PHYS 218 Chapter 11

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Equilibrium

Condition:

1. $\sum F=0$ (no acceleration)
2. $\sum \tau =0$ around ANY point

Center of Gravity

The center of gravitational pull in any gravitational field

Torque of Center of Gravity

IF gravitational field is constant: center of gravity is center of mass

Torque is cross product of position and weight $\tau =\sum r_i\times m_{i}g={\mathbf{𝐫}}_{cm}\times M\mathbf{𝐠}$

The point at which the torque due to gravity is 0.

If center of gravity extends bounds of support, then object will tip over (torque)

Faster than Gravity (FTG)

Imagine a board connected to a hinge. The tip of the board is actually accelerating faster than gravity.

Precession

Let's look back for a moment

Linear Kinematics

$\mathbf{𝐅}=\frac{d\mathbf{𝐏}}{dt}$ A force changes the linear momentum regardless of initial momentum (giving a person $5 makes that person 5$ richer)

${\mathbf{p}}_{\mathbf{𝐟}}={\mathbf{p}}_{\mathbf{𝐢}}+\mathbf{𝐅}\cdot \mathrm{\Delta }t$

Angular Kinematics

$\mathit{\tau }=\frac{d\mathbf{𝐋}}{dt}$ A torque changes the angular momentum regardless of initial momentum

${\mathbf{L}}_{\mathbf{𝐟}}={\mathbf{L}}_{\mathbf{𝐢}}+\mathit{\tau }\cdot \mathrm{\Delta }t$

Gravity makes a torque in the Z direction, but a gyroscope has an initial momentum in the X direction. adding these two vectors causes a rotation around the Y axis

Precessional velocity: $\frac{\tau }{L_0}=\mathrm{\Omega }$