You observe that at t=0.3674 s the disk has an angular velocity of 30.0 rad/s .

Question

You observe that at t=0.3674 s the disk has an angular velocity of 30.0 rad/s . At this instant you push on the disk and let it spin up with a constant angular acceleration of 28.0 rad/s2 until the time on your stopwatch is t = 1.90 s at which point you leave it (i.e. you are no longer touching it). Immediately after you let it go, you observe that friction forces are acting on it and the disk turns through an additional angle of 438 rad at constant angular acceleration before coming to a stop. By how many radians did the disk turn between t=0.3674s and the time it stopped?

Explanation / Answer

Here,

wi = 30 rad/s
a = 28 rad/s^2

total time = (1.90 - 0.3674) s

by using second equation of motin :

Theta1 = wi*t + 0.5 * a * t^2

theta1= 30 * 1.53 + 0.5 * 28 * 1.53^2

Theta1 = 78.9 rad

for the total angle rotated ,

theta = theta1 + angle rotated while stopping

theta = 78.9 + 438

theta = 516.9 rad

the angle rotated by disk is 516.9 rad

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