
#1
Apr808, 12:17 PM

P: 3

Hey there. I have a problem about rolling ball and it's from my homework.
1. The problem statement, all variables and given/known data A ball rolls down the surface of a rough fixed sphere of radius r, starting rest at the top. What's its speed when the line between the two centres makes an angle alpha with the upward vertical? Also find the normal component of the reaction force on the ball at this time, and show that the ball leaves the surface when cos(alpha)=10/17. 2. Relevant equations I=2/5ma^2 3. The attempt at a solution KE=mg(r+a)(1cos(alpha))=7/10mu^2 so u=(10/7g(r+a)(1cos(alpha))) ^1/2 this should be the answer to the first part of the question. I don't quite have a clue to do the second part. How to find the normal component of the force? It has been annoying me for days and thank anyone who helps me with it! 



#2
Apr808, 03:24 PM

P: 35

You're on the right way  conservation of energy:
[tex]E_{potential}=E_{translation}+E_{rotation} \Rightarrow mg\Delta h = \frac{mv^{2}}{2}+\frac{I \omega^{2}}{2}, \Delta h = r(1cos\alpha).[/tex] My result differs a bit from yours. I don't have the a parameter. Everything else is the same. Reaction force is the net force between one of the ball's weight component and the centrifugal force: [tex]\vec{N}=\vec{G_{y}}+\vec{F_{cf}}, G_{y}=mgcos\alpha.[/tex] Ball will leave the surface when reaction force becomes equal to 0: [tex]N=mgcos\alpha\frac{mv^{2}}{r}=0.[/tex] If you plug the velocity derived from the conservation of energy here you will get that [itex]cos\alpha=\frac{10}{17}.[/itex] 



#3
Apr808, 10:10 PM

P: 3

Thanks a lot!
By the way, a is the radius of the rolling ball which I forgot to mention. 


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