Moving at its maximum safe speed, an amusement park carousel takes 12 s to complete a...

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 3.5 rev to come to a stop. What is the magnitude of the rotational acceleration of the carousel while it is slowing down?

Solutions

Expert Solution

Maximum angular speed of the carousel = $\omega$ ₁

Final angular speed of the carousel = $\omega$ ₂ = 0 rad/s (Comes to a stop)

Time taken to complete a revolution at maximum speed = T = 12 sec

$T = \frac{2\pi }{\omega _{1}}$

$12 = \frac{2\pi }{\omega _{1}}$

$\omega$ ₁ = 0.5236 rad/s

Number of revolutions made by the carousel before coming to a stop = n = 3.5 rev

Angle through which the carousel turned before coming to a stop = $\theta$

$\theta$ = 2 $\pi$ n

$\theta$ = 2 $\pi$ (3.5)

$\theta$ = 21.991 rad

Angular acceleration of the carousel = $\alpha$

$\omega$ ₂² = $\omega$ ₁² + 2 $\alpha$ $\theta$

(0)² = (0.5236)² + (2)(21.991) $\alpha$

$\alpha$ = -6.233 x 10^-3 rad/s²

Negative as it is deceleration.

Magnitude of rotational acceleration of the carousel = 6.233 x 10^-3 rad/s²

genius_generous answered 1 year ago

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