Ball rolling down a slope ending with a loop

[Hendra]

Homework Statement

A small ball rolls down a slide to which is attached a devil's loop of radius R. What should be the height of the slide so that the ball falls off at a height of 1.5R?

Relevant Equations

NA

Hello everyone! I tried to solve this problem in a non-inertial system. Probably I should use the principle of conservation of mechanical energy in the following form
$$mgH = \frac{3mgR}{2} + \frac{mV^2}{2}.$$
So the only thing to do is to compute $V^2$. I tried to find this value using the identity below
$$\frac{mV^2}{R} \sin \alpha = mg,$$
where $\sin \alpha = 0.5$.

I think that the equation for $V$ is wrong. What should it look like?

 

[kuruman]

Try drawing a free body diagram (FBD) when the ball is at the location of interest. Does it give your, so-called, identity? How is α defined?

 

[BvU]

Hello @Hendra

$\qquad$ !
​

Can you post a drawing of the situation as you think it 'unrolls' ( ) ?

And for that matter: do you need to take the 'rolling' into account ?

In what way is your system non-inertial ?

( @kuruman : the way I read the problem statement, the ball doesn't even get there !)

 

[kuruman]

( @kuruman : the way I read the problem statement, the ball doesn't even get there !)

I interpreted this to be the standard loop-de-loop exercise in which the ball starts on a very long inclined ramp and enters the loop at its bottom. But let's wait for OP's reply.

 

[haruspex]

I think that the equation for $V$ is wrong. What should it look like?

Quite so. Please explain how you arrived at it. In what direction did you consider the balance of forces? Remember the ball will have nonzero tangential acceleration at that point.

 

[Hendra]

Thank you all for your comments. To sum up I will add some diagrams to illustrate what I am dealing with. First of all the slope and the loop.

We need to consider the position of our ball at the height of 1.5R. To my mind the distribution of the forces would look as follows

Here Q is the gravity force and F is the centrifugal force. N stands for the pressure which comes from the Newton's third law of dynamics.

It is easy to compute the angle
$$\sin \alpha = \frac{1}{2} \implies \alpha = \frac{\pi}{6}.$$

Calculating the value of H is also quite easy. Using the principle of conservation of mechanical energy we conclude that
$$H = \frac{3}{2}R + \frac{V^2}{2g}.$$

The tricky part for me was to figure out what forces need to be balanced (to compute V); I thought that would be correct
$$\frac{mV^2}{R} = mg \sin \alpha,$$
but after a while I bet it is wrong.

 

[Hendra]

Try drawing a free body diagram (FBD) when the ball is at the location of interest. Does it give your, so-called, identity? How is α defined?

I think I added all the needed information

 

[BvU]

You let @kuruman seduce you to draw the ball as a rectangle !

I don't see any relationships for the $F$, $N$ and $Q$ in your drawing ?

Your conservation of energy applies where and when ?

Note the hints by @haruspex and @kuruman : at the point of interest, what about radial and tangential accelerations ?

If you can't find a synthetic (forward) path through an exercise, you can always try reasoning backwards:
Suppose $H = 2R$. How far does the ball come ? And if $H = R$ ?

$\$

 

[Afo]

If the ball rolls (which seems to be the case in your question), you must account for rotational kinetic energy. The slide must not be smooth (in order for the ball to roll), and the ball must roll without slipping in order for the mechanical energy to be conserved. (So that the work done by the friction force is zero, so the mechanical energy of the system consisting of the Earth and ball is conserved)

Why are you trying to solve in a non-inertial reference frame? Newton's law doesn't work in a non-inertial reference frame. (Unless you introduce fictitious/fake forces like the centrifugal force)

 

[BvU]

you must account for rotational kinetic energy

'must' is a strong term. I put it as a question in post #3: since you have not been given the diameter of the ball, all you have is the term 'small'.

Perhaps you may assume the ball is small enough that you do not have to take the rotational kinetic energy into account ?

$\$

 

[kuruman]

The tricky part for me was to figure out what forces need to be balanced (to compute V); I thought that would be correct

Why do you say that the forces need to be balanced? If they were, the ball would have zero acceleration at the point of interest. To figure out the forces, you first need to ask yourself, "what can conceivably exert a force on the ball?" The answer to that is simple, there is a (assumed frictionless) track and an Earth, nothing else. "Centrifugal force" is not an item that can interact with the ball.

The next thing to decide is the direction of each of the two forces. Everybody knows that gravity is straight down. That would be force labeled "Q" in your drawing. If the track is frictionless, this means that it can only exert a force perpendicular and away from its surface. That would be force labeled "N" in your drawing. That's it. Two items that can exert a force on the ball - two arrows representing these force. Force labeled "F" does not belong. It cannot be the net force because when you add a vector that is "straight down" and a vector that is "down and to the left" there ain't no way you can get a vector that is "up and to the right". You get a vector that is "more down and to the left" which is also the direction of the (non-zero) acceleration.

Proceed from here and write F_net = ma for the diagram you have with red force F out of the picture. Pay attention to the radial direction.

 

[Lnewqban]

What does exactly mean that “the ball falls off at a height of 1.5R”?
At that point of the circle where it intersects that horizontal line 1.5 R above the ground, should the ball:
1) Reach zero forward velocity and fall down vertically?
2) Separate itself from the loop, still having forward (up and toward the left) velocity, and initiating a falling parabolic rather than circular trajectory?

 

[haruspex]

$$\frac{mV^2}{R} = mg \sin \alpha,$$

Yes, that's it. You had the sine on the wrong side previously. I assume you resolved in the vertical direction before, but overlooked that the tangential acceleration has a vertical component. Resolving radially gives your correct version.

Perhaps you may assume the ball is small enough that you do not have to take the rotational kinetic energy into account ?

The radius is irrelevant in that regard. (It only matters in that you can ignore the difference it makes to the height change.) If it rolls, a fixed proportion of the KE is rotational.
That said, it is possible the question setter overlooked this.