Having established the dynamics of rotational motion, we can now extend our study to work and energy. Given what we already know, the equations governing energetics are quite easy to derive. Finally, with the equations that we have derived, we will be able to describe the complicated situations involving combined rotational and translational motion.
Given our definition of work as W = Fs , can we generate an expression for work done on a rotational system? To derive our expression we begin by taking the simplest case: when the force applied to a particle in rotational motion is perpendicular to the radius of the particle. In this orientation, the force applied is parallel to the displacement of the particle, and would exert the maximum work. Given this situation the work done is simply W = Fs , where s is the arc length that the force acts through in a given period of time. Recall, however, that arc length can also be expressed in terms of the angle swept out by the arc: s = rμ . Our expression for work in this simple case becomes:
W = Frθ = τμ |
What if the force is not perpendicular to the radius of the particle? Let the angle between the force vector and the radius vector be θ , as shown below.
Recall that
Thus W = τμ Surprisingly enough, this equation is exactly the same as our special case when the force acted perpendicular to the radius! In any case, the work done by a given force is equal to the torque it exerts multiplied by the angular displacement.
For you calculus types, there is also an equation for work done by variable torques. Instead of deriving it, we can just state it, as it is quite similar to the equation in the linear case:
W = τdμ |
Thus we have quickly gone through deriving our expression for work. The next thing after work we studied in linear motion was kinetic energy, and it is to this topic that we turn.
Consider a wheel spinning in place. Clearly the wheel is moving, and has a kinetic energy attached to it. But the wheel is not engaged in translational motion. How do we calculate the kinetic energy of the wheel? Our answer is similar to how we calculated the result of a net torque on a body: by summing over each particle.
Given a rotating body, we state that the body is made up of n single rotating particles, each at a different radius from the axis of rotation. When each particle is considered individually, we can see that each one does in fact have a translational kinetic energy:
However, we also know from our relation between linear and angular variables that v = rσ . Substituting this expression in, we see that:
Since all particles are part of the same rigid body, we can factor our σ ^{2} :
This sum, however, is simply our expression for a moment of inertia. Thus:
K = Iσ ^{2} |
The equation for rotational power can be easily derived from the linear equation for power. Recall that P = Fv is the equation that gives us instantaneous power. Similarly, in the rotational case:
P = τσ |
F | = | ma | |
W | = | Fx | |
K | = | mv ^{2} | |
P | = | Fv |
τ | = | Iα | |
W | = | τμ | |
K | = | Iσ ^{2} | |
P | = | τσ |