Problem :
A single particle of mass 1 kg, starting from rest, experiences a torque that
causes it to accelerate in a circular path of radius 2 m, completing a full
revolution in 1 second. What is the work done by the torque over this full
revolution?
Before we can calculate the work done on the particle, we must calculate the
torque, and thus the angular acceleration of the particle. For this we turn to
our kinematic equations. The average angular velocity of the particle is given
by
= = = 2Π.
Since the particle started at rest, we can state that the final angular velocity
is simply twice the average velocity, or
4Π. Assuming acceleration is
constant, we can calculate the angular acceleration:
α = = = 4Π. With angular acceleration, we can
calculate torque, if we have the moment of inertia of the object. Fortunately
we are working with a single particle, so the moment of inertia is given by:
I = mr^{2} = (1 kg)(2^{2}) = 4. Thus we can calculate torque:
τ = Iα = (4)(4Π) = 16Π
Finally, since we know the torque, we can calculate the work done over one
revolution, or
2Π radians:
W = τφ = (16Π)(2Π) = 32Π^{2}
This quantity is measured in the same units as linear work: Joules.
Problem :
What is the kinetic energy of a single particle of mass 2 kg rotating around a
circle of radius 4 m with an angular velocity of 3 rad/s?
To solve this problem we simply have to plug into our equation for rotational
kinetic energy:
K  =  Iσ^{2} 

 =  (mr^{2})σ^{2} 

 =  (2)(4^{2})(3^{2}) 

 =  144 

Again, this quantity is also measured in joules.
Problem :
Often revolving doors have a built in resistance mechanism to keep the door from
rotating dangerously quickly. A man pushing on a door of 100 kg at a distance
of 1 meter from its center counteracts the resistance mechanism, keeping the
door moving at a constant angular velocity if he pushes with a force of 40 N.
If the door moves at a constant angular velocity of 5 rad/s, what is the power
output of the man over this time?
Because the door is moving at a constant angular velocity, we need only
calculate the torque the man exerts on the door to calculate the power of the
man. Fortunately, our torque calculation is easy. Since the man pushes
perpendicular to the radius of the door, the torque he exerts is given by: τ = Fr = (40 N)(1 m) = 40 Nm. Thus we can calculate power:
P = τσ = (40)(5) = 200
This power is measured in Watts.