Angle that makes kinetic and potential energy of simple pendulum equal

  • #1
DrBanana
35
3
Homework Statement
The max speed and the length of the string of a pendulum is given. At what angle (with regard to the vertical y axis) is the kinetic energy and potential energy the same? Assume the maximum angle displacement is small, so that the pendulum udergoes simple harmonic motion.
Relevant Equations
##E_{total}=0.5mv_{max}^2##
pendulumquestion.png


On the first attempt I used conservation of energy to get it down to a single equation involving theta and v (the speed when the angle is theta). But I had no idea how to find v.

Now since the maximum speed is given, it is possible to find the maximum angular displacement, ##\alpha##. Then, using conservation of energy, it is possible to show that ##v^2=2gL(cos\theta - cos\alpha)##. Then came my hail mary attempt, so to speak.
I don;t know if this is legal for a single particle, but I wrote ##\frac{1}{2}I \omega^2 = \int_{0}^{\beta}\tau d \beta##, where ##\tau=-Lmgsin\theta## and after solving the integral I use ##\beta=\alpha=\theta##. Of course, ##I=mL^2##. Using the fact that tangential speed=L*angular speed, it is possible to get another equation for v using this integral. Then equating the two v's: ##cos\theta-cos\alpha=1-cos(\alpha-\theta)##.
An observation is, I didn't actually use the fact that this is simple harmonic motion.
Is this way ok?

Now, I found a solution elsewhere. Though I do not understand it, it eerily yields a similar result:
##v_{max}^2=2gx##, x is the distance such that ##OBcos\theta+x=OA## (look at the figure). How it's assumed that the particle undergoes constant acceleration, I'm not sure. Using this equation, theta is calculated. At first this was the wrong approach, but the answer is the same as the one I get. Of course there is a possibility both methods are wrong.

Edit: It just occurred to me neither approach makes use of the fact that the two energies are equal. So how'd I still get an angle?

I can't believe I didn't get this before, but it should be as simple as this:
##2PE=E_{total}##
##2mgL(1-cos\theta)=mgL(1-cos\alpha)##
##\theta=9.166^{\circle}##

Still I would like to know what happened in my first attempt.
 
Last edited:
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  • #2
The snarky answer is: "Every angle that the pendulum achieves".

Potential energy depends on an arbitrarily chosen reference. Choose the right reference and potential energy is equal to kinetic energy.

For that matter, kinetic energy also depends on an arbitrarily chosen reference frame. Choose the right reference frame and kinetic energy can take on any arbitrarily chosen [non-negative] value.
 
Last edited:
  • #3
DrBanana said:
Now, I found a solution elsewhere. Though I do not understand it, it eerily yields a similar result:
##v_{max}^2=2gx##, x is the distance such that ##OBcos\theta+x=OA## (look at the figure). How it's assumed that the particle undergoes constant acceleration, I'm not sure. Using this equation, theta is calculated. At first this was the wrong approach, but the answer is the same as the one I get. Of course there is a possibility both methods are wrong.
This is not a solution to the same problem. Distance ##x## is the vertical distance from the lowest point of the motion when the pendulum is at angle ##\theta## and at rest. Note that ##v_{max}^2=2gx## does not necessarily imply constant acceleration. It is a mechanical energy conservation equation because only gravity does work on the mass, so it's as if the mass were in free fall over height ##x##. If you write ##KE+PE## at height of release (##KE=0##) and at the lowest point (##PE=0##), $$0+mgx=\frac{1}{2}mv_{max}^2\implies v_{max}^2=2gx.$$The equation that you found relates the maximum speed to the angle of release from rest.
 

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