Proof of $a = v^2/r$ using similar triangles

Question

The book states that the ‘change in velocity’ triangle and the displacement triangle has the same angle theta.

But I don’t get it? How can we prove that the two triangles will have the same angle?

Answer 1

The displacement and velocity vectors in circular motion are always perpendicular:

$$\angle \vec{r}_1 - \angle \vec{v}_1 = 90^\circ \quad \text{and} \quad \angle \vec{r}_2 - \angle \vec{v}_2 = 90^\circ$$

Subtract the above two equations:

$$\angle \vec{r}_1 - \angle \vec{r}_2 - \angle \vec{v}_1 + \angle \vec{v}_2 = 90^\circ - 90^\circ$$

With $\angle \vec{r}_1 - \angle \vec{r}_2 = \theta$ the above equation becomes:

$$\boxed{\angle \vec{v}_1 - \angle \vec{v}_2 = \theta}$$

Assuming that the circular motion is uniform (constant speed)

$$|\vec{r}_1| = |\vec{r}_2| = R, \qquad |\vec{v}_1| = |\vec{v}_2| = v_0$$

the two triangles are similar:

$$\frac{|\Delta\vec{v}|}{|\vec{v}_1|} = \frac{|\Delta\vec{r}|}{|\vec{r}_1|} \qquad \rightarrow \qquad |\Delta\vec{v}| = \frac{v_0}{R} |\Delta \vec{r}|$$

where $\Delta \vec{r} = \vec{r}_2 - \vec{r}_1$ and $\Delta \vec{v} = \vec{v}_2 - \vec{v}_1$. The average acceleration is defined as

$$a_{av} = \frac{|\Delta \vec{v}|}{\Delta t} = \frac{v_0}{R} \frac{|\Delta \vec{r}|}{\Delta t}$$

If we let the time difference go to zero $\Delta t \rightarrow 0$, the average acceleration becomes instantaneous acceleration:

$$a = \lim_{\Delta t \rightarrow 0} a_{av} = \frac{v_0}{R} \lim_{\Delta t \rightarrow 0} \frac{|\Delta \vec{r}|}{\Delta t} = \frac{v_0^2}{R}$$

The acceleration vector is perpendicular to the velocity vector and is also known as radial acceleration or centripetal acceleration:

$$\boxed{a_{\perp} = \frac{v_0^2}{R}}$$

When circular motion is not uniform, the expression for radial acceleration is the same, but there is also tangential acceleration component which is parallel to the velocity vector:

$$\boxed{a_{||} = \frac{d}{dt}v_0}$$

Here you can find detailed derivation of the expressions for radial and tangential acceleration components: https://physics.stackexchange.com/a/685423/149541

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Here I give easy-to-understand proof that the displacement and velocity vectors are always perpendicular.

The position in circular motion is defined as

$$x^2 + y^2 = R^2$$

where the radius $R$ is assumed to be constant. Take the time derivative of the above equation:

$$2 x \dot{x} + 2 y \dot{y} = 0$$

With $v_x = \dot{x}$ and $v_y = \dot{y}$ the above equation becomes:

$$x v_x + y v_y = 0$$

Now recognize that the above equation is scalar product between the displacement $\vec{r}$ and velocity $\vec{v}$ vectors:

$$\vec{r} \cdot \vec{v} = |\vec{r}||\vec{v}| \cos \phi = 0$$

The solution to the above equation is $\phi = 90^\circ$ which concludes the proof that displacement and velocity vectors are always perpendicular in circular motion.

Answer 2

The direction of vector $v_1$ is the same as vector $r_1$ but rotated 90$^\circ$ degrees clockwise and similarly vector $v_2$ is rotated $90^\circ$ with respect to vector $r_2$. So the angle between vectors $v_1$ and $v_2$ is the same as the angle between $r_1$ and $r_2$. More precisely, if you define $\text{angle}(v_1)$ as the angle that $v_1$ makes with the x-axis (or some other arbitrary axis) and $\text{angle}(v_1,v_2) $ as the angle between $v_1$ and $v_2$ then we get \begin{align} \text{angle}(v_1,v_2)&=\text{angle}(v_1)-\text{angle}(v_2)\\ &=\text{angle}(r_1)-90^\circ-\big(\text{angle}(r_2) - 90^\circ\big)\\ &=\text{angle}(r_1)-\text{angle}(r_2)\\ &=\text{angle}(r_1, r_2) \end{align}

Note that we know that the angle between $r$ and $v$ is always $90^\circ$, otherwise the length of $r$ would be increasing and we wouldn't have circular motion.

Answer 3

Okay, so I was working on this question and I think I solved it. Please tell if I made any inaccurate assumptions.

Firstly, we extend the -V_A (which originates at the tip of V_B, or in this case, E) till it intersects the line OB at D.

Then, we extend OB till it intersects V_A at C.

Since, ED (or -V_A) is parallel to AC (or V_A), angle(ACO) = angle(BDE).

Also, angle(CAO) = angle(DBE) = 90.

Therefore, angle(BED) = angle(AOC) = .