At time t=0 a grinding wheel has an angular velocity of 25.0 rad/s. It has a con

Question

At time t=0 a grinding wheel has an angular velocity of 25.0 rad/s. It has a constant angular acceleration of 28.0 rad/s^2 until a circuit breaker trips at time t=1.90s. From then on, the wheel turns through at angle of 431 rad. as it coasts to a stop at constant angular deceleration. A) through what total angle did the wheel turn between t=0 and the time it stopped? B) At what time does the wheel stop? C) What was the wheel's angular acceleration as it slowed down? I HAVE THE ANSWER TO PART A TO BE 542 RAD.

Explanation / Answer

The angle the wheel turned through is found from : ?? = ?0t + 0.5at² = (25.0rad/s)(2.20s) + 0.5(30.0rad/s²)(2.20s)² = 128rad The time for it to stop (from the point where the circuit breaker trips to when it stops is found from : ?? = (?0 + ?)t / 2 t = 2?? / (?0 + ?) The final angular speed (?), we know to be zero because it comes to rest. We don't know initial angular speed (?0), which is the speed it had when the circuit breaker tripped. This speed is : ? = ?0 + at = 25.0rad/s + (30.0rad/s²)(2.20s) = 91.0rad/s Putting this into the equation for t : t = 2?? / (?0 + ?) = (2 x 128rad) / (91.0rad/s + 0) = 2.81s Add this to the time where the circuit breaker tripped to get the time for the wheel to come to a complete stop : t(total) = 2.20s + 2.81s = 5.01s The wheel's angular acceleration (from the point where the curcuit breaker tripped) is : ?² = ?0² + 2a?? a = (?² - ?0²) / 2?? = ([0 – (91.0rad/s)²] / (2 x 433rad) = -9.56rad/s²

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