Hat height from the ground will the child become airborne? picture included

In summary, at an angle θ, the normal reaction force is zero, and the child becomes airborne at a height of m.
  • #1
gap0063
65
0

Homework Statement


A poorly designed playground slide begins with a straight section and ends with a circular arc as shown in the figure below.
[PLAIN]http://img716.imageshack.us/img716/8381/25746193.jpg
A child starts at point P and slides down both sections of the slide. At some point on the circular arc, the normal force goes to zero and the child loses contact with the ramp.
Assuming the forces of friction are negligible, at what height from the ground will the child become airborne?
Answer in units of m.


Homework Equations


Delta U= -W= - integral Fc dr


The Attempt at a Solution


I don't know where to start... is this a potential energy problem?
and if so do I use the Work formula= 1/2 mvf^2 - 1/2mvi^2+ mgh?
and if so is the h in that equation the same h I need?
 
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  • #2
hi gap0063! :smile:
gap0063 said:
I don't know where to start... is this a potential energy problem?

it's partly a conservation of energy problem, and partly a centripetal acceleration problem …

find the angle at which the normal reaction force is zero

does that help? :wink:
 
  • #3
tiny-tim said:
hi gap0063! :smile:


it's partly a conservation of energy problem, and partly a centripetal acceleration problem …

find the angle at which the normal reaction force is zero

does that help? :wink:

So for the conservation of energy, should I use the equations:
Delta K + Delta U= (Kf+Uf)-(Ki+Ui) = Delta Emech ?
where Kf= 1/2 mvf^2
and Ki= 1/2mvi^2 (which isn't this zero since its starting at rest?)
not sure what Ui equals
but Uf is zero right?


and for the centripetal acc = vtop^2/r, where vtop is at the top of the circle... is the top fo the circle where the normal reaction force is zero?
 
  • #4
(have a delta: ∆ and a theta: θ and try using the X2 and X2 icons just above the Reply box :wink:)

∆Emech is zero,

so it's just ∆KE + ∆PE = constant.

U uses the difference in height.

Vf is the speed at which the reaction force is zero.

Hint: find the speed at a typical angle θ, then find the centripetal acceleration at that point, then find the reaction force at that point, then find the angle at which that reaction force is zero. :smile:
 
  • #5
tiny-tim said:
(have a delta: ∆ and a theta: θ and try using the X2 and X2 icons just above the Reply box :wink:)

∆Emech is zero,

so it's just ∆KE + ∆PE = constant.

U uses the difference in height.

Vf is the speed at which the reaction force is zero.

Hint: find the speed at a typical angle θ, then find the centripetal acceleration at that point, then find the reaction force at that point, then find the angle at which that reaction force is zero. :smile:

Thanks for the symbol help!

So ∆KE + ∆PE= (Kf+Uf)-(Ki+Ui) <- is that what you mean by constant?

So if U uses the difference in height, like mgh=mg2r?

Vf is the speed at which the reaction force is zero?
could you explain what the reaction force is in this problem?
I would think it was friction but since it says "the forces of friction are negligible" I don't know/
 
  • #6
gap0063 said:
So ∆KE + ∆PE= (Kf+Uf)-(Ki+Ui) <- is that what you mean by constant?

That will be zero.
So if U uses the difference in height, like mgh=mg2r?

Like that, yes.
Vf is the speed at which the reaction force is zero?
could you explain what the reaction force is in this problem?
I would think it was friction but since it says "the forces of friction are negligible" I don't know/

Friction is negligible, so the reaction force must be perpendicular to the surface.

The reaction force can only ever be worked out by finding all the other forces, and putting them into F = ma.

The reaction force is the force that is needed to make that equation true.
 
  • #7
tiny-tim said:
That will be zero.


Like that, yes.


Friction is negligible, so the reaction force must be perpendicular to the surface.

The reaction force can only ever be worked out by finding all the other forces, and putting them into F = ma.

The reaction force is the force that is needed to make that equation true.

So (Kf+Uf)-(Ki+Ui)=0 =>Kf+Uf)=(K[/SUB]i+Ui) ?

then if so, 1/2 mvf2+0=0+mg2r

At what point on the slide do I draw the fbd, so I can find out all the forces?
 
  • #8
gap0063 said:
So (Kf+Uf)-(Ki+Ui)=0 =>Kf+Uf)=(K[/SUB]i+Ui) ?

Yes, that's why it's called conservation of energy … K + U before = K + U after.
At what point on the slide do I draw the fbd, so I can find out all the forces?

At an angle θ.

You will find out what θ is after you draw the fbd and write all the equations.

ok, try it … what are the forces on the fbd at angle θ, and what is the equation for the reaction force being zero?​
 
  • #9
tiny-tim said:
That will be zero.


