Pendulum velocity at any given angle

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Discussion Overview

The discussion centers on determining the velocity of a pendulum bob at various angles, specifically when released from a 90° angle relative to its equilibrium position. Participants explore equations related to pendulum motion, energy conservation, and the impact of mass on velocity calculations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant presents an equation for pendulum bob velocity at any angle, questioning its validity since it yields a higher velocity at 45° than at the equilibrium position.
  • Another participant asserts that the speed at any point is independent of mass and confirms that the maximum speed occurs at the bottom of the swing, calculated using conservation of energy.
  • A different participant suggests a formula for velocity that does not include mass and emphasizes that the maximum velocity occurs at 0° angle.
  • One participant shares their approach using energy conservation to derive the velocity at various angles, concluding with specific velocities at 90° and 0° angles.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the initial velocity equation and the role of mass in the calculations. There is no consensus on the correct approach to determining pendulum velocity at various angles, and multiple competing views remain.

Contextual Notes

Some participants note that the equations presented may depend on specific assumptions about the pendulum's motion and energy conservation principles. The discussion does not resolve the discrepancies in the equations or the interpretations of energy conservation.

lgunseor
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Trying to find an equation that will give me a pendulum bob velocity at any given angle if the pendulum is released from a 90° angle from it's equilibrium position. My parameters are as follows

9.8 (acceleration of gravity m/s^2)
1.397 (length of pendulum in meters)
.138 (mass of pendulum bob in kilograms)
90 (Θmax, maximum angle of the pendulum in degrees)
45 (Θ, the angle of interest for velocity in degrees)

Below is the equation that I found

v=√2*g*l/m*(cos(Θ) – cos(Θmax))
v=11.84

My confusion is that the equation for pendulum velocity at it's equilibrium position which should be the maximum velocity is less that at 45° in the above equation

v=√2*g*l*(1-cos(Θmax))
v=5.23

I understand that mass is in the first equation and not the second equation, but is the first equation correct for velocity at any angle? Is the equation for maximum velocity at the pendulum equilibrium position not correct? Any help would be appreciated.
 
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lgunseor said:
Trying to find an equation that will give me a pendulum bob velocity at any given angle if the pendulum is released from a 90° angle from it's equilibrium position. My parameters are as follows

9.8 (acceleration of gravity m/s^2)
1.397 (length of pendulum in meters)
.138 (mass of pendulum bob in kilograms)
90 (Θmax, maximum angle of the pendulum in degrees)
45 (Θ, the angle of interest for velocity in degrees)

Below is the equation that I found

v=√2*g*l/m*(cos(Θ) – cos(Θmax))
v=11.84

My confusion is that the equation for pendulum velocity at it's equilibrium position which should be the maximum velocity is less that at 45° in the above equation

v=√2*g*l*(1-cos(Θmax))
v=5.23

I understand that mass is in the first equation and not the second equation, but is the first equation correct for velocity at any angle? Is the equation for maximum velocity at the pendulum equilibrium position not correct? Any help would be appreciated.
I don't know where your equations are coming from. The speed at any point is independent of the mass. You are correct that the max speed is at the bottom of the swing and equal to v=√2*g*l = 5.23 m/s which comes from conservation of energy ...initial potential energy = final kinetic energy...you can do the same at 45 degrees to calculate the speed, which must be less than the max...the initial PE is mgh where h from some trig is 1.397sin45...then mgh = 1/2mv^2, m cancels, solve for v = 4.4 m/s.
 
lgunseor said:
Trying to find an equation that will give me a pendulum bob velocity at any given angle if the pendulum is released from a 90° angle from it's equilibrium position. My parameters are as follows

9.8 (acceleration of gravity m/s^2)
1.397 (length of pendulum in meters)
.138 (mass of pendulum bob in kilograms)
90 (Θmax, maximum angle of the pendulum in degrees)
45 (Θ, the angle of interest for velocity in degrees)

Below is the equation that I found

v=√2*g*l/m*(cos(Θ) – cos(Θmax))
v=11.84

My confusion is that the equation for pendulum velocity at it's equilibrium position which should be the maximum velocity is less that at 45° in the above equation

v=√2*g*l*(1-cos(Θmax))
v=5.23

I understand that mass is in the first equation and not the second equation, but is the first equation correct for velocity at any angle? Is the equation for maximum velocity at the pendulum equilibrium position not correct? Any help would be appreciated.

cos(Θmax) is zero (the maximum angle is 90 degrees, isn't it?)
So your formula should be
v=√[2*g*l*(cos(Θ)]

which is maximum for Θ=0.

Cos of 45 is √2/2.
 
There should be no mass in that equation.
 
Thanks for the feedback, ended up answering my own question. Went at it from the energy conservation standpoint.

a=[90 85 80 75 70 65 60 55 50 45 40 35 30 25 20 15 10 5 0]; // angle to calculate velocity at

g=9.8; //acceleration of gravity (meters/sec^2)
l=1.397; // length (meters)
ao=90; // maximum angle (degrees)
m = .138; // mass (kilograms)


ME=m*g*l // ME = PE + KE, PE=ME at 90 degree max bob height, KE=0

h=l-(cos(a)*l) // pendulum bob height at a given angle
PE=m*g*h // Potential Energy calculation
KE=ME-PE // Kinetic energy at a given angle
v_angle=sqrt(KE*2/m) // pendulum bob velocity at a given angle

plot of angle vs velocity (see attached JPG file)

At 90° (max pendulum bob height) 0 meters/sec velocity
AT 0° (0 pendulum bob height) equilibrium position 5.23 meters/sec velocity
 

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  • angle vs velocity.jpg
    angle vs velocity.jpg
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