Let us assume that there is a bowl fixed on a table. The radius is R. There is a small ball on the top such that it has a potential energy of $mgR$. When the ball is dropped, at its lowest point it must have a velocity of $\sqrt{2gr}$ with respect to the ground. Now let us assume that there is a car moving at velocity $v$. According to the car that ball has both potential energy and kinetic energy (total energy) is$$ mgR + 0.5mv^2$$ Since the velocity of the ball with respect to the car is the velocity of it with respect to the ground - velocity of car with respect to ground we can represent it as:$$ \sqrt {2gr}-v$$ This velocity however doesnot conserve mechanical energy with respect to the car: $$ 0.5mv^2 +mgR= 0.5m(\sqrt{2gr}-v)^2$$ This will lead to : $$ -2v\sqrt {2gr} =0 $$ What am I missing?

  • 2
    $\begingroup$ ...Since the velocity of the ball with respect to the car is the velocity of it with respect to the ground; Are you sure about that? $\endgroup$
    – khaxan
    Feb 24 at 5:08

1 Answer 1


NOTE: My initial explanation was conceptually correct but a bit imprecise. I was accounting for the work from the portion of the normal force providing the centripetal acceleration, but not the component counteracting gravity. They are both necessary. The calculation also had an error causing it to arrive at the right answer anyway. I have tightened it up and corrected all calculations as far as I can see.


In the first case, the velocity went from 0 to $\sqrt{2gR}$. The total change in KE is the negative of the change in PE, just as you say.

In the second case, the velocity goes from -$V_{c}$ (sub c for the car) to $\sqrt{2gR}-V_{c}$. The KE at the beginning and end are:

$$ KE_{i} = \frac{1}{2}V_{c}^{2} $$

$$ KE_{f}= \frac{1}{2}(\sqrt{2gR}-V_{c})^{2} $$

The total change in KE is then

$$ \Delta KE = mgR - mV_{c}\sqrt{2gR} $$

This is similar to what you say but a bit cleaner. The change in PE is $-mgR$ as before, and as you can see, the negative of that is in the $\Delta KE$, also as before. The question is, what is the remaining term in the kinetic energy change, and how does it not violate energy conservation.

If you think carefully, the normal force on the ball, which both counteracts gravity to maintain the constraint on the bowl and provides the needed resultant force (centripetal force), has a component pointing forward during the fall. We generally think of such a normal force as doing no work, but this is for when the constraining surface is stationary. The man in the car sees a moving constraint and must handle that correctly.

In the situation here, the man in the car sees the surface of contact as moving opposite to his driving, so any component of that velocity perpendicular to the surface leads to the normal force doing work. In the picture I drew for myself the car is driving right so the man in the car sees the ball moving left at a constant drift velocity while also moving in the circle. The man in the car then sees the normal reaction force on the ball (in my picture the force points to the right and up depending on where the ball is) working against the apparent leftward motion of the ball. The work the man in the car sees the normal force do accounts for the apparent missing energy, as can be proved.

Before we do, I will point out that if you know about generalized coordinates and techniques like Lagrangian mechanics, this is what the books and your professors mean by "the forces of constraint do no virtual work, but may do real work." This is why removing constraints with these techniques can be so helpful. The stationary case would hardly warrant a Lagrangian other than the book or exam saying to provide it, but in a moving case the Lagrangian removes work and energy changes from consideration.

Here is the exact calculation verifying the work-energy theorem. Note initially there was an error when I provided this, and I was not using the whole normal reaction force, but only the portion providing centripetal force.

In order to counteract gravity and provide centripetal force, the normal on the ball is

$$ F_{N} = mg\sin(\theta) + \frac{mV_{s}^{2}}{R} $$

where the first term is the ordinary normal force on an incline that counteracts gravity (in order to stop the object from falling through the surface) and the second term is the centripetal force expected by the stationary observer. This is the motivation behind explicitly referring to that velocity at $V_{s}$, it is the velocity seen by a stationary observer which is the ordinary velocity of rolling down the circle of the bowl.

