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Mass in circular motion. Draw Diagram and find/explain variables.

  • Thread starter demenius
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  • #1
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Homework Statement


A ball of mass m is held by a string of length L and swung in a horizontal circle. The string makes an angle θ with the vertical (as shown).
a. Draw a diagram clearly labeling all forces on the mass.
b. Find T, the amount of time that it takes for the ball to complete one circle in terms of m,L, g, and θ.
c. When θ increases, what happens to T? Justify your answer.

http://imageshack.us/photo/my-images/97/circularmotion.png/"

Homework Equations


None Given.


The Attempt at a Solution


I had many attempts but none seemed to work out.
 
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Answers and Replies

  • #2
PeterO
Homework Helper
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Homework Statement


A ball of mass m is held by a string of length L and swung in a horizontal circle. The string makes an angle θ with the vertical (as shown).
a. Draw a diagram clearly labeling all forces on the mass.
b. Find T, the amount of time that it takes for the ball to complete one circle in terms of m,L, g, and θ.
c. When θ increases, what happens to T? Justify your answer.

http://imageshack.us/photo/my-images/97/circularmotion.png/"

Homework Equations


None Given.


The Attempt at a Solution


I had many attempts but none seemed to work out.
Perhaps you could put any one of those attempts here.
 
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  • #3
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  • #4
PeterO
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http://imageshack.us/photo/my-images/412/freebodydiagram.png/"

I found T to be equal to 2∏√(Lcos(θ)/g). But there is no m in that equation. Is it not needed?

If that equation is right, then when θ increases, T would decrease.
You have drawn 4 forces - but two of them are actually one of the forces resolved to allow later calculation, so should not be there. there are only two forces acting, gravity, down, and Tension at an angle.

For me to comment on the accuracy of you expression for T, I would need to see the steps you used to derive it.
Certainly you interpretation that if θ increases, T would decrease is valid for the expression you gave.
 
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  • #5
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Ok. So
Fx = Fsinθ = ma_rad
Fy = Fcosθ - mg = 0 (no vertical acceleration)
a_rad = (4R∏^2)/T^2
R = Lsinθ

F = mg/cosθ Sub into Fx
(mg/cosθ)*sinθ = ma_rad
a_rad = gtanθ

gtanθ = (4R∏^2)/T^2

T = √((4R∏^2)/gtanθ) = 2∏√(R/(gtanθ)) Sub in Lsinθ for R
T = 2∏√(Lcosθ/g)
 
  • #6
PeterO
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Ok. So
Fx = Fsinθ = ma_rad
Fy = Fcosθ - mg = 0 (no vertical acceleration)
a_rad = (4R∏^2)/T^2
R = Lsinθ

F = mg/cosθ Sub into Fx
(mg/cosθ)*sinθ = ma_rad
a_rad = gtanθ

gtanθ = (4R∏^2)/T^2

T = √((4R∏^2)/gtanθ) = 2∏√(R/(gtanθ)) Sub in Lsinθ for R
T = 2∏√(Lcosθ/g)
That looks good, so you should be correct.

Certainly I know that the period decreases as the angle increases, so it is good that your formula predicts that.
 
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  • #7
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I agree strongly with PeterO.
There are only 2 forces acting on the object (if friction, air resistance can be ignored!!!)
1) the tension T in the string
2) The weight (mg) acting vertically down
The tension has a vertical component which equals the weight of the object and a horizontal component which equals the centripetal force.
It is not wise to show these component forces on a diagram..... they confuse the picture and give the appearance that there are too many forces acting.
 

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