# ECE 105: Classical Mechanics ## Motion Please see [SL Physics 1#2.1 - Motion](/g11/sph3u7/#21-motion) for more information. ## Kinematics Please see [SL Physics 1#Kinematic equations](/g11/sph3u7/#kinematic-equations) for more information. ## Projectile motion Please see [SL Physics 1#Projectile motion](/g11/sph3u7/#projectile-motion) for more information. ## Uniform circular motion Please see [SL Physics 1#6.1 - Circular motion](/g11/sph3u7/#61-circular-motion) for more information. ## Forces Please see [SL Physics 1#2.2 - Forces](/g11/sph3u7/#22-forces) for more information. ## Work Please see [SL Physics 1#2.3 - Work, energy, and power](/g11/sph3u7/#23-work-energy-and-power) for more information. ## Momentum and impulse Please see [SL Physics 1#2.4 - Momentum and impulse](/g11/sph3u7/#24-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: $$x_{cm} = \frac{1}{M}\int^M_0 xdm$$