# Example of Torque, Center of Precession

So here's the set up, we have a fence of length $2L$, and a support strut a distance $l$ from the axis (think of a railroad crossing gate). We need to find the best position for the support rod, so that the pivot takes minimum wear. We analyze the forces on the gate, and find that there are three vertical forces (we are ignoring the radial components). The force $\mathbf F _s$ is the force due to the support, $\mathbf W$, the weight, acting at center of mass (position $L$) and the force at the axis of rotation, which we will call $\mathbf F _{\rho}$

So for torques, taking the origin being the AoR (axis of rotation) are $$\tau = MgL-F_s l=I_{\rho} \ddot{\theta}.$$ Now we integrate from $t$ to $t+\Delta t$ (angular impulse?) which gives us $I_{\rho} \dot{\theta}=-F_s l \Delta t$, where we have neglected the torque due to the weight.

Next step is where we apply $\mathbf F = m\mathbf a$, and which case: $$m\frac{dV}{dt}=-(F_s + F_{\rho})+Mg$$ (This step seems to be arbitrary, what is our motivation for including both torque and Newton's law?). We have again neglected the weight, and integrate $mv=-ML\dot{\theta}=(F_s + F_{\rho})$

This step also confuses me, how in the world does this make sense $$\int ^{t+\Delta t}_{t} (F_s + F_{\rho}) dt = -(F_s + F_{\rho}).$$

Shouldn't it be $(F_s + F_{\rho})\Delta t$? The next step says we can solve our equation $$I_{\rho} \dot{\theta}=-F_s l \Delta t$$ and get $$F_s \Delta t=I_{\rho}\dot{\theta}/l$$ and plug this into our last equation, $$ML\dot{\theta}=(F_s + F_{\rho}) .$$

The only way this works out to what my book says, is if we have $(F_s + F_{\rho})\Delta t$.

We were supposed to end up with $$F_{\rho}\Delta t = \dot{\theta}(ML-\frac{I_{\rho}}{l}).$$

From this point on, it just talks about the conditions for the impulse from $F_{\rho} = 0$.

• A sketch would help. – John Alexiou Jan 8 '14 at 2:08
• Why isn't the support your axis of rotation. It is not clear for the description of the problem. – John Alexiou Jan 8 '14 at 2:18
• If you want the equations of motion about a point, not the center of mass (like the pivot) look at physics.stackexchange.com/a/80449/392 – John Alexiou Jan 8 '14 at 2:30
• I'll check out the link, I just posted a diagram – Astrum Jan 8 '14 at 2:37
• So $F_\rho$ is given and $F_s$ is the unknown reaction you are trying to minimize. Is this correct? – John Alexiou Jan 8 '14 at 3:13

Based on the diagram supplied, the equations of motion are

$$F_S - M g + F_\rho = M \left(\ell-\frac{L}{2}\right) \ddot\theta \\ L F_\rho - \left(\ell-\frac{L}{2}\right) F_S = I_G \ddot\theta$$

where $I_G = \frac{M}{3}L^2$ is the mass moment of inertia of the bar on the center of mass.

The above (ignoring gravity) is solved as

$$F_S = \frac{c M L-I_G}{L+c} \ddot\theta \\ F_\rho = \frac{I_G+M c^2}{L+c} \ddot\theta$$ where $c=\ell-\frac{L}{2}$ is the distance of the pivot to G.

The support force is zero, when $c= \frac{I_G}{M L}$ or $$\boxed{\ell = \frac{L}{2} + \frac{I_G}{M L}}$$

If gravity is to be included, then $c = \frac{I_G}{M L} - \frac{g}{\ddot\theta}$ so the design with minimum support wear depends on the rate of gate opening $\ddot\theta$.

Well, since nobody has answered yet, I've come to the conclusion that there must have been a typo in my text book, because the substitution works our perfectly if the integral gives the $\Delta t$ term, as it should.

It seems to me the reason we are using $F=ma$ is because we're looking for the point where the rotational and transnational forces cancel out, relative to some pivot.

The finer details of this question still elude me though.

If a rigid body is pivoting about a point A (not the center of mass C) then the center of percussion ( a line of percussion actually ) is found as follows:

1. Take a coordinate system with the local $\hat{z}$ axis along the rotation axis, the local $\hat{x}$ axis from the pivot to the center of mass and the origin at A. The relative location of the center of mass to the pivot is $\vec{c} = (c,0,0)$.

2. The rotation vector is then $\vec{\omega} = (0,0,\omega)$ and the velocity of the center of mass $\vec{v}_C = \vec{\omega} \times \vec{c} =(0,\omega\,c,0)$.

3. The linear momentum of the rigid body is $\vec{L} = m \vec{v}_C = (0,m\,\omega\,c,0)$.

4. The angular momentum of the rigid body about the center of mass is $\vec{H}_C = \bar{I}\vec{\omega} = (0,0,I_z \omega)$.

5. The angular momentum of the rigid body about the pivot is $\vec{H}_A = \vec{H}_C + \vec{c} \times \vec{L} = (0,0,\left(I_z+m\,c^2\right) \omega)$.

6. The direction of the percussion line is along the linear momentum vector $\vec{e} =\frac{\vec{L}}{|\vec{L}|} = (0,1,0)$.

7. The location of the percussion line relative to A is defined by $\vec{p} = \frac{\vec{L}\times\vec{H}_A}{|\vec{L}|^2} = (c+\frac{I_z}{m c},0,0)$.

Why ? This comes from the treatment of momentum as a Screw (in more detail here ) which states that just as a pure torque accelerates a body about its center of mass only, a pure force along the center of percussion yields a pure rotation about the pivot.

To prove this, apply a force along the percussion axis $\vec{F}=(0,F,0)$ which yields an equipolent moment about the center of mass C as $\vec{M}_C = \left(\vec{p}-\vec{c}\right)\times\vec{F} = (0,0,\frac{F I_z}{m c})$.

The acceleration of the center of mass is $\vec{a}_C = \frac{\vec{F}}{m} = (0,\frac{F}{m},0)$ and the rotational acceleration $\vec{\alpha} = \bar{I}^{-1} \vec{M}_C = (0,0,\frac{F}{m c})$. This is a pure rotation about the pivot if $\vec{a}_C = \vec{\alpha} \times \vec{c}$ (ignoring velocity related components). This is true since

$$(0,\frac{F}{m},0) = (0,0,\frac{F}{m c}) \times (c,0,0)$$

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