A wheel rotates with a constant angular acceleration of 3.35 rad/s2. Assume the

Question

A wheel rotates with a constant angular acceleration of 3.35 rad/s2. Assume the angular speed of the wheel is 2.50 rad/s at ti = 0. Through what angle does the wheel rotate between t = 0 and t = 2.00 s? Give your answer in radians and revolutions. What is the angular speed of the wheel at t = 2.00 s? What angular displacement (in revolutions) results while the angular speed found in part (b) doubles?
Find the angle through which the wheel rotates between t = 2.00 s and t = 3.30 s. Find the angular speed when t = 3.30 s.What is the magnitude of the angular speed five revolutions following t = 3.30 s?

Explanation / Answer

How familiar are you with derivations and integrations? Distance (position) Velocity is the derivative of position Acceleration is the derivative of velocity Going backwards, we can find the distance traveled by integrating the acceleration twice. First step a = 3.35 rad/s/s. Integrating acceleration with respect to time we get... at+V, and we must evaluate this equation at t=2 and t=0. V is a constant of integration and will be our initial velocity. We know that at t=0, a*0+V = 2.5 rad /s So our velocity equation is... 3.35t+2.5 - plug in t = 2 to solve for velocity at time = 2 seconds Taking this one step further to find the position Integrate 3.35t+2.5 (1/2) * 3.35t^2 + 2.5 t + X, where X is another constant of integration and will represent the initial position. Because we want to find the change in angular displacement, it does not matter what the initial position is. Evaluate this equation at time = 2 and time = 0. (1/2)*3.35(2)^2 +2.5(2) + X - [(1/2)*3.35*(0)^2+2.5*(0)+X] You will see that the X cancels out and everything else on the right side of the equation is 0. So our change in angular displacement = (1/2)*3.35*(2)^2+2.5(2) = 11.7 radians Because there are 2*pi radians per revolution, we can calculate the number of revolutions with the following equation. 11.7 / 2pi = 1.86 revolutions.

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