a problem on finding acceleration by differentiation

Question

The displacement of particle along the $x$ and $y$ axis is \begin{cases} x(t)=\omega t-\sin\omega t\\ y(u)=1-\cos\omega t \end{cases} Upon differentiation, the velocity is \begin{cases} v_x(t)=\omega\left(1-\cos\omega t\right)\\ v_y(t)=\omega\sin\omega t \end{cases} so $$v =\sqrt{v_x^2+ v_y^2} = 2w \sin (wt /2)$$

My problem is if I find the magnitude of acceleration by differentiation of components $v_x$ and $v_y$ followed by their sum, I get a constant acceleration $w^2$ but if I directly differentiate $v$ I got variable acceleration $w^2 \cos (wt/2)$. I don't understand why. How are the two methods different?

Answer 1

When I square $v_x$ and $v_y$, I get \begin{align} v(t)&=\sqrt{\left(\omega-\omega\cos\omega t\right)^2+\omega\sin^2\omega t}\\ &=\left[\omega^2+\omega^2\cos^2\omega t-2\omega^2\cos\omega t +\omega\sin^2\omega t\right]^{1/2}\\ &=\omega\sqrt{2-2\cos\omega t} \end{align} which, due to the square root term, is slightly different than what you have. Differentiating this returns $$ \frac{dv}{dt}=\frac{\omega^2\sin\omega t}{\sqrt{2-2\cos\omega t}} $$ which is also radically different than $a(t)=\sqrt{a_x(t)^2+a_y(t)^2}$ and your answer of $dv/dt$. The real answer is indeed the $\omega^2$ you've found via the components.

The reason the two methods do not align is because the particle's motion is along a curve. That is to say, your direction is changing. Note that the relationship for the magnitude is $$ \lvert a\rvert=\left\lvert\frac{dv}{dt}\right\rvert\neq\frac{d|v|}{dt} $$ You must treat the acceleration as a vector, take the components derivative, and then find the magnitude afterwards.