Swinging Pendulum

I've been thinking about Pendulums recently. Most people would probably already know that the highest speed reached by a pendulum is at the bottom, where all of it's potential energy has been turned into kinetic energy. But rather, for a pendulum with a string with the length 'L' , Dropped from 90, what is the highest vertical velocity reached by the weight? At which degree does this happen? How would this work if it was instead dropped from 0 < θ < 90*?

#Mechanics

Easy Math Editor

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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

Let $\theta$ be the angle with the downward vertical, and let gravity be downward. Suppose the pendulum is released from the horizontal $(\theta = \pi/2)$ . Sweep $\theta$ , and for each $\theta$ value, calculate the change in potential energy relative to the start position. Then convert this into speed $(\Delta U = m g \Delta y = m g L \cos \theta = \frac{1}{2} m v^2)$ . Then calculate the vertical component as $v_y = v \sin \theta$ . It turns out that $v_y$ is maximized when $\theta$ is approximately $54.735$ degrees.

This procedure can be generalized to use any starting position

import math

m = 1.0   # mass
g = 10.0  # gravity
L = 1.0   # string length

N = 10**5

# theta is angle with the vertical

dtheta = (math.pi/2.0)/N   # small increment in theta

theta = math.pi/2.0 - dtheta

vymax = 0.0

##########################################################

while theta >= dtheta:

    delta_y = L*math.cos(theta)   # change in vertical position....
                                   # relative to horizontal
    delta_U = m*g*delta_y         # change in grav potential energy

    # delta_U = 0.5*m*vsq        # change in grav potential converts to kinetic

    vsq = 2.0*delta_U/m

    v = math.sqrt(vsq)           # speed

    vy = v * math.sin(theta)     # vertical component of velocity

    if vy > vymax:               # store theta for max vy
        vymax = vy
        theta_store = theta

    theta = theta - dtheta

##########################################################

print (theta_store * 180.0 / math.pi)
# 54.735 degrees

Steven Chase - 1 year, 6 months ago

Yup....the exact value of the angle is $\tan^{-1}\left(\sqrt{2}\right)$

Aaghaz Mahajan - 1 year, 6 months ago

This was really helpful, thank you for the explanation! I had been trying to use derive a really really long and confusing equation but <i gave up halfway through.

Daniel Sandberg - 1 year, 6 months ago

You're welcome. Glad you found it to be helpful

Steven Chase - 1 year, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$