enter image description here A particle moves along A->B with a constant acceleration in the x direction. I'm supposed to find the velocity at B

Therefore, at $\theta$, the centripetal acceleration for some $v$ at that instant = $v^2/R$. Since acceleration along x is constant: $v^2/R cos\theta = v_0^2/R$ = initial acc along x Hence at 90 degrees, I'm getting V->infinity. Where did I go wrong?

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    $\begingroup$ Don't follow what you're doing with your equations. Think that you need to more clearly explain your reasoning step by step. $\endgroup$ – user93237 May 17 '17 at 5:24
  • $\begingroup$ Are you sure that acceleration in the X direction is constant? In circular motion, the magnitude of the acceleration is constant, but the direction is always towards the center of the circle (this is what "centripetal" means). $\endgroup$ – Mark H May 17 '17 at 5:29
  • $\begingroup$ @MarkH that is given in the question. of course, velocity & acc can and are varying throughout the motion so that it moves along AB $\endgroup$ – xasthor May 17 '17 at 5:31

In general, at every point, the acceleration of the particle has tangential as well as radial components, just as it has both x and y components. So $\ddot x$ is the sum of the x components of both radial (ie centripetal) acceleration $\frac{v^2}{R}$ and tangential acceleration. You have missed out the x component of tangential acceleration.

At A the y-component of acceleration makes no contribution to centripetal acceleration, so the x-component of acceleration equals the centripetal acceleration - ie $a_x=\frac{v_0^2}{R}$. The x-component of velocity at A is $\dot x=u_x=0$.

At B the particle has no y-component of velocity $(v_y=\dot y=0)$, only an x-component $v_x$. It has travelled a distance $R$ horizontally so
$v_x^2=u_x^2+2a_x R$
from which the velocity $v=v_x$ at B can be found.


The particle moves on a circle, so its motion can be written in polar coordinates as:

$$\begin{pmatrix} x\\y \end{pmatrix} = r \begin{pmatrix} \cos\theta\\ \sin\theta \end{pmatrix}$$

with constant $r$. To match the image, I chose the x-axis pointing to the left and y-axis up.

The velocity is then given by the derivative:

$$\begin{pmatrix}v_x\\v_y \end{pmatrix}= r \dot{\theta}\begin{pmatrix}-\sin\theta\\ \cos\theta\end{pmatrix}$$

Acceleration in the x-direction is assumed to be constant ($=a_x$) which makes $v_x$ linear in time. Taking into account that $v_x(t=0)=0$ we have: $$v_x=a_x t = -r\dot{\theta}\sin\theta$$ This equation can be readily integrated which gives you an equation for $\theta$ as a function of time: $$a_x \int_0^t t~dt = -r\int_0^\theta\sin\theta~d\theta$$ $$\frac{a_x t^2}{2} = r (\cos\theta-1)$$

If you put this back into the equation for the velocity components, you can find the velocity as a function of time. Putting $t=\sqrt{-\frac{2 r}{a_x}}$ gives you the velocity at point B.


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