Tough physics problem. Conical pendulum.

Click For Summary
SUMMARY

The discussion focuses on solving a physics problem involving a conical pendulum, where the tension in the string and the period of motion are to be determined. The user correctly identifies the components of tension as Tx = Tsinθ and Ty = Tcosθ, leading to the equations Tcosθ = mg and Tsinθ = mv²/r. The final expression for tension is derived as T = mgL/sqrt(L² - (C/2π)²), which is validated by another participant in the discussion. The approach utilizes fundamental physics concepts such as centripetal force and the Pythagorean theorem.

PREREQUISITES
  • Understanding of conical pendulum dynamics
  • Knowledge of centripetal force equations
  • Familiarity with trigonometric functions in physics
  • Ability to apply the Pythagorean theorem in problem-solving
NEXT STEPS
  • Study the derivation of the period of a conical pendulum
  • Learn about the relationship between tension and circular motion
  • Explore advanced applications of centripetal force in different contexts
  • Review examples of conical pendulum problems in physics textbooks
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators looking for examples of conical pendulum problems.

dlreynolds562
Messages
1
Reaction score
0

Homework Statement



Hey, thanks for taking a look at this.

"The figure below shows a conical pendulum, in which the bob (the small object at the lower end of the cord) moves in a horizontal circle at constant speed. (The cord sweeps out a cone as the bob rotates.) The bob has a mass m, the string has a length L and negligible mass, and the bob follows a circular path of the circumference C. What are the tension in the string and the period of the motion? (Use any variable or symbol stated above along with the following as necessary: g for the acceleration of gravity.)"

The figure is just the conical pendulum as described.


Homework Equations





The Attempt at a Solution



I start by breaking tension into x and y components.

Tx = Tsinθ & Ty = Tcosθ

Force balance in the y direction

ƩFy = Ty - mg = 0
=> Tcosθ = mg

Force balance in the x direction

ƩFx = Tsinθ

I guess this is just equal to the centripetal force?

Tsinθ = mv^2/r

... so I guess to answer the first question T = mg/cosθ, but it says not to leave it in terms of theta

cosθ = h/L

using Pythagorean theorem

h^2 = L^2 - R^2

and also C = 2∏r, -> r = C/(2∏)

h = sqrt(L^2 - (C/(2∏)^2)

Now plugging all that into the equation from above

Tcosθ = mg

gives

T = mgL/sqrt(L^2-(C/2∏)^2)

This is what I have so far and I really don't think it is right.
 
Physics news on Phys.org
Why do you think that is not right?
 
Your tension expression looks okay to me :smile:
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
2
Views
1K
Replies
8
Views
3K
Replies
1
Views
1K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
2K