13
$\begingroup$

I have a simple physics 101 problem that leads to a strange result.

A cart of mass $m$ is bound to move without friction on a rail placed in the vertical plane (as shown in the picture). The rail is formed by two circular pieces of radius $R$ glued together.

enter image description here

Initially, the cart is standing still at the bottom of the rail, and at time $t=0$, constant force $\mathbf{F}$ is applied to the cart. The force $\mathbf{F}$ is at each point of the trajectory tangent to the rail. Once the cart reaches a height of $2R$, the force $\mathbf{F}$ disappears.

Question: what is the minimum value of $F$ that makes the cart reach a height equal to $2R$?

Solution

An approach to solve the problem would be to use the work-energy theorem, which states that the variation of kinetic energy is equal to the work done

$$W_{o\to b} = \frac{1}{2}m v^2(b) - \frac{1}{2}m v^2(o) $$

The work done by the two forces (the weight and the force $\mathbf{F}$) can be computed as

$$W_{o\to b} = \int_o^b (\mathbf{F}+m\mathbf{g})\cdot \mathrm{d}\mathbf{s}=F\int_o^b\mathrm{d}\mathbf{s} - mg2R=F\pi R - 2mgR,$$

where we used the fact that the force $\mathbf{F}$ is always parallel to the displacement $\mathrm{d}\mathbf{s}$.

On the other hand, I know that the velocity in $o$ is zero, and if I want to compute the minimum value of the force $\mathbf{F}$ that allows the cart to reach the height $2R$, I assume that the velocity is zero also in the point $b$, hence

$$\frac{1}{2}m v^2(b) - \frac{1}{2}m v^2(o)= 0-0 =0$$

Therefore the work-energy theorem becomes

$$W_{o\to b} = F\pi R - 2mgR = 0,$$

which, solving for $F$, gives

$$F= \frac{2}{\pi} mg.$$

This result seems puzzling to me.

Indeed, the beginning of the rail is vertical, so the force $F$ needed to overcome the weight and allow the cart to move up must be at least $mg$.

Why does the work-energy theorem lead to the wrong result? Did I apply it in the wrong way?

$\endgroup$
3
  • 4
    $\begingroup$ Interesting problem. Simplify the problem to pure vertical movement. With $\Delta K = 0$ the result would be $\vec{F} = mg \hat{\jmath}$ which is not valid since that (external) force would never get the cart off the ground. $\endgroup$ Jan 19 at 15:15
  • 3
    $\begingroup$ As Phillip Wood points out, if $F$ is constant (along the curve), then how is it possible for the bead to reach zero velocity at point $b$? Is that a realistic possibility? At point $o$, $F$ is enough to overcome the gravitational force, yet at point $b$ that same force is not enough, even though it is of the same magnitude. $\endgroup$ Jan 19 at 16:51
  • 1
    $\begingroup$ Impose the ${\rm K.E.} \geq 0$ condition at every intermediate point between $o$ an $b$. For a point $\epsilon \bf \hat y$ just above $o$, this will require $F\geq mg$. $\endgroup$
    – Mahdiyar
    Jan 19 at 19:39

4 Answers 4

11
$\begingroup$

(a) If the force were constant neither its direction nor its magnitude would change. We are assuming here that the magnitude is constant.

(b) It isn't the work-energy theorem that has generated a paradox, but your assumption that for minimum $F$ the cart's KE is zero at b. The energy condition that we do have a right to impose is the less stringent one that $\text{KE} ≥ 0$ at b. Therefore, from your work calculations, $F ≥ \frac 2\pi mg$.

(c) As you've pointed out, the equality, $F = \frac 2\pi mg$ cannot be the right choice, because it wouldn't allow the cart to get started on the climb! We know how large $F$ does need to be.

(d) [added at the suggestion of YiFan] With $F = mg$ (the minimum value of $F$ for the journey to start) the cart will gain KE throughout the journey (except at the very beginning and the very end), so its KE at b can't be zero. The general condition for gaining KE at any point along the curve is easily shown to be $\sin \theta ≤\frac F {mg}$ in which $\theta$ is the local angle of the curve to the horizontal.

(e) Would the problem have a less trivial answer if the left hand and right hand pieces of the curve were exchanged, so that the vertical bit came in the middle of the complete curve? [I believe that it has!]

$\endgroup$
3
  • 4
    $\begingroup$ The OP assumes KE=0 at b because they thought that at the minimal F, that should be the case. (The logic being, if KE > 0 at b, then we could have decreased F a bit with the cart still reaching b; it must reach with speed zero if it only "just" reaches.) Perhaps you could elaborate more in your answer why this is incorrect. $\endgroup$
    – YiFan
    Jan 20 at 2:40
  • 3
    $\begingroup$ @YiFan The KE can't be zero at b, because if $F$ is large enough (that is $F ≥ mg$) to get the cart up the first (vertical) part of the curve, it will cause the cart to gain KE over the rest of the journey (except infinitesimally close to b). The condition for gaining KE at any point along the curve is easily shown to be $\sin \theta ≤ \frac F {mg}$ in which $\theta$ is the local angle of the curve to the horizontal. $\endgroup$ Jan 20 at 9:48
  • 2
    $\begingroup$ Yeah, perhaps it's good to edit that into the answer for completeness? $\endgroup$
    – YiFan
    Jan 20 at 10:28
8
$\begingroup$

Since this is a 1-D problem, we can define a potential energy due to the applied force as well, given by $$ U_F \equiv -\int_0^\vec{r} \vec{F} \cdot d\vec{r} = - F s $$ where $s$ is the arc length along the path. Adjusting the value of $F$ can just be thought of as adjusting the "strength" of this potential energy.

