In this post, we will find the equation of the work done by Torque. (In other words, we will work with the equation of rotational work.) Then we solve a few numerical problems using this formula of rotational work.
Equation of the work done by Torque | equation of rotational work
Here we will find out the equation of the work done by Torque. Let’s consider a scenario where we apply a force of F N to the edge of a tire to get a car moving. Then, what work do we do over s m of travel?
We will use this equation of work done: W = Fs
We can also think about this force rotationally. In the case of you applying force to the edge of a tire to get a car moving, the distance s equals the radius multiplied by the angle through which the wheel turns, s = θ r , so you get this equation:
W = Fs = F θ r
But the torque, τ , equals Fr in this case.
So we can easily write: W = Fs = F θ r = Fr θ = τ θ
work done by Torque = Torque x angular displacement
W = τ θ (equation of rotational work)
Numerical problems on work done by the torque
1 ) If you apply a torque of 500.0 N-m to a tire and turn it through an angle of 2π radians, what work have you done?
Solution:
τ = 500.0 N-m
θ = 2π rad
W = τ θ = 500 x 2π J = 3140 Joule
2 ) How much work do you do if you apply a torque of 6.0 N-m over an angle of 200 radians?
Solution:
τ = 6 N-m
θ = 200 rad
W = τ θ = 6 x 200 J = 1200 Joule
3 ) You’ve done 20.0 J of work turning a steering wheel. If you’re applying 10.0 N-m of torque, what angle have you turned the steering wheel through?
Solution:
W = 20 J
τ = 10.0 N-m
W = τ θ
=> θ = W / τ = 20/10 rad = 2 rad.
4 ) How much work do you do if you apply a torque of 75 N-m through an angle of 6 π radians?
Solution:
τ = 75 N-m
θ = 6 π rad
W = τ θ = 75 x 6 π J = 1413 Joule
5 ) You’ve done 350 J of work turning a bicycle tire. If you’re applying 150 N-m of torque, what angle have you turned the wheel through?
Solution:
W = 350 J
τ = 150 N-m
W = τ θ
=> θ = W / τ = 350/150 rad = 2.33 rad.
