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Centripetal Force 
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#1
Nov106, 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 


#2
Nov106, 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. 


#3
Nov106, 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?



#4
Nov106, 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) 


#5
Nov106, 03:08 PM

P: 41

thank you so much.



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