Question

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

Answer #1

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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