Question

A wheel rotates with a constant angular acceleration of 3.20 rad/s2. Assume the angular speed of the wheel is 2.45 rad/s at ti = 0. What is the magnitude of the angular speed three revolutions following t = 3.60 s?

Answer 1

for 0≤t≤3.6, use
\[\omega_1 = \omega_0 + \alpha t\]
\[\omega_o = 2.45, \alpha = 3.2\]for next 3 revolutions, use
\[\omega_2^2 = \omega_1^2 + 2 \alpha \theta\]
\[\theta = 6 \pi\]

Answer 2

I tried that but didn't get the correct answer

Answer 3

the answer should be 7.9 rad/sec

Answer 4

How did you get that?

Answer 5

What answer are you getting @haleykk? The process outlined by @IrishBoy123 should give you the right answer:

First find the angular speed at t=3.6 s. Using the first equation provided above and the information in the question, you should find that ω(1) = 13.97 rad/s.

You then need to use this value as the starting angular speed for the second calculation, which is for the additional 3 revolutions AFTER t=3.6 s is reached. Each revolution encompasses 2π radians of angle swept, so 3 revolutions is 6π radians. Using the second equation provided above, you should find that ω(2) = 14.67 rad/s.

If you have any questions please ask and one of us will be happy to help!

Answer 6

+1 @matt101