Values from Practice It: A wheel rotates with a constant angular acceleration of

Question

Values from Practice It: A wheel rotates with a constant angular acceleration of 3.95 rad/s2. Assume the angular speed of the wheel is 1.95 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.
11.8 rad
1.87 rev

(b) What is the angular speed of the wheel at t = 2.00 s?
9.85 rad/s

(c) What angular displacement (in revolutions) results while the angular speed found in part (b) doubles?
5.85 rev

THIS IS THE QUESTION ---> Use the values from PRACTICE IT to help you work this exercise.

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

(b) Find the angular speed when t = 3.30 s.
_________ rad/s

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

Explanation / Answer

alpha = 3.95 rad/s

at ti = 0, w = 1.95 rad/s

a. from t = 2s to t = 4 s

angular rotation = theta

theta = wt + 0.5*alpha*t^2 - (w*2 + 0.5*alpha*2^2)

theta = 1.95*4 + 0.5*3.95*4^2 - 2*1.95 - 0.5*3.95*2^2

theta = 25.7 rad

b. at t = 3.3s

w = wo + alpha*t

w = 1.95 + 3.95*3.3

w = 14.985 rad/s

c. after 5 revlusiotns after this, 5 revolutions = 10*pi radians

hence

2*alpha*10*pi = w^2 - 14.985^2

w = 21.74249 rad/s

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