Apparent weight problem (kinematics + conservation of Energy + Newton's laws)

[simphys]

Homework Statement

Show that on a roller coaster with a circular vertical loop
(Fig. 43), the difference in your apparent weight at the top of
the loop and the bottom of the loop is 6 g’s—that is, six times
your weight. Ignore friction. Show also that as long as your speed is above the minimum needed, this answer doesn’t depend on the size of the loop or how fast you go through it.

Relevant Equations

curvilinear motion + conservation of energy + newton's laws

Hello there, I have tried the problem but don't get a different of 6g's as I am supposed to. I am not sure whether I interpreted the problem in the correct way, but I would love some feedback/hints on what went wrong in my solution, thanks in advance.

Solution:
SITUATION DRAWINGS + FBDS

so based on the situations described in the situational picture:
For $F_n$ at $2$:
$F_n - mg = ma_n$ where $a_n = \frac{v_2^2}{R}$
--> $F_n = m(\frac{v_2^2}{R} + g)$ $(1)$

For $F_n$ at $3$:
$F_n + mg = m*\frac{v_3^2}{R}$
$F_n = m(\frac{v_3^2}{R} - g)$ $(2)$

Then moving on to finding those velocities we use conservation of Mechanical energy for both of the situations (2 and 3):
Note: I assumed $v_1 = 0$, I don't know if that is permitted as it doesn't say it and perhaps the mistake lies in this one?
$E_1 = E_2$
$\frac{mv_1^2}{2} + mgy_1 = \frac{mv_2^2}{2} + mgy_2$
$0 + mgh = \frac{mv_2^2}{2} + 0$
$v_2 = \sqrt{2gh}$

for $v_3$:
$E_1 = E_3$
$mgh + 0 = mg(2R) + \frac{mv_3^2}{2}$
$v_2 =\sqrt{2gh - 4gR}$

substituting these in $(1)$ and $(2)$:

$F_{n,2} = m(\frac{2gh}{R} + g) = \frac mR (gR + 2gh)$
$F_{n,3} = m(\frac{2gh - 4gR}{R} - g) = \frac mR (2gh - 4Rg - Rg) = \frac mR (2gh -5Rg)$

$\Delta F_n = F_{n,3} - F_{n,2} =\frac mR (2gh -5gR) - \frac mR (gR + 2gh)$
$= \frac mR (gh - 6gR)$

This is my answer which is definitely not 6g's (note, this is only part one of the problem) i.e. not 'Show also that as long as your speed is above the minimum needed, this answer doesn’t depend on the size of the loop or how fast you go through it.'

 

[malawi_glenn]

Where is your apparent weight maximal, at point 2 or at point 3?

 

[Orodruin]

You have an arithmetic error in the last step.

 

[simphys]

Where is your apparent weight maximal, at point 2 or at point 3?

what do you mean by that?

 

[simphys]

You have an arithmetic error in the last step.

thank you, omggg, for real... I went through the whole problem while writing and thought argh screw it, I'll just write the last part over from my paper...

and here it is ...

 

[malawi_glenn]

what do you mean by that?

Easier to do "largest minus smallest"

Also, it is easier to write $v_3^2 = v_2^2 - 4Rg$ instead of all the g's and h's

 

[simphys]

Easier to do "largest minus smallest"

Also, it is easier to write $v_3^2 = v_2^2 - 4Rg$ instead of all the g's and h's

oh yeah, you're right, thank you. Got -6mg now 6mg is more logical lol.

 

[malawi_glenn]

oh yeah, you're right, thank you. Got -6mg now 6mg is more logical lol.

$F_{N2} > F_{N3}$
$F_{N2} - F_{N3} = m\left( \dfrac{v_2^2}{R} + g \right) - m\left( \dfrac{v_2^2 - 4Rg}{R} - g \right) = 6mg$

much cleaner

 

[Orodruin]

Well … technically … (my emphasis)

Homework Statement:: … the difference in your apparent weight at the top of
the loop and the bottom of the loop is 6 g’s—

This would typically mean ”top value minus bottom value” …

 

[simphys]

$F_{N2} > F_{N3}$
$F_{N2} - F_{N3} = m\left( \dfrac{v_2^2}{R} + g \right) - m\left( \dfrac{v_2^2 - 4Rg}{R} - g \right) = 6mg$

much cleaner

oh a 100% great idea, I'll write that down like that thank you.
Btw a llittle question.
why is my assumption of $v_1 = 0$ correct? I mean what makes that assumption correct as it isn't stated in the problem.

