65 lines
1.9 KiB
Markdown
65 lines
1.9 KiB
Markdown
# ECE 105: Classical Mechanics
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## Motion
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Please see [SL Physics 1#2.1 - Motion](/g11/sph3u7/#21-motion) for more information.
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## Kinematics
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Please see [SL Physics 1#Kinematic equations](/g11/sph3u7/#kinematic-equations) for more information.
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## Projectile motion
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Please see [SL Physics 1#Projectile motion](/g11/sph3u7/#projectile-motion) for more information.
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## Uniform circular motion
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Please see [SL Physics 1#6.1 - Circular motion](/g11/sph3u7/#61-circular-motion) for more information.
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## Forces
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Please see [SL Physics 1#2.2 - Forces](/g11/sph3u7/#22-forces) for more information.
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## Work
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Please see [SL Physics 1#2.3 - Work, energy, and power](/g11/sph3u7/#23-work-energy-and-power) for more information.
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## Momentum and impulse
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Please see [SL Physics 1#2.4 - Momentum and impulse](/g11/sph3u7/#24-momentum-and-impulse) for more information.
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The change of momentum with respect to time is equal to the average force **so long as mass is constant**.
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$$\frac{dp}{dt} = \frac{mdv}{dt} + \frac{vdm}{dt}$$
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Impulse is actually the change of momentum over time.
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$$\vec J = \int^{p_f}_{p_i}d\vec p$$
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## Centre of mass
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The centre of mass $x$ of a system is equal to the average of the centre of masses of its components relative to a defined origin.
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$$x_{cm} = \frac{m_1x_1 + m_2x_2 + ... + m_nx_n}{m_1 + m_2 + ... + m_n}$$
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To determine the centre of mass of a system with a hole, the hole should be treated as negative mass. If the geometry of the system is **symmetrical**, the centre of mass is also symmetrical in the x and y dimensions.
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For each mass, its surface density $\sigma$ is equal to:
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$$
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\sigma = \frac{m}{A} \\
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m = \sigma A
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$$
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Holes have negative mass, i.e., $m = -\sigma A$.
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For a **one-dimensional** hole, the linear mass density uses a similar formula:
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$$
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\lambda =\frac{m}{L} \\
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\lambda = \frac{dm}{dx}
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$$
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This means that a hole in a rod can use a different formula:
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$$x_{cm} = \frac{1}{M}\int^M_0 xdm$$
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