1
$\begingroup$

I encountered a Physics Olympiad problem:

A ball bearing rests on a ramp fixed to the top of a car which is accelerating horizontally. The position of the ball bearing relative to the ramp is used as a measure of the acceleration of the car. Show that if the acceleration is to be proportional to the horizontal distance moved by the ball (measured relative to the ramp), then the ramp must be curved upwards in the shape of a parabola.

Attempt at a solution:

Let $A$=magnitude of acceleration of the car w.r.t a stationary observer, $a$=magnitude of acceleration of of the ball bearing w.r.t the stationary observer and $a'$=magnitude of acceleration of the ball bearing as observed in the accelerating car frame, then $$A=\mathcal{k}x' \tag{1}$$

where $x'$ is the horizontal position of the ball bearing as measured in the accelerating frame, with $x'=0$ at the bottom end of the ramp.

Since the ramp is a parabola, the position function of the ball bearing in the accelerating frame, should take the form $$y'=\alpha (x')^2 \tag{2}$$ where $\alpha$ is some constant and $y'$ is the vertical position as measured in the accelerating frame.

Using the definition of the fictitious force/acceleration $$\mathbf{a'=a-A} \tag{3}$$ The LHS of $(1)$ reads $$a+a'=kx'\tag{4}$$ (positive sense towards the direction of $\mathbf{A}$). But this doesn't seem to provide any useful information since I cannot define $a'$.

Should I try something like $\frac{dy'}{dx'}=\frac{\dot{y}}{\dot{x}}$? I am lost because I do not know whether I should analyse the ball bearing in its equilibrium (i.e. $A=constant, x'=constant$) to obtain some expressions for $y$ or it as a function $A(t)=kx'(t)$?

$\endgroup$
2
$\begingroup$

I do not know whether I should analyse the ball bearing in its equilibrium (i.e. A=constant,x′=constant) to obtain some expressions for $y$

I strongly believe that's what the question means. To be more precise, I would frame the question as follows:

"The equilibrium position of the ball bearing relative to the ramp is used as a measure of the acceleration of the car. Show that if the acceleration is to be proportional to the horizontal distance moved by the ball to its new equilibrium position (measured relative to the ramp), then the ramp must be curved upwards in the shape of a parabola."

I've provided a hint below in case you are unable to solve it after an appropriate amount of time (an amount of time which you feel is appropriate).

(1) The normal force needs to be oriented in a particular angle for the object to be in equilibrium under the action of the three forces: gravity, normal reaction, and pseudo force (I'm in the frame of the moving car). (2) The normal force provided by the ramp is perpendicular to the slope of the ramp.

$\endgroup$
2
$\begingroup$

you can start with the position vector of the ball bearing:

$$\vec{R}_B=\begin{bmatrix} x+f_x(s) \\ f_y(s) \\ \end{bmatrix}\tag 1$$

where x is the car position and $f_x(s)\,,f_y(s)$ are arbitrary function that described the position of the ball bearing relative to the car.

and the position vector of the car

$$\vec{R}_C=\begin{bmatrix} x \\ 0 \\ \end{bmatrix}\tag 2$$

with equation (1) and (2) and Euler-Lagrange method you get the first integral :

$$m\,\frac{d}{ds}\,f_x(s)\,\dot{s}+(m+M)\,\dot{x}=\text{const}\tag 3$$

where m is the mass of the ball bearing and M is the car mass.

differentiate equation (3) and solve for $\ddot{x}$ :

$$\ddot{x}=\frac{m\,\left(\frac{d\,f_x}{ds}\ddot{s}+\frac{d^2\,f_x}{ds^2}\,(\dot{s})^2\right)}{m+M}\tag 4$$

if the car acceleration should be proportional to the ball bearing acceleration then: $\frac{d^2\,f_x}{ds^2}$ must be zero, this mean that $f_x(s)=a\,s$

$$\ddot{x}=\frac{m\,a}{m+M}\ddot{s}\tag 5$$

for $f_y(s)=a^2\,s^2$ you get a parabola shape

$\endgroup$

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