Calculating Inclination & Time of Revolution for a Conical Pendulum

In summary, the string of a conical pendulum makes a horizontal circle at an initial speed of 3 meters per second. The inclination of the string to the vertical and the time of revolution are found by solving equations 3 and 2.
  • #1
haddow64
14
0
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 can't see where I'm going wrong.
 
Last edited:
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  • #2
haddow64 said:
(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
e(ho0n3 said:
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.
 
Last edited by a moderator:
  • #4
haddow64 said:
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 can't 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
haddow64 said:
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
So how can I find the angular velocity without the radius or period?
 
  • #7
You have the radius=Lsin(A) and the linear speed. That's all you need. Express w in terms of those two.
 
Last edited:
  • #8
I'm probably being really obtuse but how would I find it if I don't have angle A?
 
  • #9
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
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
No. That doesn't even make sense. Why would you do that? Dick said to rewrite the RADIUS as L sin A.
 
  • #12
There is a relation between rotational speed, radius and angular velocity. I'm really sure of it. Can you help us remember it?
 
  • #13
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 that's 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
 

What is a conical pendulum?

A conical pendulum is a type of pendulum where the bob (weight) of the pendulum moves in a circular motion instead of just back and forth. This motion is caused by the bob being attached to a string that is fixed at one end and allowed to swing freely at the other end.

What factors affect the motion of a conical pendulum?

The motion of a conical pendulum is affected by the length of the string, the mass of the bob, and the angle at which the string is released. Other factors like air resistance and friction may also have an impact on the motion.

How is the period of a conical pendulum calculated?

The period of a conical pendulum can be calculated using the formula: T = 2π√(L/g), where T is the period in seconds, L is the length of the string in meters, and g is the acceleration due to gravity in meters per second squared.

What is the purpose of a conical pendulum?

A conical pendulum can be used to demonstrate the principles of circular motion and can also be used to measure the acceleration due to gravity. It is also a common experiment in physics to study the relationship between the length of the string and the period of the pendulum.

What are some real-world applications of conical pendulums?

Conical pendulums are used in various fields such as astronomy, engineering, and medicine. In astronomy, they are used to measure the rotation of planets and stars. In engineering, they are used in the design of certain types of machinery. In medicine, they are used in the development of certain medical devices such as artificial hearts.

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