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 Author Topic: Bead rolling on hemisphere  (Read 12752 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
tanghb
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 « Embed this message on: December 08, 2009, 08:52:47 pm » posted from:Singapore,,Singapore

I solved the following problem and got 48.2o as the answer, but the provided answer is 60o. Which is correct?

A small marble resting on the top of a hemisphere starts to roll down, without slipping. At what angle will the marble become airborne?

 bead_on_hemisphere.gif (2.47 KB, 466x268 - viewed 2874 times.) « Last Edit: December 08, 2009, 08:55:06 pm by tanghb » Logged
Fu-Kwun Hwang
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 « Embed this message Reply #1 on: December 09, 2009, 12:09:16 pm » posted from:Taipei,T'ai-pei,Taiwan

The normal force for circular motion is from normal component of gravitation force, and it should be larger than the centripetal force
i.e. $N= mg cos\theta\ge m \frac{v^2}{R}$, where R is the radius.
Since it starts to roll down from the top,
the kinetic energy is coming from changes of potential energy
$\frac{1}{2}mv^2=mgR(1-cos\theta)$
So $\frac{v^2}{R}=2mg(1-cos\theta)\le mg cos\theta$
$2mg\le3mg cos\theta$
The result is $\theta\le cos^{-1}\frac{2}{3}$

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tanghb
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 « Embed this message Reply #2 on: December 09, 2009, 09:02:05 pm »

My answer and approach are the same as yours. Thanks a lot for confirming!
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mrhengrasmee
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 « Embed this message Reply #3 on: March 10, 2010, 12:56:08 am » posted from:Bangkok,Krung Thep,Thailand

Really confirm for the answer is the angle less than and equal arccos (2/3)
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Femida
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 « Embed this message Reply #4 on: March 29, 2015, 04:22:24 am » posted from:,,Satellite Provider

I can not participate now in discussion - there is no free time. But I will be released - I will necessarily write that I think on this question.
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Never underestimate others. Never overestimate oneself. ...Wisdom

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