Finding Force from acceleration at a point given constant speed

Question

Prompt: An object of mass $m$ travels along a parabola $y = x^2$ with constant speed $5$ units/sec. What is the force on the object due to acceleration at point $(\sqrt{2}, 2)$? Recall $\vec{F} = m\vec{a}$.

This was an "applied theory" problem I got incorrect on an generated exam I recently took. This question seems to be a similar idea, excepting that I need to work backwards from speed rather than velocity and that speed is constant. Here is what I understand:

1. Since we're working in 2-D, $|\vec{v}(t)| = \sqrt{v_x^2 + v_y^2} = 5 \implies |\vec{v}(t)|^2 = 25 = v_x^2 + v_y^2$. I don't know if this helps me.
2. The parabola can be reparameterized as a trajectory/displacement curve: $\vec{r}(t) = \langle t, t^2 \rangle$
3. From the reparameterization, we can differentiate $\frac{d \vec{r}}{dt} = \vec{v}(t) = \langle 1, 2t \rangle$ and once more $\frac{d \vec{v}}{dt} = \vec{a}(t) = \langle 0, 2 \rangle$.
4. If I compute the supposed magnitude of $\vec{v}$ at point $(\sqrt{2}, 2) \Longleftrightarrow t = \sqrt{2}$, I find $$ |\vec{v}(t=\sqrt{2})| = \sqrt{1^2 + (2\sqrt{2})^2} = \sqrt{1 + 8} = \sqrt{9} = 3 \neq 5 $$ A contradiction of sorts.

Point ($3$) above would lead one to naively assume that acceleration is $2$ units/sec$^2$ upwards for all times $t$. From there, one could conclude $\vec{F} = m \langle 0, 2 \rangle = \langle 0, 2m \rangle$. Alas the automated grader reports the correct answer is:

$$\displaystyle \vec{F} = \langle-\frac{-100\sqrt{2}}{81}m, \frac{50}{81}m\rangle$$

The connection between 3 (or 9) and 81 is not lost on me ($3^4 = 9^2 = 81$), but I don't see how to bridge the gap. Any help would be greatly appreciated. I've already got my grade and the "answer". What I seek now is understanding.

Many thanks in advance.

Answer 1

The curvature of the parabola $y=x^2$ is:

$$k=\dfrac{2}{(4x^2+1)^{3/2}}$$

So the centripetal force is:

$$F_c=V^2k$$

Answer 2

Here is an alternate method. We will approximate the parabola as a circle at $(\sqrt{2}, 2)$. The radius of curvature is given by:

$$R = \bigg\vert\frac{(1 + y'^{2})^{\frac{3}{2}}}{y''}\bigg\vert$$

Plugging in $y = x^{2}, y' = 2x, y'' = 2$:

$$R = \frac{(1 + 4x^{2})^{\frac{3}{2}}}{2}$$

At $x = \sqrt{2}$:

$$R = \frac{9^{\frac{3}{2}}}{2} = \frac{27}{2}$$

Now, the centripetal force on the object is given by:

$$\vert F\vert = \frac{mv^{2}}{R} = \frac{50m}{27}$$

The force is directed towards the center of the circle. From $(\sqrt{2}, 2)$, the direction towards the center of the circle is normal to the tangent of the parabola. Thus, the line between the center of the circle and $(\sqrt{2}, 2)$ has slope $-\frac{\sqrt{2}}{4}$. Also, the center of the circle will be "inside" the curve of the parabola. Using this information, we find the force to be:

$$\boxed{\vec{F} = \bigg(-\frac{100m\sqrt{2}}{81}, \frac{50m}{81}\bigg)}$$

Answer 3

This is how you can proceed without using the notion of curvature, which is not necessary. I write some hints.

Write $\mathbf x(t)=(\alpha(t),\alpha^2(t))$

In general terms:

$\dot{ \mathbf x}=(\dot \alpha(t),2\alpha(t) \dot \alpha(t)) [1]$

Can you get the analog of [1] for $\ddot{ \mathbf x}$ ? Hint: Just differentiate [1]

Now imposing the constrain on the velocity:

$v^2=\dot{ \mathbf x}^2(t)=\dot \alpha^2(t)+4\alpha^2(t)\dot \alpha^2(t)$

So that:

$\dot \alpha^2(t)=v^2/(1+4\alpha^2(t))$ [2]

Can you get the analog of [2] for $\ddot \alpha(t)$ ? Hint: just differentiate [2].

If you did all the steps you could substitute the analog of [2] into the analog of [1] and obtain your answer :) ?