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xdrgnh
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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