Moving at its maximum safe speed, an amusement park carousel takes 12 s to complete a revolution. At the end of the ride, it slows down smoothly, taking 2.3rev to come to a stop.What is the magnitude of the rotational acceleration of the carousel while it is slowing down
maximum safe speed is initial speed of the carousel before going to stop
time period is T = 12 S
but time period is T = 2*pi/wi
wi is the initial speed = 2*pi/T = 2*3.142/12 = 0.523 rad/s
wf is teh final speed = 0 rad/s
no.of revolutions = 2.3 rev = 2.3*2*3.142 = 14.4532 rad
thne apply wf^2-wi^2 = 2*alpha*theta
theta = 14.4532 rad
alpha is negative since speed is decreasing
0^2-0.523^2 = -2*alpha*14.4532
alpha = rotational accelaration = 0.523^2/(2*14.4532) = 0.00946 rad/s^2
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