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AlexFokin [52]
3 years ago
14

A wheel with a 0.10-m radius is rotating at 35 rev/s. it then slows uniformly to 15 rev/s over a 3.0-s interval. what is the ang

ular acceleration of a point on the wheel?
Physics
1 answer:
telo118 [61]3 years ago
8 0
The initial angular speed of the wheel is (keeping in mind that 1 rev=2 \pi rad):
\omega _i=35 rev/s \cdot ( 2 \pi \frac{rad}{rev})=219.8 rad/s
The final angular speed instead is:
\omega _f=15 rev/s \cdot ( 2 \pi \frac{rad}{rev})=94.2 rad/s
Therefore we can find the value of the angular acceleration:
\alpha =  \frac{\omega _f-\omega _i}{t}= \frac{94.2 rad/s-219.8rad/s}{3.0 s}=-41.87 rad/s^2
and the negative sign is due to the fact the wheel is decelerating.
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