Using centripetal acceleration to find back the velocity magnitude at $t+dt$

Question

Considering a circular motion with no angular acceleration. How can you find the same magnitude for the velocity vector at different time using the formula $v_{t} = v_0 + a.t$ with vectors?

The acceleration vector $\vec{a} = \vec{a_c} + \vec{a_t}$ where $\vec{a_c}$ is the centripetal acceleration $a_c$ = $\dfrac{-v^2}{r}$ and $\vec{a_t}$ is the tangential acceleration if the tangential speed change.

On every diagram I've seen so far, the radial acceleration vector $\vec{a_r}$ and the tangential velocity $\vec{v_t}$ are perpendicular and their vector tails share the same point, then how could $\vec{v_{(t_0+dt)}}$ be the same magnitude than $\vec{v_{(t_0)}}$ ? How can I prove a hypothenuse be the same length as one of its component? In my book they speak of radial acceleration instead of centripetal but if I'm right, I tried this : $$\vec{v_{(t_0+dt)}} = v_{r_{(t_0)}}\vec{r} + v_{t_{(t_0)}}\vec{\theta} + (a_r.dt)\vec{r} + (a_t.dt)\vec{\theta} $$

$\vec{r}$ is the unit vector in the radius vector's direction which goes to the point p at $t_0$

and $\vec{\theta}$ is the unit vector of the tangent at the point p and in the positive direction counterclockwise and perpendicular to $\vec{r}$

the scalar $v_r$ is zero and with no angular acceleration $a_t$ is zero. Is this equation correct? $$\mid\mid\vec{v_{(t_0+dt)}}\mid\mid = (v_{t_{(t_0)}}^2 + (a_r.dt)^2)^{\frac{1}{2}} $$ I've just seen the derivation leading to $a_c$ = $\dfrac{-v^2}{r}$ but in the equation above any magnitude for $a_r$ would give $v_{(t_0+dt)} = v_{t_{(t_0)}}$ since $dt$ is small enough.

Answer 1

It's very easy to do: \begin{align} |\mathbf{v}(t+dt)|^2&= \left|\mathbf{v}(t)+\frac{d\mathbf{v}}{dt}dt\right|^2\\ &=\left|v(t)\boldsymbol{\hat\theta}+\mathbf{a}(t)dt\right|^2\\ &=\left|v(t)\boldsymbol{\hat\theta}+a(t)dt\mathbf{\hat r}\right|^2\\ &=v^2(t)+2v(t)a(t)dt \boldsymbol{\hat\theta}\cdot\mathbf{\hat r}+a^2(t)(dt)^2 \end{align} Since $\boldsymbol{\hat\theta}\cdot\mathbf{\hat r}=0$ and $(dt)^2$ is negligible, $$|\mathbf{v}(t+dt)|^2=v^2(t)$$ It works for any acceleration as long as it's perpendicular to velocity.

Answer 2

There is a general argument that is beautifully simple. The (squared) speed is given by $\vec{v}\cdot\vec{v}$. Now, let's consider the rate of change of (squared) speed:

\begin{align} \frac{d}{dt} (\vec{v}\cdot\vec{v})&= \frac{d\vec{v}}{dt}\cdot\vec{v}+ \vec{v}\cdot\frac{d\vec{v}}{dt} \\ &= 2 \frac{d\vec{v}}{dt}\cdot\vec{v}\\ &= 2 \vec{a}\cdot\vec{v} \end{align}

Thus, if the acceleration is perpendicular to the velocity, the change in (squared) speed would vanish because the scalar product of perpendicular vectors vanish. In other words, acceleration that is perpendicular to the velocity will only contribute to change in the direction of the velocity.