A thin wire of length (2 m) and (1 mm2 ) cross-section area is clamped horizontally between two walls, a weight of (10 kg) is hu
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a. The disk starts at rest, so its angular displacement at time is
It rotates 44.5 rad in this time, so we have
b. Since acceleration is constant, the average angular velocity is
where is the angular velocity achieved after 6.00 s. The velocity of the disk at time
is
so we have
making the average velocity
Another way to find the average velocity is to compute it directly via
c. We already found this using the first method in part (b),
d. We already know
so this is just a matter of plugging in . We get
Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that
Then for we would get the same
.