Pendulum: Generic Solution for Angle

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Homework Help Overview

The discussion revolves around a pendulum system involving an inelastic collision with a ball, where the original poster seeks to determine the generic solution for the angle (theta) that the pendulum swings after the collision. The context includes concepts of momentum and energy conservation, particularly in non-simple harmonic motion scenarios.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the conservation of momentum before and after the collision, as well as the conservation of energy during the pendulum's ascent. Questions arise regarding the definitions of initial and final conditions for potential and kinetic energy, and how to relate these to the angle of swing.

Discussion Status

The discussion has progressed with participants breaking down the problem into manageable parts, focusing on the collision and subsequent motion of the pendulum. Some guidance has been provided regarding the use of energy equations and the need to express height in terms of the angle theta. There is an ongoing exploration of the relationships between variables involved.

Contextual Notes

Participants are navigating the complexities of inelastic collisions and the implications for energy conservation, questioning how to accurately represent the height in relation to the angle of the pendulum's swing. There is an acknowledgment of the need for diagrams to clarify relationships between variables.

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[SOLVED] Pendulum: Generic Solution for Angle

Hey guys, I'm a little lost on this one, not sure where to start.

I'm doing a lab involving a pendulum bob being hit by a ball and them sticking together (inelastic collision). Almost like a ballistic pendulum right? Anyway, I need to find the generic solution for the angle (theta) that the pendulum swings.

From what I understand this is not simple harmonic motion in general, and can only use that as a model for very small angular displacements. Also momentum is only conserved right before the collision and right after the collision, right? Other than that I really don't know where to start. I figure that the angle has to relate to the velocity of the bullet, the force exerted on the bob by the bullet, and angular velocity of the two together.

Any help to set me on track would be greatly appreciated!
 
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Momentum conservation is useful before to immediately after the collision, after which energy conservation takes you the rest of the way home.
 
But isn't energy not conserved because the collision is inelastic?
 
Attack this problem in two steps:
(1) The collision. What's conserved?
(2) The rising of the pendulum after the collision. What's conserved?
 
1. Ok at the collision, momentum is conserved p(i) = p(f)
mv(i)(bullet) = [m(bullet) + m(bob)]v(f)

2. As the pendulum is rising, energy is conserved? E(i) = E(f)
so...PE(i) + KE(i) = PE(f) + KE(f) --> ok here's where I am a little confused, the initial conditions for PE and KE, they are for the instant after the collision has taken place? and the final KE and PE are for when the pendulum reaches its maximum height?

the arc the pendulum makes equals (r)(θ) right? So angular velocity would be (r)(θ)/t? This all makes sense to me for the most part, but I don't understand how to put it all together.


lol thank you though for making me think of it in two parts, it makes way more sense this way.
 
what factors do I include in the equation of theta? can I use kinematics equations?
 
Hi Return,

Return said:
1. Ok at the collision, momentum is conserved p(i) = p(f)
mv(i)(bullet) = [m(bullet) + m(bob)]v(f)

2. As the pendulum is rising, energy is conserved? E(i) = E(f)
so...PE(i) + KE(i) = PE(f) + KE(f) --> ok here's where I am a little confused, the initial conditions for PE and KE, they are for the instant after the collision has taken place? and the final KE and PE are for when the pendulum reaches its maximum height?

That's right; and you have formula for PE and KE. Use them in the energy equation; once you see what you have you'll know what you still have to find. What do you get for the energy equation?
 
ok so i end up getting

V^2/2 = gh

so I need to express h in terms of theta right? Lsin(theta)?
 
Return said:
ok so i end up getting

V^2/2 = gh
Right, where V is your v(f) from the momentum equation.

so I need to express h in terms of theta right?
Yes.
Lsin(theta)?
No, that gives you the horizontal distance; you want the vertical. (Draw yourself a diagram.)
 
  • #10
phew, ok thanks for the help, i got the generic solution earlier today!
 

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