The equation of motion for a constant acceleration, x(t) = x(0)+v(0)t+0.5at^2, has an angular equivalent: ?(t) = ?(0)+?(0)t+0.5?t^2. For the uninitiated, ?(t) refers to the measurement of some angle at time \"t\" while ?(0) refers to the angle at time zero. ?(0) refers to the initial angular speed, at time zero. ? is the constant angular acceleration.
An example of when you might want to find a revolution count after a certain time \"t,\" given a constant angular acceleration, is when a constant torque is applied to a wheel.
For nonconstant angular momentum, use calculus to integrate the formula for the angular acceleration twice with respect to time to get an equation for ?(t).
Suppose you want to find the number of revolutions of a wheel after 10 seconds. Suppose also that the torque applied to generate rotation is 0.5 radians per second-squared, and the initial angular velocity was zero.
Plug these numbers into the formula in the introduction and solve for ?(t). Use ?(0)=0 as the starting point, without loss of generality. Therefore, the equation ?(t) = ?(0)+?(0)t+0.5?t^2 becomes ?(10) = 0 + 0 + 0.5x0.5x10^2 = 25 radians.
Divide ?(10) by 2? to convert the radians into revolutions. 25 radians / 2? = 39.79 revolutions.
Multiply by the radius of the wheel, if you also want to determine how far the wheel traveled.
- Fundamentals of Physics; David Halliday and Robert Resnick; 1991
- University of Winnipeg: Angular Momentum Problems