After rolling down height $h$, that stationary frame ball velocity is $V_{s} = \sqrt{2gh}$ as you know from before, where the height of fall is given by $h = R\sin(\theta)$, $\theta$ measured from the line to the starting position. The net normal reaction force force then cleans up to just:

$$ F_{N} = 3g\sin(\theta) $$

where I would recommend understanding how the equation ends up so clean (because otherwise the integration analysis is a pain and subject to easy to make errors). In essence, the velocity squared is $2gR\sin(\theta)$, and centripetal force divides by $R$, so the $R$ cancels. All that remains is $2g\sin(\theta)+g\sin(\theta)$ which obviously adds to $3$.

The differential work done by the normal force is then

$$ dW_{N} = \vec{F}_{N} \cdot d\vec{r} $$

where the differential displacement $d\vec{r}$ is in the moving frame of the car. You can do the analysis carefully yourself, but what you will find is the intuitive result that dotting the normal with the apparent motion of the ball still has the circular portion of the motion cancelling, just as in the stationary frame. This is because that motion is perpendicular to the normal.

The only portion of the apparent motion $d\vec{r}$ that causes the normal to perform work is the apparent motion coming from the moving frame itself (the motion of the car). The car is moving in the negative $x$ direction, so $dW_{N} = -F_{N}\cos(\theta)dx_{c}$, the differential motion of the car. This differential motion is given by $dx_{c} = V_{c}dt$, and we can remove time from consideration and make the entire problem angular by solving for $dt$ from the angular velocity:

$$ dt = \frac{R}{V_{s}}d\theta $$

We know how to substitute out $V_{s}$ from above, so the whole problem can be integrated:

$$ \begin{align} dW_{N} &= -F_{N}\cos(\theta)dx \\ dW_{N} &= -3mg\sin(\theta)\cos(\theta)V_{c}\frac{R}{V_{s}}d\theta \\ dW_{N} &= \frac{-3V_{c}mgR\sin(\theta)\cos(\theta)}{\sqrt{2gR\sin(\theta)}}d\theta \\ dW_{N} &= -3mV_{c}\sqrt{\frac{Rg\sin(\theta)}{2}}\cos(\theta)d\theta \\ W_{N} &= -3mV_{c} \sqrt{\frac{Rg}{2}} \int_{0}^{pi/2}(\sin(\theta))^{1/2}\cos(\theta)d\theta \\ W_{N} &= -3mV_{c} \sqrt{\frac{Rg}{2}} \int_{0}^{1}(u)^{1/2}du \\ W_{N} &= -mV_{c}\sqrt{2Rg} \end{align} $$

This is exactly the extra term in $\Delta KE$ above, and therefor the "missing" energy. The work energy theorem is verified.

  • $\begingroup$ isn't the energy transferred by the normal force irreversible. Then how, does the ball reach the same height as before after passing the lowest point? $\endgroup$
    – Ash
    Feb 25 at 7:15
  • $\begingroup$ No, energy transferred by something like friction is irreversible. Energy transfer by the normal is not irreversible. Just lift up a plate with a block on it and set it back down. The normal from the plate to the block passes work in both directions with no mechanical energy lost. In a frame moving with the plate, the normal does no work, but otherwise it does. In the "stationary frame" (to the earth), you see the plate move in both directions by the same amount. $\endgroup$ Feb 25 at 10:23
  • $\begingroup$ Think of it this way, in the frame where the ground is stationary, the normal force does no work at any time. In a moving frame, because the ground appears to move, the normal can do work along the path. However, over the course of the path, that work must add up to 0, because in the stationary frame the ground is not transferring energy to the object. In the case here, the object rolls down the bowl and up the other side where it stops and reverses direction. In the moving frame, the normal works one way on the way down (for this direction of car it was against) and the other way as it rises. $\endgroup$ Feb 25 at 22:38

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