What's the Velocity at the Highest Point of a Swing?

[Oliver919]

So currently I'm stuck in a topic which I don't understand because either my teacher is doing wrong or I'm just blind to see my mistake

I want to know the velocity of a object at the highest point in a swing (looping is the German word, I'm not sure if it's the same in English) (I don't know, plane, bird, superman whatever can do a swing;D)

So far I equated the centrifugalforce and the weightforce

Fz = Fg which I substituded in Fz = Mv^2 / r

mg = mv^2/r which resulted in v = sqrt gr
(mg = mv^2/r (m's cancel out)
g = v^2/r
v = (sqrt(gr))
therefore the minimum speed of the swing must be: v = (sqrt(gr))

I honestly don't know what I did wrong but my teacher keeps telling me I made a mistake and it starts to bother me :D

Does someone realize the mistake or did I do everything correct?

I'm usually the type of guy who just calculates things and normally doesn't scrutinize things, so don't waste your time doing huge physical explanations, I'd be thankful but I can assure you I'm not really going to understand them anyway :D

 

[Charles Link]

Usually, by swing, they are referring to the motion of a pendulum. At its highest point it is reversing direction from a very small positive velocity to going the other way, or visa versa. So, assuming it reverses direction, what must the velocity be a precisely the instant it turns around? (Incidentally, a swing is (one definition) a small chair held by two ropes that a child swings back and forth in, like a pendulum. I did a google of swingset in German, and it came up as "Schaukel".)

 

[Oliver919]

In this case it is not a pendulum, I can't really find an example... Like it's a military pilot and he's joking around and doing fun stuff then he's looping the loop (is that the correct phrase?) so it doesn't really have to do anything with the velocity given at the instant turnaround

 

[Charles Link]

In this case it is not a pendulum, I can't really find an example... Like it's a military pilot and he's joking around and doing fun stuff then he's looping the loop (is that the correct phrase?) so it doesn't really have to do anything with the velocity given at the instant turnaround

That's called a loop, but in that case there is no precise answer. In that case as long as his centripetal acceleration at the top of the arc is greater than or equal to g, the passenger won't fall out of their seat when they are upside down. I think they might simply be looking for the answer $v=0$ at the highest point in a pendulum.

 

[Oliver919]

Yes that's what I thought too (In that case as long as his centripetal acceleration at the top of the arc is greater than or equal to g, the passenger won't fall out of their seat.) because Fz has to be greater or the same as Fg but to calculate the needed velocity for DOING this loop (and here I realize my writing mistake, sorry for that!) and not just to accomplish the loop (because then v = 0 at highest point would be correct, obviously) and thus I ask if my formula is correct for the wanted calculation

To be honest, I don't know if I can state my problem properly I hope you understand what I'm asking for since there has to be a way to calculate the needed velocity

 

[Charles Link]

Yes that's what I thought too (In that case as long as his centripetal acceleration at the top of the arc is greater than or equal to g, the passenger won't fall out of their seat.) because Fz has to be greater or the same as Fg but to calculate the needed velocity for DOING this loop (and here I realize my writing mistake, sorry for that!) and not just to accomplish the loop (because then v = 0 at highest point would be correct, obviously) and thus I ask if my formula is correct for the wanted calculation

In doing a loop, the vertical v is zero at this point, but the horizontal v can be quite large, making for a large $a=v^2/r$.

 

[Oliver919]

Okey, I got that but what do I calculate with this: v = (sqrt(gr)) ? And there comes my other question to your formula: a = v^2/r, well I have to calculate a for my v because that's not given, so that doesn't really help, does it?

I equated Fz and Fg but for which purpose, do you know that?

 

[Charles Link]

Okey, I got that but what do I calculate with this: v = (sqrt(gr)) ?

I equated Fz and Fg but for which purpose, do you know that?

By adjusting the flaps (how fast the airplane turns), or even in an amusement park ride, there is no requirement that the force is exactly zero between the passenger and the seat at the top of the arc. If you have the $a=v^2/r$ larger than g, the person will get pushed against the seat, so making this force zero is not a requirement. In general, $F_{net}=F_g+F_{seat}=ma=mv^2/r$ with $F_g=mg$ so that $F_{seat}=m(v^2/r-g)$.

 

[Oliver919]

Aaah I see, this case comes to the same solution afterall as I thought it would be

 

[Charles Link]

Alright but then comes to my mind: How do I calculate a if v is not given (because 0 can't be in this case, and horizontal is unknown)

(Please read the edited part of my last post.) I think you understand the physics of the entire topic for whatever they might ask, but the instructor needs to be more clear on exactly what it is that they want you to compute.

 

[Oliver919]

Yes, he's quite confused and confusing, well I'll take it, thanks!