Solving a Hard Energy Physics Problem: Pendulum Cut Question

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Homework Help Overview

The discussion revolves around a physics problem involving energy conservation in a pendulum system, specifically focusing on the scenario where the pendulum string is cut. Participants are exploring the relationships between potential energy (Ep) and kinetic energy (Ek) at various points in the pendulum's motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the conservation of energy, questioning how to relate the potential energy at the highest point of the pendulum's swing to the kinetic energy just after the string is cut. There are attempts to derive equations for energy at different positions and to calculate velocities and heights.

Discussion Status

Some participants have provided guidance on setting up energy equations and calculating velocities, while others are exploring different interpretations of the problem. There is an ongoing dialogue about the relevance of horizontal motion and how it affects the energy calculations.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available and the methods they can use. There is a focus on understanding the implications of energy conservation without reaching a definitive conclusion.

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Homework Statement




http://img694.imageshack.us/img694/1115/pendulumcutquestion.jpg


Please ignore the blue pen.

Homework Equations


Ek=1/2 mv2
Ep=mgh


The Attempt at a Solution



Basically I looked at the problem firstly as the pendulum has reached the lowest point, therefore having an Ep of 0 and an Ek . Then I took a final point, where the string is cut. At this point there is Ek and Ep is there not?

The height would therefore be from the bottom of the swing, to the place where it's cut, which equals

6.4 - 6.4cos(30)

So now I have this

Ek= Ek2 + Ep
0.5mv2 = 0.5m(vsin30)2 + mgh


Am I on the right track?

What do I do after this? I don't know what my next move should be...

Should I use the point where the string is cut , and set it equal to the ball at the top of the height's energy?
 
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It's hard to say if you're on the right track or not, since you haven't provided a clear statement of what some of your formulas are meant to represent. Also you've used the variable 'v' on both sides of an energy equation, so what does the formula mean?

I think if you can put a number to the velocity of the bob at position Y then you'll be making excellent progress. What's the KE of the bob at point Y?
 
You're almost there, but your energy equation is a little off. Total energy is conserved, so find the speed at the height 6.4 - 6.4cos(30). (You start off with all potential energy, and end up with potential and kinetic energy.) Once you know the speed, you can find the vertical velocity, because you know that the path of the ball with be orthogonal to a line drawn from the ball to the point where the string is fixed.
 
Okay so this is what I've done.

I've taken the Ep (energy potential) when the ball is at the top (position X) and set it equal to the Ek (energy kinetic) and the new Ep2 at position Y, Thus obtaining the following:

Ep = Ep2 + Ek
mg(6.4) = mg(6.4-6.4cos(30)) + 0.5mv2

Solving for v, I get 9.9m/s


Now, I am looking at the problem starting at position Y and resulting at the max point of h.

So

Ep2 + Ek = Ep3 + Ek2


Ek2 = .5(9.9sin30)2

And solving for h I get 4.6

Is this right? Or does Ek2 not exist since the ball is no longer traveling in the y direction?!
 
You can find the energy when the pendulum is in position X since the only energy is potential, the diagram also gives you the length of the pendulum and the angle at which the rope is cut therefore you can find the potential energy at this point; the difference in these potentials is the kinetic energy at this point, from this and the angle you have you can find the vertical velocity and a kinematics equation will give you your max height (or another energy conservation application if that's more your style)
 
JHamm said:
You can find the energy when the pendulum is in position X since the only energy is potential, the diagram also gives you the length of the pendulum and the angle at which the rope is cut therefore you can find the potential energy at this point; the difference in these potentials is the kinetic energy at this point, from this and the angle you have you can find the vertical velocity and a kinematics equation will give you your max height (or another energy conservation application if that's more your style)

Okay so I did this, and it makes perfect sense to me! I used the V'2 = Vi2 -2ad formula and I got 3.43m which I believe is correct.

How would I use another energy conservation application though to solve for the answer? I don't really know what to have as an equation.
 
Just ignore any horizontal displacement. Imagine that you have a ball that is tossed up with some vertical velocity. How high does it go above where it is released? You might be making this problem more complicated than it is.
 
gnulinger said:
Just ignore any horizontal displacement. Imagine that you have a ball that is tossed up with some vertical velocity. How high does it go above where it is released? You might be making this problem more complicated than it is.

I know the concept, I mean I just don't really understand it written out in an equation.

When the ball is cut from the string, and it's at the highest point, it has potential energy and kinetic energy does it not? I mean it is still moving? In the X at least... How can we not consider this part of the question?

I apologize if I sound stupid, maybe I'm missing something...
 
You don't sound stupid at all, in fact what you said is entirely true :)
Try this, if I have a ball with a vertical velocity of 1ms^{-1} and it reaches a height h then I take this same ball and fire it with a velocity of 2ms^{-1} at an angle of 30^o how high will it go?
 
  • #10
It will go the same height as the last one, as 2sin(30) = 1 m/s !

So basically, I understand that. It's just, now, what do we take from this into the question?
 

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