Question with normal force

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the question is:

The figure below shows the plan for a proposed rollercoaster track. Each car will start from rest at point A and will roll with negligible friction. For safety, it is important that there be at least some small positive normal force (a push) exerted by the track on the car at all points. What is the minimum safe value for the radius at point B?

the picture shows a cart at the top of the track with a heigh of 15.0meters signified by point A. it then goes down and back up to a point B which is 5.0meters high.

if someone can just start me off in the right direction i would greatly appreciate it.
 

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  • #2
dav2008
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Edit: I misread the question.
 
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  • #3
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I know Normal force will be consistent throughout...so should I use the Work Energy Theorem for the external forces...because I know it cant just be mg...
 
  • #4
dav2008
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Edit: NM I thought it was a circular loop.
 
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  • #5
Astronuc
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Think of centripetal force.

Fcentripetal = mv2/r, which must be less than the weight of the car in order to assure + force down onto the track.

http://hyperphysics.phy-astr.gsu.edu/Hbase/cf.html#cf

Then one has to determine the speed (magnitude of velocity) at B.
 
  • #6
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if its a circular motion problem shouldnt the net force be 0?..and there is no loop in the diagram...its just point a at 15.0 meters that goes down to 0 meters then back up another hump 5 meters tall...
 
  • #7
dav2008
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d0rk said:
if its a circular motion problem shouldnt the net force be 0?..and there is no loop in the diagram...its just point a at 15.0 meters that goes down to 0 meters then back up another hump 5 meters tall...
Oh my bad. I misunderstood the problem. For some reason I assumed that it was a circular loop.
 
  • #8
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without any mass how would i be able to incorporate the centripetal force...
 
  • #9
Astronuc
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One does not need mass if all equations can be written on a per mass basis or specific energy. For example,

gravitational potential energy = mgh, but specific GPE = gh.

kinetic energy = 1/2 mv2, but specific KE = v2/2,

graviational force (weight) = mg, but in specific terms = g (which is just acceleration), and

centripetal force F = ma = mv2/r, is simply v2/r with the mass divided out, and actually one determines when centripetal acceleration = gravitational acceleration, or

g > v2/r for a net downward force, or use = for minimum radius.
 
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  • #10
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Please check if this is right

Astronuc said:
Think of centripetal force.

Fcentripetal = mv2/r, which must be less than the weight of the car in order to assure + force down onto the track.

http://hyperphysics.phy-astr.gsu.edu/Hbase/cf.html#cf

Then one has to determine the speed (magnitude of velocity) at B.


this is what i have so far

v^2/R<g
K=U
1/2mv^2=mgh
v^2=2gh
g>2gh/r

i put my reference line at the bottom which = 0
the car starts at the top at 15m high then goes down to 0 then back up another hump, point B, at 5.0meters. for h do i subtract 15-5=10m if i do this is what i get...

g>2g(10)/r
r=20m?

could that be the minimum safe value for the radius at point B??
 
  • #11
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ok...now after thinking about it...i solved for the velocity at the bottom of the track right before it hits the hump B...

mgh=1/2mv^2
mg(15)=1/2mv^2
30g=v^2
v=sqrt(30g)

is this the velocity at the bottom of the track for the first part?
 
  • #12
Astronuc
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mgh=1/2mv^2
mg(15)=1/2mv^2
30g=v^2
v=sqrt(30g)

is this the velocity at the bottom of the track for the first part?
That would be the velocity at an elevation of 0.

You seem to be headed in the right direction.

Basically at 15 m, and v=0, the energy for this system is all potential. Then at 0 m elevation, the energy is all kinetic.

The cart loses some KE, while it regains some GPE as it increases elevation again to 5 m.

From conservation of energy, one can determine the change in KE from the change in PE, and from KE, one obtains the velocity (at the elevation of interest), and from velocity on uses the balance of force (gravity and centripetal) to get the corresponding radius.
 
  • #13
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so with the velocity at sqrt(30g)...

with that when the cart starts to go over the second hump (B) the velocity needs to equal:

mg(5)=1/2mv^2
10g=v^2
v=sqrt(10g)

I'm assuming that this equals the velocity at the top of hump B...

conservation of energy states that PE(final)+KE(final)=PE(initial)+KE(initial)

Fcen=mv^2/r

so at the top of point B would it be correct to say that it is all potential? and that i would just need to take the difference of the velocity at 0 meters and at 5 meters and substitute v in the the Centripetal force equation and solve for r? but then again for the mass in the equation i would need to use g>v^2/r??
 
  • #14
Astronuc
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so at the top of point B would it be correct to say that it is all potential?
The total energy at B is a combination of potential and kinetic, since the cart has increased elevation, but is still moving. The minimum potential is at 0 m.

To find the velocity at B, the KE at B should be just the difference in potential between the start (where KE = 0) and the potential energy at B.

So find the corresponding velocity.

Then mg = mv2/r.

Here is an example of the principle with which we are working - http://hyperphysics.phy-astr.gsu.edu/hbase/flobj.html

Gravitational potential energy (a linear function of elevation - mgh) is converted to kinetic energy. The course simply affects the direction in which the vehicle travels - assuming no dissipative forces like air resistance or friction with the surface.
 
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  • #15
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This is what I got:

PE(final)+KE(final)=PE(initial)+KE(initial)
1/2mv^2 + mg(5) = 0 + mg(15)
1/2v^2 + 5g = 0 + 15g
1/2v^2 = 10g
v^2 = 20g

mg = mv^2/r
g = 20g/r
r = 20???

Is that correct??
 
  • #16
Astronuc
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20 m is correct - and I am assuming that the track at point B has a horizontal tangent. This is important because the centripetal force acts perpendicular to the tangent on a curve, which can have any orientation with respect to horizontal (or vertical), but gravity is always downward.

Is there a diagram of the track? Is the tangent of the curve at point B horizontal?

One has to remember that forces and acceleration are vector quantities.
 

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