# Total Angular Momentum of 2 connected falling bodies.

 P: 4 I have two 2 rigid bodies with masses m1 and m2 and Moments of Inertia I1 and I2, they are connected by a free rotational joint at some point, their coms lie at c1 and c2. There's gravity. In the beginning both have some angular velocity $\omega_i$ Questions: - Total angular momentum of the system - COM trajectory Attempt: - Calculating L of the two bodies is no Problem with $L=I\cdot\omega$, Also if they're unconnected L is simply $L_1+L_2$. However my attempts at finding the total angular Momentum don't seem to work out. - I thought that the COM trajectory should simply be gravity dependent, i.e. $\frac{1}{2}gt^2$, but somehow it wobbles in my physics simulator. (I'm using PhysX 2.8.3) (The COM trajectory is what it should be :) ) - Also I thought to calculate the total Angular Momentum I would have to shift the two Momenta to the COM via the parallel axis thm, and then build the sum. So I'm stuck - any help would be very much appreciated. Best, xtin
P: 1,135
Hi xtinch!!
Welcome to PF !!

 Quote by xtinch I have two 2 rigid bodies with masses m1 and m2 and Moments of Inertia I1 and I2, they are connected by a free rotational joint at some point, their coms lie at c1 and c2. There's gravity.
I cant aacually understand how they are connected but whatever axis that may be, you need to find moment of inertia's along a common axis,

Use $I = I_{COM} + md^2$

Here d is the distance b/w COM of body and axis along which you need to find I
 P: 4 I see, $I=I_{COM}+md^2$ is the parallel axis thm i mentioned, also I use $R_{ig}*I*R_{ig}^T$ to rotate the MoI. So my final attempt was to rotate the MoI so they coincide with the global coordinate frame, then translate them to the global COM. (Let $T(v)$ be the Matrix that shifts the MoI by the vector v) $L_{total} = R^T_1*I_1*R_1+T(c_{total}-c_1)+R^T_2*I_2*R_2+T(c_{total}-c_2)$ I also tried $R^T_1*(I_1+T(c_{total}-c_1))*R_1$ to no avail. =\
P: 1,135
Total Angular Momentum of 2 connected falling bodies.

 Quote by xtinch I see, $I=I_{COM}+md^2$ is the parallel axis thm i mentioned, also I use $R_{ig}*I*R_{ig}^T$ to rotate the MoI. So my final attempt was to rotate the MoI so they coincide with the global coordinate frame, then translate them to the global COM. (Let T(v) be the Matrix that shifts the MoI by the vector v) $L_total = R^T_1*I_1*R_1+T(c_{total}-c_1)+R^T_2*I_2*R_2+T(c_{total}-c_2)$ I also tried $R^T_1*(I_1+T(c_1-c_{total}))*R_1$ to no avail. =\
Do you have any pic related to the ques?
I can't really imagine the case !!
 P: 4 Here's an image, m3 is simply a mass that puts the system into rotation. I'm only interested in the Momentum after the collision. Below is a plot of the L calculated with my above suggestion - it's clearly not constant. So m1+m2 fall together as it's basically a box with an attached, freely rotateable rod (there are no collisions between the rod and the box), hit m3 which is fixed in space, start to rotate and then AngVel should be constant. (Note: ignore that the plot is labeled AngVel it's the momentum ;) )
 P: 4 Ok, I found the solution. I forgot to account for the rotation of the masses around the com. The solution is simply: $L_{total} = I_1\cdot\omega_1+m_1*(c_1-c_{total})\times(v_{com}-v_1)+I_2\cdot\omega_2+m_2*(c_2-c_{total})\times(v_{com}-v_2)$ Thanks for your help and I hope this will help others who search for this simple but annoying thing in vain ;)

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