Like that, yes.


Friction is negligible, so the reaction force must be perpendicular to the surface.

The reaction force can only ever be worked out by finding all the other forces, and putting them into F = ma.

The reaction force is the force that is needed to make that equation true.

tiny-tim said:
Yes, that's why it's called conservation of energy … K + U before = K + U after.


At an angle θ.

You will find out what θ is after you draw the fbd and write all the equations.

ok, try it … what are the forces on the fbd at angle θ, and what is the equation for the reaction force being zero?​

Well [tex]\sum[/tex] Fy= N-mg+Fsin[tex]\theta[/tex]
[tex]\sum[/tex] Fx= Fcos[tex]\theta[/tex]
maybe?
 
  • #10
nooo :redface:

i] i assume N is the normal reaction force, but what's F? :confused:

ii] don't do the x and y components, you don't need them, you only need the normal (radial) component

ok, how many forces are there (at angle θ)?

and what is the F = ma equation in the normal (radial) direction? :smile:
 
  • #11
tiny-tim said:
nooo :redface:

i] i assume N is the normal reaction force, but what's F? :confused:

ii] don't do the x and y components, you don't need them, you only need the normal (radial) component

ok, how many forces are there (at angle θ)?

and what is the F = ma equation in the normal (radial) direction? :smile:

okay, so Ncos[tex]\theta[/tex]=mg => N= mg/cos[tex]\theta[/tex]

and the normal radial force is Nsin[tex]\theta[/tex] right?

so Nr= mg/cos[tex]\theta[/tex]*sin[tex]\theta[/tex]= tan[tex]\theta[/tex]mg

which can be tan[tex]\theta[/tex]= v2/gr


but now I'm stuck because there are two unknowns...
 
  • #12
Sorry, I don't understand any of this. :confused:

There are two forces … mg vertically down, and N radially outward.

We don't know what N is until we write the Ftotal = ma equation.

ok, so what is the Ftotal = ma equation (in the radial direction) in this case?​
 
  • #13
tiny-tim said:
Sorry, I don't understand any of this. :confused:

There are two forces … mg vertically down, and N radially outward.

We don't know what N is until we write the Ftotal = ma equation.

ok, so what is the Ftotal = ma equation (in the radial direction) in this case?​

So the radial direction is perpendicular to the normal force?

Ftotal = Ncos[tex]\theta[/tex]-mg ?
 
  • #14
gap0063 said:
So the radial direction is perpendicular to the normal force?

No, the radial direction is the normal direction.

Normal means perpendicular to the surface. The surface is tangential to the circle. The normal is radial (towards the centre of the circle).
Ftotal = Ncos[tex]\theta[/tex]-mg ?

Ncosθ - mg is the vertical component of the total force.

You need the radial component.
 
  • #15
tiny-tim said:
No, the radial direction is the normal direction.

Normal means perpendicular to the surface. The surface is tangential to the circle. The normal is radial (towards the centre of the circle).


Ncosθ - mg is the vertical component of the total force.

You need the radial component.

I have no idea how to find the radial component

i know that radial acc= v2/r

Fc=mac= m v2/r
 

Related to Hat height from the ground will the child become airborne? picture included

1. How does the height of the hat from the ground affect a child's ability to become airborne?

The height of the hat from the ground can affect a child's ability to become airborne as it determines the height at which the child will jump from. The higher the hat is from the ground, the more force and momentum the child will need to generate in order to become airborne.

2. At what hat height from the ground will the child be able to become airborne?

The hat height from the ground needed for a child to become airborne will vary depending on factors such as the child's age, weight, and physical abilities. However, on average, a hat height of at least 12 inches from the ground is recommended for a child to have enough momentum to become airborne.

3. Is there a specific age range for children to be able to become airborne at a certain hat height from the ground?

No, there is no specific age range for children to be able to become airborne at a certain hat height from the ground. Each child's physical development and abilities will vary, so the hat height from the ground needed for them to become airborne may also vary.

4. How can the hat height from the ground affect the safety of the child?

If the hat height from the ground is too low, the child may not have enough force and momentum to become airborne, resulting in a failed attempt and potential injury. On the other hand, if the hat height from the ground is too high, the child may have a harder landing and risk injury as well. It is important to find a safe and appropriate hat height for the child based on their abilities.

5. Are there any other factors besides hat height from the ground that can affect a child's ability to become airborne?

Yes, there are other factors that can affect a child's ability to become airborne, such as the surface they are jumping from, their physical strength and coordination, and the type of hat they are wearing. It is important to consider all of these factors when determining an appropriate hat height for the child.

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