Question

At time t=0 a grinding wheel has an angular velocity of 23.0 rad/s . It has a constant angular acceleration of 35.0 rad/s2 until a circuit breaker trips at time t = 2.10 s . From then on, the wheel turns through an angle of 436 rad as it coasts to a stop at constant angular deceleration

Part A) Through what total angle did the wheel turn between t=0 and the time it stopped?

Part B) At what time does the wheel stop?

Part C) What was the wheel's angular acceleration as it slowed down?

Answer 1

a)
angular displacement during first 2.10s,

theta1 = wo*t + 0.5*alfa*t1^2

= 23*2.1 + 0.5*35*2.1^2

= 125.5 rad

total angular displcament, theta = theta1 + thea2

= 125.5 + 436

= 561.5 rad <<<<<<<<-------------Answer

b) angular acceleration after 2.1s, alfa = (wf^2 - wo^2)/(2*theta2)

= (0^2 - 23^2)/(2*436)

= -0.6067 rad/s^2

time taken to stop, t = (wf - wo)/alfa

= (0 - 23)/(-0.6067)

= 37.9 s

so, at time = 2.1 + 37.9

= 40 s the wheel stops <<<<<<<<-------------Answer

c) alfa = -0.6067 rad/s^2 <<<<<<<<-------------Answer