Rotational Work & Power (AQA A Level Physics)

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Work Done & Torque

Work Done by a Rotating Object

  • Work has to be done on a rigid body when a torque turns in through an angle about an axis

    • For example, rotating cranes and fairground rides

  • In systems with linear acceleration, work W is the product of the force and the distance moved

  • Therefore, the work done for a rotating object is defined by the equation

W space equals space tau theta

  • Where:

    • W = work done (J)

    • tau = torque (N m)

    • θ = angular displacement (the angle turned through by the rotating object) (rads)

  • Work can also be calculated by finding the area under a torque-angular displacement graph

Torque-angular displacement graph

11-1-8-torque-displacement-graph-1

The work done is the area under the torque-angular displacement graph

Power Output of a Rotating Object

  • Power is the rate of doing work, and is defined by

P space equals space fraction numerator straight capital delta W space over denominator straight capital delta t end fraction space equals space fraction numerator straight capital delta straight tau straight theta space over denominator straight capital delta t end fraction space equals space tau fraction numerator straight capital delta straight theta space over denominator straight capital delta t end fraction

P space equals space tau omega

  • Where:

    • P = power (W)

    • ω = angular velocity (rad s–1)

  • This equation is the angular version of the linear equation PFv

Examiner Tips and Tricks

Don't forget that θ is always in radians when you're doing conversions from revs s–1 or rev min–1.

Frictional Torque

  • In rotational mechanics, frictional forces produce a specific torque called frictional torque

    • This is the torque caused by the frictional force when two objects in contact move past each other

  • Frictional torque can be defined as:

    The difference between the applied torque and the resulting net, or observed, torque

  • This means that the net torque is the sum of the applied and frictional torque

    Net torque = applied torque + frictional torque

  • In rotating machinery, power has to be expended to overcome frictional torque

    • This is due to resistive forces within the machinery

  • In most cases, frictional torque is minimised to reduce the kinetic energy losses transferred to heat and sound

  • However, sometimes a frictional torque is applied, such as when using a screwdriver

    • When a screwdriver is gripped and turned, this increases its rotational kinetic energy

  • The frictional force must always be added to the torque resulting from a force to get the total torque in the system 

  • Frictional torque is calculated using the same equations as torque

tau space equals space F r space equals space I alpha

  • The only difference is F is the frictional force instead of an externally applied force

Worked Example

The figure below shows a type of circular saw. The blade is driven by an electric motor and rotates at 3100 rev min–1 when cutting a piece of wood.

A constant frictional torque of 2.7 N m acts at the bearings of the motor and axle. 

11-1-8-rotational-power

A horizontal force of 45 N is needed to push a piece of wood into the saw. The force acts on the blade at an effective radius of 22 cm.

Calculate the output power of the motor when the saw is cutting the wood. 

Answer:

Step 1: Calculate the torque on the saw blade

tau space equals space F r space equals space 45 space cross times space 0.22 space equals space 9.9 space straight N space straight m 

Step 2: Calculate the total torque

Total torque = torque on the saw blade + frictional torque

Total torque = 9.9 + 2.7 = 12.6 N m

Step 3: Calculate the angular velocity

1 revolution = 2π radians

3100 rev min–1 = 3100 × 2π

min–1 → sec–1  = ÷ 60

3100 space rev space min to the power of negative 1 end exponent space cross times space fraction numerator 2 straight pi over denominator 60 end fraction space equals space 324.63 space rad space straight s to the power of negative 1 end exponent space

Step 4: Calculate the output power

P space equals space tau omega
P space equals space 12.6 space cross times space 324.63 space equals space 4090.338 space equals space 4100 space straight W

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Ashika

Author: Ashika

Expertise: Physics Project Lead

Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.