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a Secure Ihttps//session masteringphysics.com/myct/itemView?assignmentProblemiD 87742886 Exercise 9.16 Part A Through what total angle did the wheel turn betweent 0 and the time it stopped? At t-0 a grinding wheel has an angular velocity of 25.0 rad/s. It has a constant angular acceleration of 32.0 rad/s2 until a circuit breaker trips at time 1.70 s . From then on, it turns through an angle 438 rad as it coasts to a stop at constant angular acceleration 485 rad Submit My Answers Give Up Incorrect: Try Again; 5 attempts remaining Part B At what time did it stop? Submit My Answers Give Up

Explanation / Answer

a) Using kinematic equations

theta = theta_o+(wi*t)+(0.5*alpha*t^2)

wi = 25 rad/s

alpha = 32 rad/s^2

t = 1.7 sec

then

theta = 438+(25*1.7)+(0.5*32*1.7^2)

theta = 526.74 rad

B) after t = 1.7 sec

speed is w = wo+(alpha*t) = 25+(32*1.7) = 79.4 rad/s

theta = (w*t)+(0.5*alpha*t^2)

438 = (79.4*t)-(0.5*alpha*t^2)

and also

wf = w+(alpha*t)

0 = 79.4-(alpha*t)

alpha*t = 79.4 rad

then

438 = (79.4*t)-(0.5*alpha*t^2)

438 = (79.4*t)-(0.5*79.4*t)

t = 11 sec

so total time taken is t = 11+1.7 = 12.17 sec

C) alpha*t = 79.4

alpha*12.17 = 79.4

alpha = 6.62 rad/s^2

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