A ball is swung in a circular path

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234jazzy2
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Homework Statement


A 0.5 kg ball is swung in a circular path from a 80 cm long rope, attached to a vertical pole.
A. What is the speed of the ball theta (between the pole and rope) is 40 deg?
B. What is the KE of the ball at theta = 40 deg?
C. The ball slows down and the KE drops to 50% of the value in (B). What is the new value of theta?

Homework Equations


F = ma
centripetal acceleration = V^2/r

The Attempt at a Solution



A.
Fy = 0
Tcos(theta) = mg
Fx = mv^2/r
Tsin(theta) = mv^2/r - > V = sqrt(g*r*tan(theta)) = 2.056 m/s
B. KE = 0.5 *m*v^2 = 1.057151179
C. I get all the have to new velocity but i don't know how to get the angle... I need some pointers.

Also, i am not sure if this is the right approach.

Thanks
 
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Hello jazzy, :welcome:

Looks like the right approach. A few remarks: KE = 1.06 J (don't forget the units and don't give many more digits than the given variables have -- but if the first digit is a one, then give one more).

For C, you have the same equilibrium equation (##\ v^2 = g\, r \tan\theta\ ##), only now v is given and ##\theta## has to be determined. Your problem is then the goniometric equation when you put in ##r = L \sin\theta## (L is the length of the rope).

If you have no way to solve this, perhaps you are supposed to find the answer with trial and error ?
 
Yea, i get suck at the trig. And, it's definitely not trial and error. Trying different reference frame to see if I can get rid of a trig.
 
234jazzy2 said:
Yea, i get suck at the trig. And, it's definitely not trial and error. Trying different reference frame to see if I can get rid of a trig.
What trig equation do you get? Something like sin(θ)tan(θ)=value? There is an analytic way to solve that.
 
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(Sin^2(theta))/( cos(theta)) = some number. I tried using some trig identities but nothing seemed to work. As I write this, I think I could've solved it, because that also equals (1 - cos^ 2(theta))/ cos(theta) = something and set x = cos(theta) and sove the quadratic. But that will give two answers... I'll solve it later. But if you have any othersuggestions, please let me know.