2
$\begingroup$

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?

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.