Calculating the Angle of a Pendulum Swing into a Peg

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The discussion revolves around calculating the angle θ of a pendulum swing into a peg using the equation cosθ = r/L * cosα - √(3)/2 * (1 - r/L). The user expresses difficulty in progressing from their current solution attempt, specifically regarding the application of Newtonian position formulas, which are deemed unsuitable for non-uniform acceleration scenarios. Suggestions are made to consider conditions for angle β and the implications of the pendulum's trajectory on its motion. The conversation highlights the need for a deeper understanding of changing acceleration and its effects on the pendulum's behavior. Overall, the focus is on resolving the mathematical aspects of the pendulum problem rather than the physics fundamentals.
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Homework Statement


With this problem I have to get the answer: cosθ = r/L * cosα - √(3)/2 * (1 - r/L)
which in other words mean I need to find angle θ with arccos[r/L * cosα - √(3)/2 * (1 - r/L)].

Here's the picture:
11d636180c.jpg

Lcosθ is the vertical length of the string at its lowest point.

rcosα is a fraction of that same vertical string in terms of displacement "r" (which is from the start of the string to the peg)

∠β is the angle between the peg and the horizon.

(L-r) sinβ is the height from the end of the peg and the horizontal

(L-r) cosα is the horizontal length of that same peg.

(L-r)cosα is the vertical length of the string from the ball to the peg.

So, this is not really a physics issue but more like a math issue but since this is a physics problem I've decided to put it under here.

My problem is that I am unable to continue from this point as shown on the picture of my attempt. I don't know where to continue from here on out. I am trying to find "t" for the equation but I am unsure how. Where do I continue from now?

Homework Equations


[/B]
Newtonian Position Formula:
yf = yi +viyt + .5gt2
xf = xi +vixt + .5gt2

Energy Equation:
Work of hand - force of friction * displacement = delta Kinetic Energy + delta Potential Energy

Wh - fF*d = [.5*mvf2 - .5*mvi2] - [mghf - mghi]

The Attempt at a Solution


Picture of Attempt:
e4bfe92db3.jpg
[/B]
 
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Hello Stranger, :welcome:

Your position formula is only valid for uniform acceleration. You don't have that here !
Can you think of a condition you can impose on ##\beta## ?
 
BvU said:
Hello Stranger, :welcome:

Your position formula is only valid for uniform acceleration. You don't have that here !
Can you think of a condition you can impose on ##\beta## ?

Thanks for the welcome :).
The only method I can think of when dealing with changing acceleration by breaking it into parts. Each part for every time the value of acceleration changes. I have not yet learned ho to derive very well but I know it exists. As for angle β I am clueless on what to impose.
 
Well, then perhaps you can conquer this one without dealing with changing acceleration ?
The sketch suggests a trajectory, but is it realistic ? Where must the mass run out of sped to fall on the peg ? What would happen if it ran out of speed at e.g. 85 degrees ?
 
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?

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