Use the worked example above to help you solve this problem. A wheel rotates wit

Question

Use the worked example above to help you solve this problem. A wheel rotates with a constant angular acceleration of 3.95 rad/s2. Assume the angular speed of the wheel is 2.10 rad/s at ti = 0. Through what angle does the wheel rotate between t = 0 and t = 2.00 s? Give your answer in radians and revolutions. What is the angular speed of the wheel at t = 2.00 s? 10 What angular displacement (in revolutions) results while the angular speed found in part doubles? Your response differs from the correct answer by more than 10%. Double check your calculations, rev Use the values from PRACTICE IT to help you work this exercise. Find the angle through which the wheel rotates between t = 2.00 s and t = 3.50 s. Sketch the angular speed of the wheel as a function of time. What is its initial value? What does the slope represent? What does the area underneath it represent? rad Find the angular speed when t = 3.50 s. rad/s What is the magnitude of the angular speed five revolutions following t = 3.50 s? rad/s

Explanation / Answer

a) alpha = 3.95 rad/s^2

wi = 2.10 rad/s

using theta = wi*t + alpha*t^2 /2

theta = ( 2.10 x 2) + ( 3.95 x 2^2 /2 ) = 12.1 rad

revolution = theta / 2pi = 1.93 rev

b)
wf = wi + alpha*t = 2.10 + ( 3.95 x2) = 10 rad/s

c) wf = wi + alpha*t

20 = 2.10 + alpha*2

alpha = 8.95 rad/s^2

using wf^2 - wi^2 = 2 x alpha x theta

20^2 - 2.10^2 = 2 x 8.95 x theta

theta = 22.1 rad

revolutions = 22.1 / 2pi =3.52 rev.

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a) using theta = wi*t + alpha*t^2

theta = 10(3.50 - 2) + 3.95(3.50-2)^2 /2 = 19.44 rad

b) wf = wi + alpha*t

wf = 10 + 3.95(3.50 - 2) = 15.93 rad/s

c) theta = 5 x 2pi rad = 31.42 rad

using wf^2 - wi^2 = 2 x alpha x theta

wf^2 - 15.93^2 = 2 x 3.95 x 31.42

wf = 22.40 rad/s

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