Title: Bead rolling on hemisphere Post by: tanghb on December 08, 2009, 08:52:47 pm I solved the following problem and got 48.2^{o} as the answer, but the provided answer is 60^{o}. 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: FuKwun 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 $\frac{1}{2}mv^2=mgR(1cos\theta)$ So $\frac{v^2}{R}=2mg(1cos\theta)\le mg cos\theta$ $2mg\le3mg cos\theta$ 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.
