Work and power in rotating systems Revision Notes for AQA A-Level Physics

Work and power in rotating systems

Work done by a torque

When a force causes a rigid body to rotate about an axis, work is done on the system. To understand this, consider a force F applied at a perpendicular distance r from the axis of rotation. This force creates a constant torque given by $T = Fr$ .

As the rigid body rotates through an angle θ, the point where the force is applied moves along a circular arc. The length of this arc is $s = r\theta$ . Since work done equals force multiplied by distance moved in the direction of the force, we can write:

$W = F \times r\theta$

Rearranging this expression:

$W = (Fr)\theta = T\theta$

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Fundamental Equation for Work in Rotating Systems

This gives us the key relationship between work, torque, and angular displacement:

$W = T\theta$

Where:

W is the work done, measured in joules (J)
T is the constant torque applied, measured in newton metres (Nm)
θ is the angular displacement, measured in radians (rad)

This equation is the rotational equivalent of the linear work equation $W = Fs$ , where torque replaces force and angular displacement replaces linear displacement.

The role of frictional torque

When objects rotate, work done can increase their rotational kinetic energy. However, real rotating systems must also overcome resistive forces. Friction inevitably occurs when surfaces are in contact, even with measures to minimize it such as lubrication or ball bearings.

In a rotating system, such as a flywheel on an axle, the frictional force acts at the edge of the rotating object. Since this force acts at a distance from the axis, it creates a torque about the axis, which we call frictional torque.

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Two Types of Frictional Torque

Frictional torque can act in two ways:

Resistive frictional torque: This opposes the rotation and reduces the rotational kinetic energy of the system. For example, friction in the bearings of a spinning wheel gradually slows it down.
Applied frictional torque: This is when friction is deliberately used to apply torque, such as when gripping and turning a screwdriver. In this case, the rotational kinetic energy increases.

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Key Principle

Whenever a torque causes an object to rotate through an angle, work is done, regardless of whether the torque increases or decreases the rotational motion.

Power in rotating systems

Power is defined as the rate at which work is done. For a rotating system, we can express this mathematically by considering how work changes with time.

Starting with the definition of power:

$P = \frac{\Delta W}{\Delta t}$

Since $W = T\theta$ for rotational systems, we can write:

$P = \frac{\Delta(T\theta)}{\Delta t} = T\left(\frac{\Delta\theta}{\Delta t}\right)$

The term $\Delta\theta/\Delta t$ represents the rate of change of angular displacement with time, which is the angular velocity, ω. Therefore:

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Power in Rotating Systems

$P = T\omega$

Where:

P is the power, measured in watts (W)
T is the torque, measured in newton metres (Nm)
ω is the angular velocity, measured in radians per second ( $\text{rad} \cdot \text{s}^{-1}$ )

This equation is the rotational equivalent of the linear power equation $P = Fv$ . It allows us to calculate the power required to overcome resistive torques while maintaining a constant angular velocity, or the power needed to change the angular velocity of a rotating object.

Practice calculations

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Practice Question 26

Calculate the power needed to maintain an angular velocity of 15 $\text{rad} \cdot \text{s}^{-1}$ in a disc where the frictional torque is 25 Nm.

Solution:

Using $P = T\omega$ : `$$P = 25 \times 15 = 375 \text{ W}$$`

Answer: 375 W

lightbulbExample

Practice Question 27

A braking force of 200 N is applied to the edge of a disc with diameter 20 cm. Calculate the braking torque. The power supplied by a motor keeps the disc rotating at 45 $\text{rad} \cdot \text{s}^{-1}$ . Calculate the power delivered by the motor.

Solution:

Step 1: Find the braking torque:

Radius $r = \frac{20 \text{ cm}}{2} = 10 \text{ cm} = 0.10 \text{ m}$
Torque $T = Fr = 200 \times 0.10 = 20 \text{ Nm}$

Step 2: Calculate the power: ` $P = T\omega = 20 \times 45 = 900 \text{ W}$

Answer: Braking torque = 20 Nm, Power = 900 W

lightbulbExample

Practice Question 28: Potter's Wheel Problem

A potter's wheel system has these specifications:

Potter's wheel diameter: 70.0 cm, mass: 40.0 kg
Clay disc diameter: 20.0 cm, mass: 2.40 kg
Initial rotation rate: 3.50 $\text{rev} \cdot \text{s}^{-1}$
Applied frictional force: 20 N for 10 s on the clay edge

(a) Calculate the torque applied:

Solution:

Radius of clay disc $= \frac{20.0 \text{ cm}}{2} = 10.0 \text{ cm} = 0.100 \text{ m}$
Torque $T = Fr = 20 \times 0.100 = 2.0 \text{ Nm}$

Answer: 2.0 Nm

(b) Calculate the angular velocity after 10 s:

This requires calculating the moment of inertia of the system and using angular impulse concepts (beyond the scope of work and power alone).

(c) Calculate the work done:

This requires knowing the final angular velocity from part (b), then using $W = T\theta$ or energy considerations.

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Key Points to Remember:

Work done by a torque is given by $W = T\theta$ , where T is the torque in Nm and θ is the angular displacement in radians.
Frictional torque opposes rotation in real machines, reducing rotational kinetic energy by converting it to thermal energy.
Power in rotating systems is given by $P = T\omega$ , which is the rotational equivalent of $P = Fv$ for linear motion.
To maintain constant angular velocity against friction, the applied power must equal the rate at which work is done against the frictional torque.

Work and power in rotating systems (AQA A-Level Physics): Revision Notes