(I just need part C) A wheel rotates with a constant angular acceleration of 3.5

Question

(I just need part C)

A wheel rotates with a constant angular acceleration of 3.50 rad/s2. If the angular speed of the wheel is 2.00 rad/s at t=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=2.00s? (c) What angular displacement (in revolutions) results while the angular speed found in part (b) doubles?

(I know that the answer for part a is 11 radians pr 1.75 revs and for part b it's 9rad/s)

If you could explain how to do part C I would appreciate it. I know the answer is 5.52 revs I just don't know how to get there.

Explanation / Answer

angular velocity at t=2.00s is u=9rad/s.

doubled angular velocity, v=18 rad/s.

angular acceleration, a=3.5 rad/s^2.

angular displacement "s" will be given as---

v^2 = u^2 - 2*a*s.

324= 81- 7*s

s=34.714285radians.

1revolution =2*pi rad.

so in terms of revolutions "s" = 34.714285/(2*pi) revolutions = 5.52 revolutions...

thank you

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