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

[gap0063]

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?

 

[tiny-tim]

hi gap0063!

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? ​

 

[gap0063]

hi gap0063!


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? ​

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?

 

[tiny-tim]

(have a delta: ∆ and a theta: θ and try using the X² and X₂ icons just above the Reply box )

∆E_mech is zero,

so it's just ∆KE + ∆PE = constant.

U uses the difference in height.

V_f 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.

 

[gap0063]

(have a delta: ∆ and a theta: θ and try using the X² and X₂ icons just above the Reply box )

∆E_mech is zero,

so it's just ∆KE + ∆PE = constant.

U uses the difference in height.

V_f 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.

Thanks for the symbol help!

So ∆KE + ∆PE= (K_f+U_f)-(K_i+U_i) <- is that what you mean by constant?

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

V_f 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/

 

[tiny-tim]

So ∆KE + ∆PE= (K_f+U_f)-(K_i+U_i) <- 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.

V_f 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.

 

[gap0063]

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 (K_f+U_f)-(K_i+U_i)=0 =>K_f+U_f)=(K[/SUB]i+U_i) ?

then if so, 1/2 mv_f²+0=0+mg2r

At what point on the slide do I draw the fbd, so I can find out all the forces?

 

[tiny-tim]

So (K_f+U_f)-(K_i+U_i)=0 =>K_f+U_f)=(K[/SUB]i+U_i) ?

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?​

 

[gap0063]

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.

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 $$\sum$$ Fy= N-mg+Fsin$$\theta$$
$$\sum$$ Fx= Fcos$$\theta$$
maybe?

 

[tiny-tim]

nooo

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

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? ​

 

[gap0063]

nooo

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

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? ​

okay, so Ncos$$\theta$$=mg => N= mg/cos$$\theta$$

and the normal radial force is Nsin$$\theta$$ right?

so N_r= mg/cos$$\theta$$*sin$$\theta$$= tan$$\theta$$mg

which can be tan$$\theta$$= v²/gr


but now I'm stuck because there are two unknowns...

 

[tiny-tim]

Sorry, I don't understand any of this.

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

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

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

 

[gap0063]

Sorry, I don't understand any of this.

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

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

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

So the radial direction is perpendicular to the normal force?

F_total = Ncos$$\theta$$-mg ?

 

[tiny-tim]

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).

F_total = Ncos$$\theta$$-mg ?

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

You need the radial component.

 

[gap0063]

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

F_c=ma_c= m v²/r