Classical Mechanics

Author

John Robin Inston

Published

September 25, 2026

0.0.1 References

0.0.2 Contents

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1 Elementary Mechanics

1.1 Newton’s Second Law

We define the following functions of time for a particle: (1) the position \(\vec{r}(t)\); (2) the velocity \(\vec{v}(t)= \frac{d}{dt}\vec{r}(t)\) (the speed is \(\lvert \vec{v} \rvert\)); (3) the acceleration \(\vec{a}(t)=\frac{d}{dt}\vec{v}(t)=\frac{d^2}{dt^2}\vec{r}(t)\); and (4) the momentum \(\vec{p}(t)=m(t)\vec{v}(t)\) where \(m(t)\) denotes the mass of the particle.

In inertial frames it holds that the force \(\vec{F}(t)\) acting on a particle is given by \[ \vec{F}(t)= \frac{d}{dt}\vec{p}(t)=\frac{d}{dt}(m(t)\vec{v}(t)). \] If the mass is not time-dependent then we obtain the familiar \[ \vec{F}(t)=m \frac{d}{dt}\vec{v}(t)=m\vec{a}. \]

From Newton’s Second Law we also Law of Conservation of Linear Momentum. Supposing the force on a particle is \(\vec{F}\) and that there is a vector \(\vec{s}\) such that the force has no component along \(\vec{s}\), that is \[ \vec{F}\cdot \vec{s}=0. \] Then, from Newton’s Second Law we have that \[ \frac{d}{dt}\vec{p}\cdot \vec{s}=0 \implies \vec{p}\cdot \vec{s}=\alpha, \] where \(\alpha\) is constant, that is there is conservation of the component of linear momentum along the direction \(\vec{s}\) in which there is no force.

Example (Atwood’s Machine Problem):

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1.2 Retarding Forces

We introduce forces that are not static but rather depend on the velocity of the moving object. Such a force can frequently be written as a power law in the velocity \[ \vec{F}_{r}=\vec{F}_{r}(v)=-kv^n \frac{\vec{v}}{v}, \] where \(k\) is a constant that dependent on the details of the problem. We note that the force is always directed opposite to the velocity of the object.

Example (Retarding Forces):

1.3 Angular Motion

Next we derive analogues of linear momentum, force and Newton’s Second Law for angular motion. The angular velocity of a particle as a function of time with respect to a particular origin \[ \vec{v}(t)=\vec{\omega}(t) \times \vec{r}(t). \] This can be written mathematically as \[ \partial \vec{r}= \partial \vec{\theta} \times \vec{r} \] where \(\partial \vec{\theta}\) points along the axis of the motion and \(\times\) indicates a vector cross-product. The cross-product gives the correct direction for displacement \(\partial \vec{r}\) (perpendicular to the axis and \(\vec{r}\)) and the correct amplitude (\(\lvert \partial \vec{r} \rvert = R \partial \theta = r \partial \theta \sin \alpha\)). If we then divide by the time \(\partial t\) required to make this displacement we have \[ \frac{\partial\vec{r}}{\partial t} = \frac{\partial\vec{\theta}}{ \partial t} \times \vec{r} \implies \vec{v} = \vec{\omega} \times \vec{r}. \]

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