Solving Motion Around a Dome: Time for Complete Circuit

[ova5676]

Homework Statement

"You are on a dome, which is a hemisphere, with a 90 m radius. You hold a rope to point A at the peak of the dome. The rope remains in contact with the outer surface of the dome and is long enough to make an angle of 25° as shown. You are given an initial horizontal velocity tangent to the dome and you move in a horizontal circle, keeping the rope tight. You have a maximum speed above which you cannot maintain contact with the frictionless dome surface.

What's the time that it takes you to complete one circuit around the dome at this speed?"

Homework Equations

Force equations, centripetal acceleration equations

The Attempt at a Solution

Okay so I'm thinking this is a centripetal acceleration question. Since the motion is in a horizontal circle I'm thinking I'll need the radius of the circle. I think the angle and dome radius form a right angle triangle so I can get the opposite side of the angle which would be the radius right? So now I have the radius.

I'm really confused on the initial velocity and maximum velocity thing, can someone explain that? No velocities were given, just the angle and the radius?

Is the force of gravity acting here or is it canceled by the normal force?

 

[ova5676]

Anyone there?

edit: I think I got it. There has to be a vertical (or would it be horizontal) force of gravity acting on this guy as he's moving around this dome, correct? (I assume no normal force since there is no friction and Ff = uk * Fn) However, Fg must be equal to the centripetal force applied as he's moving across the dome to balance, correct? In other words:

Fnet = Fc = Fg

m*a = m*g

a = g

v^2 / r = mgcostheta

v = sqrt(mgcostheta/r)

that's how i would get v

to get d, i would get the circumference of the circle, but I'm not sure if it's 2*pi*r where r is the 90 m or if it's the opposite side of the angle?

with v and d, i would have v = d/t and rearrange for t = d/v.so where did i go wrong?

 

[ova5676]

Anyone there?

 

[ova5676]

So I'm guessing it's right?

 

[ova5676]

Anyone there?

 

[tiny-tim]

hi ova5676!

(have a theta: θ and a square-root: √ and a pi: π )

edit: I think I got it. There has to be a vertical (or would it be horizontal) force of gravity acting on this guy as he's moving around this dome, correct? (I assume no normal force since there is no friction and Ff = uk * Fn) However, Fg must be equal to the centripetal force applied as he's moving across the dome to balance, correct? In other words:

Fnet = Fc = Fg

m*a = m*g

a = g

v^2 / r = mgcostheta

v = sqrt(mgcostheta/r)

sorry, i don't understand this at all

there are two forces not one, the weight vertically down, and the tension along the tangent …

start again ​

 

[ova5676]

hi ova5676!

(have a theta: θ and a square-root: √ and a pi: π )


sorry, i don't understand this at all

there are two forces not one, the weight vertically down, and the tension along the tangent …

start again ​

true, there should be tension... but how would I calculate tension? am i looking for the verticle component, as in Ty? if so, how do I get that angle?

 

[ova5676]

i don't really understand how one incorporates tension?

 

[tiny-tim]

true, there should be tension... but how would I calculate tension?

nobody's asked you for the tension, so ignore it …

take components (of force and acceleration) perpendicular to it ​

 

[ova5676]

nobody's asked you for the tension, so ignore it …

take components (of force and acceleration) perpendicular to it ​

so would it be like this:

If so, how would I get the forces perpendicular to the tension?

 

[tiny-tim]

the components of forces?

there's only one candidate, and that's gravity …

just use cos as usual ​

 

[mike2732]

the components of forces?

there's only one candidate, and that's gravity …

just use cos as usual ​

Can you be more confusing? I'm trying to solve the same problem.

Will splitting the force of gravity into it's x-component, then using that as the centripetal acceleration to solve for speed using the following formula work?

ac=v^2/r

 

[ova5676]

the components of forces?

there's only one candidate, and that's gravity …

just use cos as usual ​

So Fnet = Fg where Fg = mgcos25?

Or Fnet = Fc = mv^2/r = mgcos25?

I just solve for v, then v = d/t and rearrange for t?

 

[tiny-tim]

hi mike2732!

Will splitting the force of gravity into it's x-component, then using that as the centripetal acceleration to solve for speed using the following formula work?

no, because F = ma horizontally (i assume that's what you mean by "x") must also include a component of the tension

 

[mike2732]

hi mike2732! no, because F = ma horizontally (i assume that's what you mean by "x") must also include a component of the tension

You just said:

there's only one candidate, and that's gravity …

:S

Let me give this another shot, and if I'm wrong can you please try to be more helpful? :)

So incorporating tension: Ty= -Fg
then solve for Tx and use that acceleration as the centripetal acceleration?

Edit: wait.. that would just give you the same answer.. :|

 

[tiny-tim]

You just said:

there's only one candidate, and that's gravity …

yes … "perpendicular to the tension", not horizontal

So incorporating tension: Ty= -Fg
then solve for Tx and use that acceleration as the centripetal acceleration?

yes, that would do it …

but why bother with two equations when the one equation perpendicular to the string will solve it, without finding T?

 

[mike2732]

yes … "perpendicular to the tension", not horizontal


yes, that would do it …

but why bother with two equations when the one equation perpendicular to the string will solve it, without finding T?

what force is perpendicular to the tension?

 

[ova5676]

yes … "perpendicular to the tension", not horizontal

but why bother with two equations when the one equation perpendicular to the string will solve it, without finding T?

So it's Fgy, right? If you're looking at the diagram as a rotated coordinate system, Tension would be in the horizontal and Fg would have x- and y- components, correct?

 

[tiny-tim]

If you're looking at the diagram as a rotated coordinate system …

"a rotated coordinate system" ?

why make things so complicated?

we can take components in any direction (in this case, perpendicular to the tension) without mucking about with the coordinate system

 

[mike2732]

"a rotated coordinate system" ?

why make things so complicated?

we can take components in any direction (in this case, perpendicular to the tension) without mucking about with the coordinate system

Why are you making things so complicated? XD

Can you explain the solution in more detail so we can actually understand?

Thanks

 

[tiny-tim]

Why are you making things so complicated? XD

Can you explain the solution in more detail so we can actually understand?

erm … "we" ?

how am i supposed to help two different people with two different levels of understanding at the same time?​

 

[Minion]

Tiny-Tim, can the equation Fc= Fgy be used to solve the problem?

so it becomes mv^2/r=mgcos25

the masses cancel each other out.
and you get the v

then you can use v=2piR/T

 

[ova5676]

erm … "we" ?

how am i supposed to help two different people with two different levels of understanding at the same time?​

I think if you just listed the forces involved in the x- and y- components and you explain why the component section with forces perpendicular to the Tension should be used, we would all be on the same page.

 

[mike2732]

I think if you just listed the forces involved in the x- and y- components and you explain why the component section with forces perpendicular to the Tension should be used, we would all be on the same page.

Agreed.

Basically just fully explain your solution.

 

[ova5676]

Anyone got anything?