At t = 0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30 rads2 until a circuit breaker trips at t = 2.00 s. From then on, it turns through 432 rad as it coasts to a stop at constant angular acceleration. (a) Through what total angle did the wheel turn between t = 0 and the time it stopped? (b) At what time did it stop? (c) What was its acceleration as it slowed down?

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At t = 0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30 rads2 until a circuit breaker trips at t = 2.00 s. From then on, it turns through 432 rad as it coasts to a stop at constant angular acceleration. (a) Through what total angle did the wheel turn between t = 0 and the time it stopped? (b) At what time did it stop? (c) What was its acceleration as it slowed down?

Pademagk71 · Accepted Answer

b) We will take the interval from when the circuit breaker trips until th wheel comes to a stop ω0z→ the wheel will stop at the end θ−θ0=12)ω0z+ωz)t 432=12(84+0)t t=10.3 s Adding t of the first interval to t of the second interval: tT=2+10.3 =12.3 s The wheel will stop after 12.3 s

RizerMix · Accepted Answer

a) In order to calculate the total angle, we will divide the entire interval from t=0 to the wheel stops into two intervals. From t=0 to t=2 s θ−θ0=12(ω0+ωz) We will calulate ωz first: ωz=ω0z+αz ωz=24+(30)(2) ωz=84 rad/s Substitute ωz into equation θ−θ0=12(ω0z+ωz) θ−θ0=12(24+84)(2) θ−θ0=108 rad We can calculate it directly from the following equation: θ−θ0=w0zt+12azt2 θ−θ0=(24)(2)+12(30)(2)2 θ−θ0=108 rad Thus, the total angle the wheel turned between t=0 and the time it stopped: θT=108+432 540 rad

alenahelenash · Accepted Answer

c) We will take the interval from when the circuit breaker trips untill the wheel comes to a stop&amp;nbsp;ω0z=84&amp;nbsp;rad/sωz=0&amp;nbsp;ωz=ω0z+azt0=84+aa(10.3)az=−8.16&amp;nbsp;rad/s2

At t = 0 a grinding wheel has an angular velocity of 24.0 ra

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