Vectors play an extremely important role in both geometry and physics. Now, let us see how to use vectors to study circular motion in physics.
First, we must understand some basics:
In vector analysis, a single "position vector" (\vec{r}) is sufficient to describe the motion of an object without the need to establish a coordinate system. This is why vectors are applied in physics: physical laws should not depend on the coordinate system, and vectors happen to be independent of coordinate systems!
Newton’s Second Law: \vec{F}=m\vec{a}
Vector calculus operations, etc. (refer to Wikipedia or relevant materials).
In the following descriptions, for the convenience of the reader, vectors are written in the form \vec{r} rather than bold letters. Generally, in the descriptions on this site, we have |\vec{r}|=r, |\dot{\vec{r}}|=v, |\ddot{\vec{r}}|=a. However, \dot{r} = \frac{d|\vec{r}|}{dt} \neq |\dot{\vec{r}}|.
The figure above is a simple schematic diagram of circular motion, from which we obtain: \begin{aligned} \vec{F} &= m\vec{a} \\ \vec{a} &= \ddot{\vec{r}} \\ \vec{v} &= \dot{\vec{r}} \end{aligned}
Where \vec{F} is the net external force. Since the direction of velocity is always perpendicular to the direction of the position vector, we have: \vec{r} \cdot \dot{\vec{r}} = 0 \tag{1} And \frac{d}{dt}(\vec{r} \cdot \dot{\vec{r}}) = \dot{\vec{r}}^2 + \vec{r} \cdot \ddot{\vec{r}} = 0 Resulting in \begin{aligned} m\dot{\vec{r}}^2 + \vec{r} \cdot m\ddot{\vec{r}} &= 0 \\ mv^2 + \vec{r} \cdot \vec{F} &= 0 \end{aligned} \tag{2}
Where \vec{F} is the net external force (please remember, centripetal force is provided by the net external force and is a component of it; one cannot say that an object is acted upon by a centripetal force). Combining (1) and (2) gives the basic equations describing circular motion.
If the motion is uniform circular motion, v is constant, then we should have \dot{\vec{r}}^2 = \text{const}, i.e., \frac{d}{dt}(\dot{\vec{r}}^2) = 2\dot{\vec{r}} \cdot \ddot{\vec{r}} = 0 That is \dot{\vec{r}} \cdot \vec{F} = 0 \tag{3} This means the direction of the net external force is perpendicular to the direction of velocity. Combining this with (2), we have \begin{aligned} \vec{F} &= -m\left(\frac{\dot{\vec{r}}^2}{r^2}\right)\vec{r} = -m\omega^2 \vec{r} \\ F &= \frac{mv^2}{r} \end{aligned}
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