 

[kuruman]

why is my assumption of $v_1=0$ correct? I mean what makes that assumption correct as it isn't stated in the problem.

Any expression for $v_1$ would be correct as long as the speed $v_2$ at the bottom is sufficient for the mass to loop the loop. You can see that in the "much cleaner" approach. There is no $h$ and no $v_1$.

 

[Orodruin]

why is my assumption of v1=0 correct?

It is not. It is just independent of $v_1$ as you would see if you included it. In fact, you do not even have to consider $h$.

 

[malawi_glenn]

why is my assumption of v1=0 correct? I mean what makes that assumption correct as it isn't stated in the problem.

If you use v1 = 0 then v2 = (2gh)^½
What I wrote in my previous post is always true provided v3 is big enough to make F_N3 > 0

This should help you answer the second question in your problem.

 

[Orodruin]

If you use v1 = 0 then v2 = (2gh)^½
What I wrote in my previous post is always true provided v3 is big enough to make F_N3 > 0

Actually, it is sufficient that $v_3$ is big enough for the coaster to go around. Negative $F_3$ would imply hanging from the top of the loop instead of being pressed into the track. Now, I don’t know if any roller coasters are actually constructed to provide negative gs in loops, but I believe roller coaster cars should be able to stay on track even with negative g for safety reasons.

 

[malawi_glenn]

”top value minus bottom value”

Then the question is wrong, because the difference is negative xD
The difference of 2 and 5 is -3 in my world

Actually, it is sufficient that v3 is big enough for the coaster to go around

True, I assumed "sliding point particle"

 

[simphys]

Any expression for $v_1$ would be correct as long as the speed $v_2$ at the bottom is sufficient for the mass to loop the loop. You can see that in the "much cleaner" approach. There is no $h$ and no $v_1$.

yep, cleaner 100% and more correct I presume as I assumed something I wasn't really sure of tbh
I think I could've also done just $E_2 = E_3$ and relating the velocities that way instead doing this long work around, haven't tried tough..

 

[simphys]

It is not. It is just independent of $v_1$ as you would see if you included it. In fact, you do not even have to consider $h$.

oh and I forgot that there was a second part, thanks for the reminder

 

[Orodruin]

I think I could've also done just E2=E3 and relating the velocities that way instead doing this long work around, haven't tried tough..

This was done in post #8

$F_{N2} > F_{N3}$
$F_{N2} - F_{N3} = m\left( \dfrac{v_2^2}{R} + g \right) - m\left( \dfrac{v_2^2 - 4Rg}{R} - g \right) = 6mg$

much cleaner

 

[simphys]

This was done in post #8

for real? how come?
I thought that the equations $(1) and (2)$ were only used here aka the ones from Newton's law which gives an equation for these forces

 

[malawi_glenn]

for real? how come?

Look in post #6

 

[erobz]

What do I forget here?

If we choose $\uparrow^+$ for position 3:

$$ -N -mg = m\frac{v^2}{R} $$

If we let $N$ go to zero, to find the minimal velocity at (3) such that the car stays on the track, I seem to run into a bit of an issue with a non-real under the root:

It needs to be this:

$$-N -mg = -m\frac{v^2}{R}$$

But what justifies needing to manually manipulate the sign on the RHS - What am I forgetting?

 

[malawi_glenn]

What do I forget here?

The direction of the resulting force (acceleration) should be pointing towards the centre of the circle.
Positve : directed towards the centre of the circle
Negative: directed away from the centre of the circle

 

[erobz]

The direction of the resulting force should be pointing towards the centre of the circle.
Positve : directed towards the centre of the circle
Negative: directed away from the centre of the circle

But I don't recall messing with the RHS of Newtons 2^nd based on convention? I assume it must be because of $\frac{v^2}{R} > 0$ that the direction is "lost"?

