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)

$\mathbf {p_{f}} =\mathbf {p_{i}} +\mathbf {F} \cdot \Delta t$

Angular Kinematics

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

$\mathbf {L_{f}} =\mathbf {L_{i}} +\mathbf {\tau } \cdot \Delta t$

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: ${\frac {\tau }{L_{0}}}=\Omega$