Rotational Acceleration of an Amusement Park Carousel

[kolua]

Homework Statement

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

Homework Equations

w²=w_o²+2alpha theta

The Attempt at a Solution

plug in and get -pi/144 rad/s² is this right?

 

[kolua]

or -pi²/504 s^-2

 

[Daeho Ro]

When it moves with maximum speed, the angular velocity is $\omega_i = \pi / 12.$ You want to stop it within $3.5$ revolution, that is $\theta_f - \theta_i = 7\pi$ and it have to be $\omega_f = 0.$ Then, you can get the angular acceleration, $\alpha.$

 

[kolua]

When it moves with maximum speed, the angular velocity is $\omega_i = \pi / 12.$ You want to stop it within $3.5$ revolution, that is $\theta_f - \theta_i = 7\pi$ and it have to be $\omega_f = 0.$ Then, you can get the angular acceleration, $\alpha.$

shouldn't the angular velocity be pi/6?

 

[Daeho Ro]

Oh, yes I missed the factor $2$.

 

[kolua]

Oh, yes I missed the factor $2$.

Can you also help me check this one? You hold a small ice cube near the top edge of a hemispherical bowl of radius 100 mm. You release the cube from rest. What is the magnitude of its acceleration at the instant it reaches the bottom of the bowl? Ignore friction. is a=2g=19.6 m/s²

 

[ehild]

or -pi²/504 s^-2

Which one?

Always show your work in detail.