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Conical pendulum- angle? help please!

  • Thread starter fruitl00p
  • Start date
  • #1
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Homework Statement


A mass m= 4.7 kg is suspended from a string of length L=1.19m. It revolves in a horizontal circle. The tangential speed of the mass is 2.97 m/s. What is the angle between the string and the vertical (in degree.)


Homework Equations


r= Lsin(theta)
Tension is broken up into x and y components, so Tsin theta=ma(of radius)
and Tcos theta=mg
sin^2theta =1-cos^2theta

The Attempt at a Solution



I'm having a hard time solving this because of the unknown radius along with the unknown angle!

first I tried T=mg/cos theta; then plugged that into the tension in the x axis, making it tan theta=a(of radius)/g

But the equation for a rad is 4pi^2R/t. And the peiod is another unknown.

This is as far as a I got: tan theta= 4pi^2 Lsin theta/ gt^2

The assignment was due a couple of weeks ago and I didn't get the answer correct. But I have a test coming up and on the practice exam the same problem is on it.

I was told to use quadratic equation to solve for angle, which makes sense but for the life of me, I just can't do it. I don't understand how to use the quadratic equation to solve it. Please someone help me. :frown:

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Dick
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You need an expression for your "a(radius)". It's moving in circle with a given velocity. So?
 
  • #3
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The correct answer was 46.3 degrees.

Dick, I thought I did type down the expression for a(rad).
a(rad) =4pi^2r/t. I realize that there is another way to express a(rad): v^2/r
But the r is still unknown and I change a(rad) to equal v^2/l sin theta, I still do not understand how to use the quadratic equation to solve for the angle.
(I am not great at math :frown: )
 
  • #4
Dick
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Ah, yes you did. But you meant 4*pi^2*r/t^2, right? I would just use the v^2/r form (since v is given). You should now be able to write down two equations for the balance of the horizontal forces and vertical forces. There should be exactly two unknowns, the theta and the tension. Did you get that far?
 
  • #5
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Edgardo, I am not sure what you mean. I did make a force vector diagram.
since the circular motion is horizontal, the acceleration in the x axis goes toward the circle and so there will be force only on the x axis since the vertical experiences no acceleration.
And I think I am supposed to assume that the angle does not alter.
 
  • #6
703
13
Sorry, I just deleted my post. But I repost it again:

Do you have an answer given (formula or numeric) so I can check my calculations? I just want to make sure my solution is correct and I can give you some correct advise.

However, I would start by making a drawing and drawing some force vectors.

Which condition do you find for the forces (note that the angle is static)?
Is it said in the problem that you have to use those relevant equations for the tension in x- and y-axis? I'm asking because I somehow used another approach (I could be wrong though).
 
  • #7
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ok, right now I have for Fx: Tsin theta =ma(rad)
Tsin theta= m v^2/r = mv^2/l sin theta
for Fy: Tcos theta=mg

then T= mg/cos theta
and putting Fx and Fy together, I got (mg/cos theta)*sin theta= (mv^2)/(l sin theta)

tan theta=v^2/(lsin theta)g

is this right so far?

Edgardo, how did you approach the problem?
 
  • #8
Dick
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It's perfect. But leave tan(theta) as sin(theta)/cos(theta) and multiply both sides by sin(theta). Now replace the sin(theta)^2 by (guess).
 
  • #9
703
13
Hello frootl00p,

I have the same result [tex]\rm{tan} \theta = \frac{mv^2}{gR}[/tex].
But I have another angle as result, 86.2°. I will check again.
 
  • #10
Dick
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Hello frootl00p,

I have the same result [tex]\rm{tan} \theta = \frac{mv^2}{gR}[/tex].
But I have another angle as result, 86.2°. I will check again.
That's not even dimensionally correct. fruitl00p is almost there. He just needs to realize he has a quadratic equation in cos(theta).
 
  • #11
703
13
That's not even dimensionally correct. fruitl00p is almost there. He just needs to realize he has a quadratic equation in cos(theta).
Yes, thx for the correction. It should be

[tex]\rm{tan} \theta = \frac{a}{g}=\frac{v^2}{Rg}[/tex]

With Dick's correction I also get the correct angle of 46.3°.

But you should follow Dick's instruction in post #8 and #10.
 
Last edited:
  • #12
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ok...since sin^2theta= 1-cos^2theta, then would it be:
1-cos^2theta/cos theta=v^2/lg

so is it cos^2theta + (v^2/lg)cos theta -1

I tried to use the quadratic equation formula, but I keep getting error on my calculator. Is that the correct formula and how do I solve it?
 
  • #13
703
13
Your equation is correct. Where exactly do you get an error (what did you type in your calculator)?
 
  • #14
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Well, just putting cos without any value in the parenthesis doesn't work.
Now I am really confused because sin^2 theta= (1-cos2 theta)/2 due to double agents.

then it changes to 2sin^2theta - 1 = cos2theta =(v^2/lg)cos theta

How did you plug in the numbers in your calculator because I guess I have no idea how to use my calculator!:blushing:
 
  • #15
Dick
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You are SO CLOSE.
Let x=cos(theta). So your equation is:
x^2+(v^2/lg)*x-1=0. v^2/lg is just a number. So this is a quadratic equation in x. Look up the quadratic formula to find the two roots. Then take the inverse cosine of the result. Don't be afraid of quadratics and don't fear the reaper.
 
  • #16
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Thank you so much, I got the answer!!
 

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