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Centripetal Force

by Touchme
Tags: centripetal, force
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Touchme
#1
Nov1-06, 02:20 PM
P: 41
A 0.50 kg ball that is tied to the end of a 1.0 m light cord is revolved in a horizontal plane with the cord making a 30 angle, with the vertical (See Fig. P7.52.)

I managed to get my answer down to solving for theta however I dont know how to do the math:

I solved for T = mg / cos(x)
The Fc is = Tsin(x) = (mg)[tan(x)]
(mg)[tan(x)] = (mv^2) / r
R (radius) = 1sin(x)

The following equation is where I am stumped...
sin^2(x) / cos(x) = 8 / 4.9
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p7-52.gif  
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rsk
#2
Nov1-06, 02:25 PM
P: 147
Are you trying to find the speed?

Calculate Fc from Tsin(x) = (mg)[tan(x)]

Then calculate r, and you can find v.

I don't understand why you're stumped at that point, since you know what x is.
Touchme
#3
Nov1-06, 02:27 PM
P: 41
sorry for not clarifying my question. I am trying to solve for x. The question is: If, instead, the ball is revolved so that its speed is 4.0 m/s, what angle does the cord make with the vertical?

rsk
#4
Nov1-06, 03:04 PM
P: 147
Centripetal Force

OK - sorry for delay - site wouldn't let me post

sin^2(x) / cos(x) = 8 / 4.9

Gives you 4.9 sin^2 x = 8 cos^2 x

Remember sin^2 x = (1 - cos^2 x) - substitute this and you should be able to solve for cos x (might be a quadratic you have to solve)
Touchme
#5
Nov1-06, 03:08 PM
P: 41
thank you so much.


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