1.9 KiB
ECE 105: Classical Mechanics
Motion
Please see SL Physics 1#2.1 - Motion for more information.
Kinematics
Please see SL Physics 1#Kinematic equations for more information.
Projectile motion
Please see SL Physics 1#Projectile motion for more information.
Uniform circular motion
Please see SL Physics 1#6.1 - Circular motion for more information.
Forces
Please see SL Physics 1#2.2 - Forces for more information.
Work
Please see SL Physics 1#2.3 - Work, energy, and power for more information.
Momentum and impulse
Please see SL Physics 1#2.4 - Momentum and impulse for more information.
The change of momentum with respect to time is equal to the average force so long as mass is constant.
\[\frac{dp}{dt} = \frac{mdv}{dt} + \frac{vdm}{dt}\]
Impulse is actually the change of momentum over time.
\[\vec J = \int^{p_f}_{p_i}d\vec p\]
Centre of mass
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.
\[x_{cm} = \frac{m_1x_1 + m_2x_2 + ... + m_nx_n}{m_1 + m_2 + ... + m_n}\]
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.
For each mass, its surface density \(\sigma\) is equal to:
\[ \sigma = \frac{m}{A} \\ m = \sigma A \]
Holes have negative mass, i.e., \(m = -\sigma A\).
For a one-dimensional hole, the linear mass density uses a similar formula:
\[ \lambda =\frac{m}{L} \\ \lambda = \frac{dm}{dx} \]
This means that a hole in a rod can use a different formula: \[x_{cm} = \frac{1}{M}\int^M_0 xdm\]