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I need help to calculate the forces F1 and F2 of this physical pendulum consisting of a rigid and homogenous rod with length l. enter image description here

I have found the following equation using the law of angular momentum:

enter image description here

How can I then calculate the forces F1 and F2 on the axis of rotation?

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  • $\begingroup$ Please do not post images of math, but use math formatting instead. Try $$\ddot{\theta} + \frac{3g}{2\ell}\sin \theta = 0$$ $\endgroup$ – John Alexiou Mar 20 at 17:08
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The forces on the axle (from the pendulum) must be equal and opposite to the forces on the pendulum from the axle. I would start with the radial and tangential forces. Radial: T – mg cos(θ) = m(L/2)$ω^2$. (The tension,T, acts on the axle and the center of mass.) Tangential: F(L/2) = Iα. Where F is the force (from the axle and perpendicular to T) which is causing the angular acceleration about the center of mass. I, is the rotational inertia of the rod about the center of mass, and, ω, and, α, are the instantaneous angular velocity and acceleration of the pendulum. If you want horizontal and vertical forces, break, T, and, F, into components. Note: For, α: mg sin(θ) (L/2) = I' α. Where I' is the rotational inertia about the axle. The instantaneous, ω, depends on the starting angle.

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  • $\begingroup$ Thank you! One question though, is F equal to the tangential force of the weight of the pendulum? $\endgroup$ – Tanamas Mar 20 at 18:31
  • $\begingroup$ No. F is the force from the axle which causes the rod to rotate about its center of mass $\endgroup$ – R.W. Bird Mar 21 at 20:38
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Problems like these are always solved with the following steps

  1. Description - Decide which DOF parameters to use to fully describe all the configurations of the system. In this case DOF=1 and the angle $\theta$ suffices.

  2. Kinematics - Track the position of the center of mass as a function of the DOF parameters and evaluate position, velocity and acceleration vectors. In this case $$ \mathrm{\vec{pos}}=(\ell/2\sin\theta, \;-\ell/2\cos\theta,\;0)$$ $$ \mathrm{\vec{vel}} = (\ell/2 \dot{\theta}\cos\theta, \,\ell/2\dot{\theta}\sin\theta,0) $$ $$ \mathrm{\vec{acc}} = \ldots$$ $$ \vec{\omega} = (0,0,\dot{\theta}) $$ $$ \vec{\alpha} = (0,0,\ddot{\theta}) $$

  3. Equations of motion - Use the sum of forces and sum of torques to write out the equations of motion. Here you have 3 unknowns, the two pin forces $F_1$ and $F_2$ and the angle acceleration $\ddot{\theta}$ and three equations.

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