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Conical Pendulum - help?

  • Thread starter haddow64
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  • #1
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1. The string of a conical pendulum is 1m long. An initial speed of 3m/s causes the bob to describe a horizontal circle. Find the inclination of the string to the vertical and the time of revolution.
The answers at the back give 50.1 and 1.6s

I'm trying to find the angle that the string makes with the vertical (I've called it A)

I have constructed 2 equations after making a fbd or the pendulum.

(1) TcosA=mg
(2) T=mLw^2

These give me equation 3, by replacing Tension in equation 1 then the masses cancel out.

(3) cosA=g/Lw^2

I put my numbers in

cosA=9.81/1*3^2
cosA=9.81/9
cosA=1.09

I cant see where I'm going wrong.
 
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Answers and Replies

  • #2
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(1) TcosA=mg
(2) T=mLw^2
What are w and L in the last equation above? In any case, the last equation is wrong. Please check your diagram again.

These give me equation 3, by replacing Tension in equation 1 then the masses cancel out.
Three equations? I count only two.
 
  • #3
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What are w and L in the last equation above? In any case, the last equation is wrong. Please check your diagram again.
w is the angular speed of the mass. (3ms)
L is the length of string (1m)

Heres the digram i drew (not very good with paint)

http://img2.freeimagehosting.net/uploads/e1b4cfa5c1.jpg [Broken]

Ho would I use my first 2 equations to form the third one and then solve? I'm not too sure what do do to resolve the forces.
 
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  • #4
Dick
Science Advisor
Homework Helper
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1. The string of a conical pendulum is 1m long. An initial speed of 3m/s causes the bob to describe a horizontal circle. Find the inclination of the string to the vertical and the time of revolution.
The answers at the back give 50.1 and 1.6s

I'm trying to find the angle that the string makes with the vertical (I've called it A)

I have constructed 2 equations after making a fbd or the pendulum.

(1) TcosA=mg
(2) T=mLw^2

These give me equation 3, by replacing Tension in equation 1 then the masses cancel out.

(3) cosA=g/Lw^2

I put my numbers in

cosA=9.81/1*3^2
cosA=9.81/9
cosA=1.09

I cant see where I'm going wrong.
Good job, actually. Except that when you finally put your numbers in w (omega, right?) is not 3m/s. w is the angular velocity in radians per second. If you carry your units to the end you can catch stuff like this. Cos(A) should be dimensionless.
 
  • #5
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w is the angular speed of the mass. (3ms)
L is the length of string (1m)
Wrong, the angular speed is not 3 meters / second. (Hint: What is the SI unit for angular velocity?) Your diagram also has an error: You have an arrow pointing to the right. I pressume that arrow was supposed to be the centripetal force vector. Where should that arrow be pointing?
 
  • #6
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So how can I find the angular velocity without the radius or period?
 
  • #7
Dick
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You have the radius=Lsin(A) and the linear speed. That's all you need. Express w in terms of those two.
 
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  • #8
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I'm probably being really obtuse but how would I find it if I don't have angle A?
 
  • #9
Dick
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You'll have to leave the radius as L*sin(A). Rewrite your final solution in terms of v rather than w and then think about solving for A. You'll have both sin(A) and cos(A) hanging around.
 
  • #10
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I think i need another nudge (or shove) in the right direction, I'm not entirely sure what you mean.

Replace w^2 in T=mLw^2 with LsinA?

is this what you mean or am i heading in the entirely wrong direction?
 
  • #11
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No. That doesn't even make sense. Why would you do that? Dick said to rewrite the RADIUS as L sin A.
 
  • #12
Dick
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There is a relation between rotational speed, radius and angular velocity. I'm really sure of it. Can you help us remember it???
 
  • #13
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Thanx for your help, you both nudged me in the right direction.

I finally managed to get it, on the angle I got 50.7 (.6 out but thats probably just rounding) And I got the period.

It was easier than I thought I was trying all sorts of weird and wonderful things when I wasn't thinking properly.

Thanx again for your help :D
 

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