A wheel rotates with a constant angular acceleration of 3.30 rad/s2.

[KTANNE]

A wheel rotates with a constant angular acceleration of 3.30 rad/s2. Assume the angular speed of the wheel is 2.45 rad/s at ti = 0.
(a) Through what angle does the wheel rotate between t = 0 and t = 2.00 s? Give your answer in radians and revolutions.
(b) What is the angular speed of the wheel at t = 2.00 s?

Answers: a) 11.50 rad and 1.83 rev
b) 9.05 rad/s

Use the values from above to help you work this exercise.
(a) Find the angle through which the wheel rotates between t = 2.00 s and t = 3.00 s.

I understood the first part and got the correct answers, however, I have tried this problem a few times and came up the wrong answer... Here are my attempts with wrong answers:

(2.45rad/s)(3.00s) + 1/2(3.30rad/s^2)(3.00s)^2 = 22.2rad (wrong)
AND
(9.05rad/s)(3.00s) + 1/2(3.30rad/s^s)(3.00s)^2 = 42rad (wrong again)

Can someone tell me where I am doing this equation incorrectly? I thought on the second attempt I was spot on, but I'm not.

Thanks!

 

[tiny-tim]

Welcome to PF!

Hi KTANNE! Welcome to PF!

A wheel rotates with a constant angular acceleration of 3.30 rad/s2. Assume the angular speed of the wheel is 2.45 rad/s at ti = 0.
(a) Through what angle does the wheel rotate between t = 0 and t = 2.00 s?
…
(2.45rad/s)(3.00s) + 1/2(3.30rad/s^2)(3.00s)^2 = 22.2rad (wrong)

erm …

2 seconds ! ​