A bit of geometry shows that we can relate $s$ to the height $y$ by $$ s = R \begin{cases} \arcsin(y/R) & 0 \leq y \leq R \\ \arccos(2 - y/R) + \pi / 2 & R < y \leq 2R\end{cases} $$ and so the total potential energy for the object is $$ U = m g y - F R \begin{cases} \arcsin(y/R) & 0 \leq y \leq R \\ \arccos(2 - y/R) + \pi / 2 & R < y \leq 2R \end{cases}. $$

For $F = mg$, the potential energy looks like this:

enter image description here

It is easy to see that a particle at $y = 0$ with negligible initial kinetic energy will make it to $y = 2R$ (and will have a substantial amount of KE when it gets there.) On the other hand, if we plot the potential for $F = 2 m g /\pi$, we get the following graph: enter image description here

If the cart did get to $y = 2R$ in this potential, it would arrive there with the same KE it started with; we would have $\Delta K = -\Delta U = 0$ between these points. But it would have to have an initial velocity that's sufficient to get it "over the hump". On the other hand, if the cart starts with zero KE, it can't get over the hump and so will never arrive at $y = 2R$.

$\endgroup$
4
$\begingroup$

The equation of motion is:

$$m\,\ddot s+F-\cos\left(\frac sR\right)\,m\,g=0$$

form here with $~s=\pi\,R~$ and $~\ddot s=0~$ you obtain that $$F=-m\,g$$

now your problem

multiply the EOM with $~\dot s~$ and integrate you obtain

$$\frac m2\dot s^2=-\int F\,ds-\int \cos\left(\frac sR\right)\,m\,g\,ds=-F\,s-m\,g\,R\sin\left(\frac sR\right)\quad\Rightarrow\\ W=-F\,s-m\,g\,R\sin\left(\frac sR\right)$$ but with $~s=\pi\,R~$ you obtain that $~F=0$ thus your ansatz is probably wrong


but if the work is

$$W=m\,g\,y+F\,y$$ you obtain for y=$~2\,R~,W=0$ ,$~F=-m\,g~$

$\endgroup$
3
  • $\begingroup$ In the first equation, don't we have a minus sign in front of the $\cos(s/R)$? The component of the weight parallel to the rail is always opposite to the displacement. Otherwise, great insight! $\endgroup$ Jan 20 at 10:59
  • $\begingroup$ But then you will obtain that F=mg ? $\endgroup$
    – Eli
    Jan 20 at 12:43
  • 1
    $\begingroup$ no because cos(π) = -1 $\endgroup$ Jan 20 at 19:20
1
$\begingroup$

The way that the problem is stated, $F >= mg$ for the cart starts to move. But afterwards, the force will always be bigger than the tangential component of mg. So the cart will accelerate all the time. So the minimum is $F = mg$

And if not only the magnitude but also the direction of $\mathbf F$ is constant, but the cart is constrained to follow the trajectory?

$\mathbf F_{net} = m\mathbf a$

During the trajectory, there are normal forces. As they are always orthogonal to the allowable displacement, we can get rid of them by making a dot product with the elementary displacement:

$(\mathbf F - m\mathbf g)\mathbf{.dr} = m\mathbf a\mathbf{.dr}$

Until $x = R$, naming $cos(\theta) = 1 - \frac{x}{R}$: $$(F - mg)drcos(\theta) = m\frac{\mathbf {dv}}{dt}\mathbf{.dr}\implies (F - mg)Rd\theta cos(\theta) = m\mathbf v\mathbf {.dv} = d\left(\frac{1}{2}mv^2\right)$$

Integrating from $0$ to $\frac{\pi}{2}$, and supposing $v = 0$ for $\theta = 0$: $$(F - mg)R = \frac{1}{2}mv_1^2 $$ It is necessary that $F >= mg$ for this equation be fulfilled

The second path is similar, except that $cos(\theta) = \frac{x}{R} - 1$. When $\theta = -\frac{\pi}{2}$, $v = v_1$ and the integral is from $-\frac{\pi}{2}$ to $0$ $$(F - mg)R = \frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2 $$

If $F = mg$, $v_1 = 0$, and $v_2 = 0$. So, also in this case, $F = mg$

But the problem makes more sense. The speed in this limit situation is always zero. The projection of F and mg will always be equal, so given a small initial velocity, it keeps the same.

$\endgroup$

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.