A hula hoop (modeled as a thin-walled hollow cylinder) is rolling on a horizonta

Question

A hula hoop (modeled as a thin-walled hollow cylinder) is rolling on a horizontal surface at speed 2.8 m/s when it reaches a 25 incline. The hula hoop rolls without slipping on the incline.

How long will it take for the hula hoop to roll up the incline and then come back down to the bottom?
(include units with answer)

Hint #1: Constant acceleration equations Hint #2: Time up and time down are the same A hula hoop (modeled as a thin-walled hollow cylinder) is rolling on a horizontal surface at speed 2.8 m/s when it reaches a 25½ incline. The hula hoop rolls without slipping on the incline. How long will it take for the hula hoop to roll up the incline and then come back down to the bottom? (include units with answer) Hint #1: Constant acceleration equations Hint #2: Time up and time down are the same

Explanation / Answer

Note that

Torque = I(alpha)

-->alpha = Torque / I

Here, the torque is

Torque = M g R sin 25

Also,for a hoop,

I = M R^2

Thus,

alpha = g sin 25 / R

As

vo = 2.8 m/s

wo = vo/R

Then

wf = wo + alpha*t

as alpha is negative as it goes up,. alpha = -g sin 25 /R, and wf = 0 at the top, solving for t,

t = 0.676 s, moving up

as moving down takes the same time,

t total = 2t = 1.35 s   [ANSWER]

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