A wheel rotates with a constant angular acceleration of 3.45 rad/s 2 . Assume th

Question

A wheel rotates with a constant angular acceleration of 3.45 rad/s2. Assume the angular speed of the wheel is 2.20 rad/s at ti = 0.

(a) Find the angle through which the wheel rotates between t = 2.00 s and t = 3.30 s.

I got 11.34 rad, but it says it's wrong.
You have successfully computed the angular displacement between t = 0 and t = 2.00 s. What is the question? Have you answered it?

(c) What is the magnitude of the angular speed three revolutions following t = 3.30 s?
------rad/s

This one I am completely confused...

Explanation / Answer

(a) First, you need to know the initial angular velocity initial when ti = 2.00 s (I will call t = 2.00 "initial", because it is the initial time in the time interval considered in our problem. The motion actually starts at tstart = 0 s). You know that it started with an angular velocity start of 2.20 rad/s from tstart = 0 s, and that it has been accelerating at a rate of 3.45 rad/s2 for 2.00 seconds between tstart = 0 s and tinitial = 2.00 s. Let's solve for initial using rotational kinematics:

initial = start + t = (2.20 rad/s) + (3.45 rad/s2)(2.00 s) = 9.10 rad/s

We have now found the initial angular velocity when t = 2.00 s. Next, we want to find the angular displacement 1 if angular acceleration is still 3.45 rad/s2, except this time, motion is between ti = 2.00 s and tfinal = 3.30 s (t = 1.30 s is the time elapsed during this motion).

1 = initialt + (1/2)t2 = (9.10 rad/s)(1.30 s) + (1/2) (3.45 rad/s2)(1.30 s)2

1 = 14.74525 rad 14.7 radians

(c) If 3 revolutions passed after tfinal = 3.30 s, then 3 (2 rad) = 6 radians is the new angular displacement 2. First, let's find the new initial angular velocity 2 = final, which happens to be the final angular velocity of the problem we just solved. Using initial = 9.10 rad/s, t = 1.30 s, and = 3.45 rad/s2, which were all information known from before, we can solve for 2 = final:

final = initial + t = (9.10 rad/s) + (3.45 rad/s2) (1.30 s) = 13.85 rad/s

Now that we know the angular velocity 2 at which the new motion starts, we can use rotational kinematics to find the magnitude |3| of the final angular velocity after a displacement of 6 radians:

32 = 22 + 2 = (13.85 rad/s)2 + 2(3.45 rad/s2)(6 rad)

|3| = ((13.85 rad/s)2 + 2(3.45 rad/s2)(6 rad))

|3| = 17.9411380... rad/s 17.9 rad/s

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