Question #104553
a tire 0.500m in radius rotates at a constant rate od 200 rev/mon. Find the speed and acceleration of a small stone lodged in the tread of the tire (on its outer edge)
1
Expert's answer
2021-01-31T19:45:57-0500

Given:

R=0.500m;R=0.500\:\rm m;

n=200rev/min.n=200\:\rm rev/min.

The angular velocity


ω=2πn=2π×200=1256rad/min=20.9rad/s.\omega=2\pi n=2\pi\times 200=1256\:\rm rad/min=20.9\:\rm rad/s.

The linear speed


v=ωR=20.9×0.500=10.45m/s.v=\omega R=20.9\times 0.500=10.45\:\rm m/s.

The acceleration


a=ω2R=20.92×0.500=218m/s2.a=\omega^2R=20.9^2\times 0.500=218\:\rm m/s^2.

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Comments

Assignment Expert
01.02.21, 02:46

Dear Bocaj Klontz, thank you for comment

Bocaj Klontz
30.01.21, 03:26

The linear velocity is incorrect on this, it's actually 10.45m/s, but that's the only thing wrong that I've found

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