Part A At time t0 a grinding wheel has an angular velocity of Through what total

Question

Part A At time t0 a grinding wheel has an angular velocity of Through what total angle did the wheel turn between t 0 and the time it stopped? 30.0 rad/s. It has a constant angular acceleration of 31.0 rad/s2 until a circuit breaker trips at time t 1.80 s. From then on, the wheel turns through an angle of 433 rad as it coasts to a stop at constant angular deceleration. Express your answer in radians Hints rad Submit My Answers Give Up Part B At what time does the wheel stop? Express your answer in seconds Hints Submit My Answers Give Up Part C What was the wheel's angular acceleration as it slowed down? Express your answer in radians per second per second

Explanation / Answer

Part A

w0 = 30 rad/s. t1 = 1.80 s, alfa1 = 31 rad/s^2

use the expression -

theta2 = w0*t1 + (1/2)*alfa1*t1^2 = 30*1.8 + 0.5*31*1.8^2 = 54 + 50.22 = 104.22 rad.

So, the total angle of running of the wheel = 104.22 + 433 = 537.22 rad.

Part B

Velocity of the wheel at t = 1.8 s -

w = w0 + alfa*t = 30 + 31*1.8 = 85.8 rad/s

use the expression -

w'^2 = w^2 + 2*alfa2 *theta2

=> 0 = 85.8^2 + 2*433*theta2

=> theta2 = -85.8^2 / (2*433) = -8.5 rad/s^2

again use the expression -

w' = w + alfa*t2

=> 0 = 85.8 - 8.5*t2

=> t2 = 85.8/8.5 = 10.09 s

So the time at which the wheel stops = t2 + 1.8 = 10.09 + 1.8 = 11.89 s

Part C

As calculated above -

angular acceleration as the wheel slows down = -8.5 rad/s^2.

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