Question

Can someone please help me with this: A wheel rotates with a constant angular acceleration of 3.30 rad/s2. Assume the angular speed of the wheel is 2.40 rad/s at ti = 0. (a) Find the angle through which the wheel rotates between t = 2.00 s and t = 3.50 s. (b) Find the angular speed when t = 3.50 s. (c) What is the magnitude of the angular speed three revolutions following t = 3.50 s?

Answer 1

\[\theta (t) = \omega _{i}t +\frac{ 1 }{2}\alpha t^2\]\[\theta(3.5) - \theta (2) = \omega _{i}(3.5)+\frac{ 1 }{2}\alpha (3.5)^2 -(\omega _{i}(2)+\frac{ 1 }{2}\alpha (2)^2)\]

Answer 2

\[\omega _{f} = \omega _{i}+\alpha (3.5)\]

Answer 3

Can you help me with where to fill in the numbers? I must be making some small error because the answers aren't matching up.

Answer 4

\[\omega _{i} =2.4 \alpha =3.3\]

Answer 5

Thank you! You are a life saver!!!!

Answer 6

sure:)

Answer 7

Do you by chance know how to get part c?

Answer 8

use the omega from (b) in :\[\omega^2 _{f} = \omega^2 _{i} +2 \alpha \theta\]

Answer 9

where theta = 6pi (three rev.s)

Answer 10

alpha is the same as in all the other parts... find final omega...