Past Cambridge, conservation of energy problem.

[xdrgnh]

Homework Statement

http://www-teach.phy.cam.ac.uk/dms/dms_getFile.php?node=5735 problem B8, the one with the with finding h>11(R-a)/4

Homework Equations

(1/2)mv^2=mgh
ac=(v^2/R)
w=V/R
(1/2)mv^2+(1/2)Iw^2

The Attempt at a Solution

At the top of the circle, the velocity squared is GR. And because the a is above ground, conservation of energy is written as. MG(h+a)=(3/4)MGR+MG(2R-2a). When I solve it out I get h=(11/4)R-a

Help would be greatly appreciated

 

[PhanthomJay]

You are leaving out some steps so I can't see all your workings, but possibly your error is in forgetting to include the ball's radius when determining critical speed and potential energy. Please write out the conservation of energy equation and show your numbers term by term.

 

[xdrgnh]

You are leaving out some steps so I can't see all your workings, but possibly your error is in forgetting to include the ball's radius when determining critical speed and potential energy. Please write out the conservation of energy equation and show your numbers term by term.

well I got the 3/4 because of the rolling condition.

(1/2)(mv^2)+(1/2)(ma^2*.5*(V^2/a^2) so when that is added you get (3/4)mv^2.

the critical speed squared at the top of the circle is just GR, because I just need to know the min speed, which is just GR. So v^2 at the top of the circle can just be replaced by GR. The M and G's cancel out of the mgh equations. The problem first states, that a is above h initial. So MG(h+a)= (3/4)(mGR)+MG(2R-2a). Does the a, affect what the radius would be. Maybe the radius is (R-a)?

 

[PhanthomJay]

Yes, the radius is R-a, and the critical speed is not root RG, but rather, ___?

 

[xdrgnh]

Yes, the radius is R-a, and the critical speed is not root RG, but rather, ___?

root (R-a)G

 

[PhanthomJay]

yes..