i can do part a)

and for b) i started with T-mgsin60-mw^2Lsin30=0

but didnt end up with the answer.

i got w^2 = (2-rt3)g/L

instead of w^2 = 2(2-rt3)g/L

thanks.Attachment 19396

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- October 20th 2010, 05:57 AMBabyMiloCircular Motion
i can do part a)

and for b) i started with T-mgsin60-mw^2Lsin30=0

but didnt end up with the answer.

i got w^2 = (2-rt3)g/L

instead of w^2 = 2(2-rt3)g/L

thanks.Attachment 19396 - October 20th 2010, 07:17 AMmathaddict
Firstly, note that T=mg.

Then, draw the forces acting on the ball outside the cone(Reaction force, gravity, tension and the centripetal force).

If the system is in equilibrium, R sin 30 + T cos 30 = mg

0.5 R + mg (root3 /2)= mg

R= (2- root(3)) mg

For horizontal components, T sin 30 - R cos 30 = mrw^2

Substitute T=mg and R. - October 20th 2010, 09:25 AMBabyMilo
Thanks for ur answer.

but i want to understand where i went wrong with T-mgsin60-mw^2Lsin30=0.

thanks