Rotational Dynamics

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)(2²) = 4 . Thus we can calculate torque:

Finally, since we know the torque, we can calculate the work done over one revolution, or 2Π radians:

This quantity is measured in the same units as linear work: Joules.

To solve this problem we simply have to plug into our equation for rotational kinetic energy:

Again, this quantity is also measured in joules.

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 N-m. Thus we can calculate power:

This power is measured in Watts.