 

[malawi_glenn]

But I don't recall messing with the RHS of Newtons 2^nd based on convention? I assume it must be because of $\frac{v^2}{R} > 0$ that the direction is "lost"

in a fixed cartiesian coordinate system in the ground frame, the acceleration would be $- \dfrac{v^2}{R} \mathbf{\hat j}$ at point 3

Thus $-F_N \mathbf{\hat j} + -mg \mathbf{\hat j} = -\dfrac{v^2}{R} \mathbf{\hat j}$

 

[Orodruin]

But I don't recall messing with the RHS of Newtons 2^nd based on convention? I assume it must be because of $\frac{v^2}{R} > 0$ that the direction is "lost"?

Nobody is messing with Newton’s laws. It is just that you chose your positive direction to be away from the center instead of towards the center. The centripetal force has magnitude $mv^2/r$ and points to the center of the circular motion.

 

[kuruman]

Nobody is messing with Newton’s laws. It is just that you chose your positive direction to be away from the center instead of towards the center. The centripetal force has magnitude $mv^2/r$ and points to the center of the circular motion.

And at points 2 and 3 all external forces on the mass are collinear in which case the centripetal force and the net force coincide.

 

[erobz]

Nobody is messing with Newton’s laws.

I wasn't making accusations, I just realized that I was making that adjustment to get a solution, without knowing the justification.

 

[malawi_glenn]

@erobz there is a curse... if you mess with Newton, he will summon a giant apple from the sky which will fall upon thy head

 

[erobz]

@erobz there is a curse... if you mess with Newton, he will summon a giant apple from the sky which will fall upon thy head

I'd be hit by about a ton of apples by now with how often I blunder!

 

[erobz]

in a fixed cartiesian coordinate system in the ground frame, the acceleration would be $- \dfrac{v^2}{R} \mathbf{\hat j}$ at point 3

Thus $-F_N \mathbf{\hat j} + -mg \mathbf{\hat j} = -\dfrac{v^2}{R} \mathbf{\hat j}$

Yeah, and at point 2 it would be:

$F_N \mathbf{\hat j} -mg \mathbf{\hat j} = \dfrac{v^2}{R} \mathbf{\hat j}$

I still feel as though we are manually intervening a bit? to Oroduins point it seems like we must know or surmise the center of rotation.

 

[malawi_glenn]

Yeah, and at point 2 it would be:

$F_N \mathbf{\hat j} -mg \mathbf{\hat j} = \dfrac{v^2}{R} \mathbf{\hat j}$

I still feel as though we are manually intervening a bit?

One should probably write F_N2, v₂ and F_N3, v₃ to assert we are not dealing with the same F_N and v at all points.

 

[Orodruin]

@erobz there is a curse... if you mess with Newton, he will summon a giant apple from the sky which will fall upon thy head

Luckily Einstein’s theory showed that free falling objects have zero proper acceleration and he was saved as he adopted a local inertial feame where there was no ”up” for the apple to fall from.

 

[simphys]

Yeah, and at point 2 it would be:

$F_N \mathbf{\hat j} -mg \mathbf{\hat j} = \dfrac{v^2}{R} \mathbf{\hat j}$

I still feel as though we are manually intervening a bit?

Erobz do you know about the tangential and normal components aka n-t axes(for curvilinear motion)
it states that the positive normal axis (perpendicular to the tangential axis) is always towards the concave inside and thus towards center of curvature where in this case this center of curvature is constant as the motion itself is a circle.

 

[simphys]

Nobody is messing with Newton’s laws. It is just that you chose your positive direction to be away from the center instead of towards the center. The centripetal force has magnitude $mv^2/r$ and points to the center of the circular motion.

as orodruin actually indicated.

 

[Orodruin]

I wasn't making accusations, I just realized that I was making that adjustment to get a solution, without knowing the justification.

That was not the point or the important part of my post. The important point came after.