- #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##

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.

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