Conical pendulum- angle? help please

In summary: Yes, cos^2theta + (v^2/lg)cos theta -1 is the correct equation. To solve for the angle, use the standard quadratic equation:x^2+y^2=r^2.
  • #1
fruitl00p
94
0

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:
 
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  • #2
You need an expression for your "a(radius)". It's moving in circle with a given velocity. So?
 
  • #3
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
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
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
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
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
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
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
Edgardo said:
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
Dick said:
That's not even dimensionally correct. fruitl00p is almost there. He just needs to realize he has a quadratic equation in cos(theta).

Yes, thanks 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
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
Your equation is correct. Where exactly do you get an error (what did you type in your calculator)?
 
  • #14
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
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
Thank you so much, I got the answer!
 

1. What is a conical pendulum?

A conical pendulum is a type of pendulum where the bob, or weight, is attached to the end of a string or rod and swings in a circular motion rather than back and forth. The string or rod is fixed at the top, creating a cone-shaped path for the bob to follow.

2. How is the angle of a conical pendulum measured?

The angle of a conical pendulum is measured as the angle between the string or rod and the vertical line passing through the fixed point at the top. This angle is typically denoted as θ.

3. What factors affect the angle of a conical pendulum?

The angle of a conical pendulum is affected by the length of the string or rod, the mass of the bob, the velocity of the bob, and the strength of the gravitational force acting on the bob. Other factors that can affect the angle include air resistance and the shape of the bob.

4. How does the angle of a conical pendulum change over time?

The angle of a conical pendulum will change as the bob swings around the fixed point. The angle will be at its maximum when the bob is directly below the fixed point, and it will decrease as the bob swings towards the sides of the cone. The angle will then increase again as the bob swings back towards the bottom of the cone.

5. How does the angle of a conical pendulum affect the period of its motion?

The angle of a conical pendulum does not directly affect the period of its motion. The period is primarily determined by the length of the string or rod and the strength of the gravitational force. However, a larger angle can result in a larger distance traveled by the bob, which can affect the period of the motion. Additionally, a larger angle can also result in a larger tension in the string or rod, which can affect the period as well.

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