Item 6 Part A At time t 0 a grinding wheel has an angular velocity of 30.0 rad/s

Question

Item 6 Part A At time t 0 a grinding wheel has an angular velocity of 30.0 rad/s. It has a constant angular acceleration of 26.0 rad/s2 until a circuit breaker trips at time t = 2.50 s. From then on, the wheel turns through an angle of 440 rad as it coasts to a stop at constant angular deceleration Through what total angle did the wheel turn between t = 0 and the time it stopped? Express your answer in radians. Hints 56.25 rad Submit My Answers Give Up Incorrect; Try Again; 29 attempts remaining

Explanation / Answer

during constant angular acceleration

theta1 = wo*t1 + (1/2)*a1*t1^2

theta1 = (30*2.5) + (1/2)*26*2.5^2

theta1 = 156.25 rad

during deceleration theta2 = 440 rad

total angle rotated = theta1 + theta2 = 596.25 rad

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during deceleration

velocity at time t = 2.5 s

w = wo + a1*t1 = 30 + (26*2.5) = 95 rad/s

theta2 = (w + wf)*(t-2.5)/2

sf = final velocity = 0

440 = (95+0)*(t-2.5)/2

at time t = 11.76 s <<<<------ANSWER

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part C

wf^2 - 2^2 = 2*alpha*theta2

0 - 95^2 = 2*alpha2*440

acceleration = -10.3 rad/^2

angular acceleration = 10.3 rad/s^2

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