# NTNUJAVA Virtual Physics LaboratoryEnjoy the fun of physics with simulations! Backup site http://enjoy.phy.ntnu.edu.tw/ntnujava/

## 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.