Proving that apparent weight is 5 times actual weight (no numbers)

[testme]

Homework Statement

Without numbers, identify how you could find the apparent weight of motorcycle to be 5 times that of the actual weight. The motorcycle is going up a loop the loop with a velocity of v.

Homework Equations

Fnet = mv^2/r
Fg = mg

The Attempt at a Solution

Fnet = mv^2/r
Fg + Fn = mv^2/r
Fn = mv^2/r - Fg
Fn = mv^2/r - mg
Fn = mv^2/r - mgr/r
Fn = (mv^2 - mgr)/r

I have no idea of this is right or if I'm completely misinterpreting the question.

 

[Dadface]

This seems to be an open ended question and does not require you to consider a specific position for the bike.May I suggest that you consider the bike at its bottom most position(that's where its apparent weight will be biggest for a constant speed).What are the two relevant forces on the bike,in what directions do they act and what is an expression for the resultant force?

 

[tiny-tim]

hi testme!

Without numbers, identify how you could find the apparent weight of motorcycle to be 5 times that of the actual weight. The motorcycle is going up a loop the loop with a velocity of v.
…
I have no idea of this is right or if I'm completely misinterpreting the question.

they're asking for an explanation in words

start by deciding:

at which point in the loop is the apparent weight greatest? ​

 

[testme]

Thats what I was assuming. The bike was at the bottom, the two forces acting upon it are the normal force (apparent weight) and the gravitational force (actual weight).

using that I came up with the equation

Fnet = mv^2/r
Fg + Fn = mv^2/r
Fn = mv^2/r - Fg
Fn = mv^2/r - mg
Fn = mv^2/r - mgr/r
Fn = (mv^2 - mgr)/r

Though I'm not sure if that'd be the right way to find out. Also, that value would have to be 5 times your mass times gravitational acceleration.

When the questions asks without numbers they mean as in no numbers given, so just theoretically the process. That's what my teacher said.

 

[Dadface]

At the bottom the normal force(apparent weight) acts upwards and the weight acts downwards.

 

[tiny-tim]

words only:

"as soon as it goes into the loop, if will have a centripetal acceleration which can only be supplied by an increased normal force, which (for fixed v) will be proportional to … , which can be made large enough by making … sufficiently … " ​

 

[testme]

Fnet = mv^2/r
-Fg + Fn = mv^2/r
Fn = mv^2/r + Fg
Fn = mv^2/r + mg
Fn = mv^2/r + mgr/r
Fn = (mv^2 + mgr)/r

Would that be it then?

 

[Dadface]

Looks good but remember Fn=5mg

 

[testme]

Would it be fine that we find Fn using that and then we multiply Fg we would find by 5 at the very end?

 

[Dadface]

Fg is the "true weight" of the bike and that remains constant.What you are finding is an equation giving the necessary speed(v) for a given radius r for the bike to have an "apparent weight"(at the bottom) of 5mg.

 

[testme]

I get that, but what I mean is does it matter if I divide Fn by 5 at the end or multiply Fg by 5? Could I just say 5Fg must equal Fn or Fn/5 must equal Fg for this to be true?

 

[Dadface]

I'm not sure that I understand you.At the bottom the resultant force is Fn-mg.
(Fn represents the apparent weight in other words the weight that would be measured,by say,a set of scales over which the bike rides).
We can write: Fn-mg=mv^2/r
If the apparent weight is to be 5 times the real weight then Fn=5mg so we can write:
5mg-mg=4mg=mv^2/r

 

[testme]

Nevermind, I got my answer, thanks for the help!