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Pendulum initial condtions

  • Thread starter EvLer
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The problem:
L = 0.5. Pendulum of L is displaced angle of 0.1 rad and then released from rest. Determine resulting motion.

Ok, so I know it's going to be a second order DE of this kind which I will need to solve:

theta''(t) + (g/L)theta(t) = 0;

so theta(0) = 0.1? then what about theta'(t)(0)? is it just = 0? and what is it exactly anyway? if derivative is a rate of change with respect to time? rate of change of angle?

Thanks for any explanation/source.
 

dextercioby

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"Released from rest".Does that ring a bell...?Remember that it's a circular movement,linear velocity is ~ to the angular one.

Daniel.
 

OlderDan

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EvLer said:
The problem:
L = 0.5. Pendulum of L is displaced angle of 0.1 rad and then released from rest. Determine resulting motion.

Ok, so I know it's going to be a second order DE of this kind which I will need to solve:

theta''(t) + (g/L)theta(t) = 0;

so theta(0) = 0.1? then what about theta'(t)(0)? is it just = 0? and what is it exactly anyway? if derivative is a rate of change with respect to time? rate of change of angle?

Thanks for any explanation/source.
[tex] \dot{\theta}(t) = \omega [/tex]

is the angular velocity of the pendulum, and the initial value is zero, as you have observed
 
when you solve this problem for a variable. theta in this case
the first initial condition is the initial state of the variable of interest
the second initial condition is the first order derivative of the variable
the third initial condition is the 2nd order derivative of the variable and so on.

This is provided we have a single variable of n-th order.
If this is so then there are:
n initial conditions
The first is always the zeroth derivative of the variable of interest
the last is always the n-1 derivative of the variable of interest.

This our IC's are
theta0 = 0.1 rad
thetadot0 = 0 rad/s
 

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