\= 100 × 2π\\ = 628.3 J

The work done by the tangential force is 628.3 J.

Torque and Rotational Equilibrium

Torque plays a crucial role in determining rotational equilibrium. An object is in rotational equilibrium when the net torque acting on it is zero. This concept is essential in engineering and practical applications, such as balancing structures or designing mechanical systems.

Practice Question 3

A see-saw has a length of 4 m and pivots at its centre. A child weighing 300 N sits 1.5 m from the pivot on one side. Where should a 400 N adult sit on the other side to balance the see-saw?

For rotational equilibrium, the net torque must be zero. Let’s choose counterclockwise as positive:

τ_{child} + τ_{adult} = 0\\(300 ×1.5 ) + (400 ×x) = 0\\450  + 400x  = 0\\400x =− 450\\x =− 1.125 m

The negative sign indicates that the adult should sit $1.125 m$ from the pivot on the opposite side of the child.

Quantitative Analysis of Uniform Circular Motion

Return to Module 5: Advanced Mechanics