**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 = = = 2**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 N-m. Thus we can calculate power:

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