# Swinging rod

i have a mechanics problem. i want to represent the motion of a swinging rod (given initial conditions) as a function of time, but instead i can only find it as a function of the angle.

this rod has uniform density and freely pivots at one end.

Q = angle from 0 counter-clockwise to the rod
L = length
M = mass
Fgrav = force due to gravity

I = integral(r^2 dm, 0,L)
= integral(r^2 M*dl/L, 0,l)
= (M*L^2)/3
Fgrav = Mg -> Mg*cos(Q) tangentially
torque = F*L/2 = I*a
a = (3g/2L)*cos(Q)

how am i able to convert this a(Q) into a(t) ? (that's rotational acceleration by the way)
unfortunately, the following website was of no help. it simply took the exact same steps to get the exact same result
http://www.myphysicslab.com/pendulum1.html

this is for a graphical physics simulation i'm about to finish writing in Java. now i just need the formula

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I'm not sure if this thinking is correct, but is the acceleration of the rod not constant? ie. a(t)=0. Just a thought...

Otherwise, you have to find Q in terms of t.

Sorry, that's about as much as I can offer.

Fgrav = Mg -> Mg*cos(Q) tangentially?
You sure that isn't -Mg*sin(Q)?

Anyway, if you have an expression of the form I*a = -torque, you can set up the differential eq. I*a + torque = 0, or a + torque/I = 0, where a equals d^2Q/dt^2 (angular acceleration=second derivative of Q with respect to time).

You have chosen an awkward way to set this problem up with you choice of the angle Q.

The usual way to set up a pendulum problem is to measure an angle theta from the downward vertical to the rod. Then the equation of motion is
Sum of Torques = - M*g*L/2*sin(theta) = I * ddtheta

Now what you have at this point, and what you had in your original formulation by the way, is a differential equation, that has to be solved, either in closed form or numerically. The closed form solution involves elliptic integrals and is a pretty hairy affair, so unless you are really up to speed on such things, I would suggest that you try a numerical Runge-Kutta solution.