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Information about this web site => Question related to Physics or physics related simulation => Topic started by: tanghb on December 08, 2009, 08:52:47 pm



Title: Bead rolling on hemisphere
Post by: tanghb on December 08, 2009, 08:52:47 pm
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?

(http://www.phy.ntnu.edu.tw/ntnujava/index.php?action=dlattach;topic=1378.0;attach=1546)


Title: Re: Bead rolling on hemisphere
Post by: Fu-Kwun Hwang on December 09, 2009, 12:09:16 pm
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 $\theta\le cos^{-1}\frac{2}{3}$
 


Title: Re: Bead rolling on hemisphere
Post by: tanghb on December 09, 2009, 09:02:05 pm
My answer and approach are the same as yours. Thanks a lot for confirming!


Title: Re: Bead rolling on hemisphere
Post by: mrhengrasmee on March 10, 2010, 12:56:08 am
Really confirm for the answer is the angle less than and equal arccos (2/3)


Title: Bead rolling on hemisphere
Post by: Femida on March 29, 2015, 04:22:24 am
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.