When peddaling a bicycle, forces are applied to a rotating body and work is done on it. Other similar events are a rotating motor shaft driving a power tool or a car engine propelling the vehicle. Work can be expressed in terms of torque and angular displacement.
Suppose a tangential force Ftan acts at the rim of a pivoted disk — for example, a child running while pushing on a playground merry-go-round. The disk rotates through an infinitesimal angle dθ about a fixed axis during an infinitesimal time interval dt. The work dW done by the force Ftan while a point on the rim moves a distance ds is dW = Ftands. If dθ is measured in radians, then ds = R dθ and:
dW = FtanR dθ
Now FtanR is the torque τ due to the force Ftan, so:
dW = τ dθ
The total workW done by the torque during an angular displacement from θ1 to θ2 is:
W =
∫
θ2
θ1
τ dθ
This is the equation for work done by a torque
If the torque is constant while the angle changes by a finite amount Δθ = θ2 − θ1, then:
W = τ(θ2 − θ1) = τ Δθ
This is the equation for work done by a constant torque.
The work done by a constant torque is the product of torque and the angular displacement. If torque is expressed in newton-meters [Nm] and angular displacement in radians, then the work is in joules. Eq3 is the rotational analog of Eq1 of the lesson Work, and Eq2 is the analog of Eq2 of the lesson Work, for the work done by a force in a straight-line displacement.
If the force in the figure had an axial or radial component, that component would do no work because the displacement of the point of application has only a tangential component. An axial or radial component of force would also make no contribution to the torque about the axis of rotation, so Eq2 and Eq3 are correct for any force, no matter what its components.
When a torque does work on a rotating rigid body, the kinetic energy changes by an amount equal to the work done. This can be proved by using exactly the same procedure that was used in the lesson Work Energy Theorem for Straight Line Motion for a particle. First, τ is used to represent the net torque on the body so that from Eqx from the lesson Torque, Inertia, and Angular Acceleration for a Rigid Body, τ = Iα. By using this equation, it is being assumed that the body is rigid so that the moment of inertia I is constant. The integrand in Eq1 is then transformed into an integral on ω as follows:
Since τ is the net torque, the integral in Eq1 is the total work done on the rotating rigid body. This equation then becomes:
Wtot =
∫
ω2
ω1
Iω dω =
1
2
Iω22 −
1
2
Iω12
The change in the rotational kinetic energy of a rigid body equals the work done by forces exerted from outside the body. This equation is analogous to the equation for the work-energy theorem for a particle.
With regards to the power associated with work done by a torque acting on a rotating body, when dividing both sides of Eq1 by the time interval dt during which the angular displacement occurs, it is found that:
dW
dt
= τ
dθ
dt
But dW/dt is the rate of doing work, or power P, and dθ/dt is angular velocity ω, so:
P = τω
When a torque τ (with respect to the axis of rotation) acts on a body that rotates with angular velocity ω, its power (rate of doing work) is the product of τ and ω. This is the analog of the relation P = Farrow ⋅ varrow that was developed in the lesson for particle motion.