Painting a Dome with rollerskates

[~christina~]

[SOLVED] Painting a Dome with rollerskates

Homework Statement

Two prolific painters concoct a hair brain scheme for painting a spherical shaped

observatory dome. One painter has invented a pair of roller skates that will eject paint

only when the wheels of the skates are in contact with a surface. The daredevil of the

two, wearing the roller skates, starts with negligible velocity at the top of the dome and

coasts down over the dome's surface as shown in the figure below. As long as the

wheels are in contact, that surface of the dome will receive a coat of paint. The more

sensible painter is positioned with a net some distance from the base of the observatory

to break the daredevil's fall. The mass of the painter, including the skates, is 80 kg and

the radius of the dome is 8 m. A test run is made to demonstrate how much of the dome

can be painted in one pass. Assume that the painter's mass is constant over this one

run. Neglect friction and air resistance.


(a) Describe in angles (relative to the vertical) the portion of the dome that the daredevil would paint and the portion that would remain unpainted.

ABSOLUTELY NO IDEA HOW I WOULD GET THIS (doesn't help since I need to find this to do the rest of the problem

(b) If the net is removed, what velocity would the painter have when he impacts the ground?

I think I would need to find the painter's velocity as he leaves the dome and then just use
the uniform acceleration equations (projectile motion)
but I would I assume need to find the painter's angle relative to the horizontal when he leaves the dome for that. Then I would find distance to the floor, time it takes to get to zero and then velocity at that time

(c) What impulse would the ground deliver to the painter in order to stop him?

Hm..would it be$$p_f= p_i thus I= \Delta p= p_f-p_i$$

(d) A person can just survive a full body collision if the average force is less than 72,000 Newtons. If it takes the ground 0.05 seconds to stop the painter, is it likely he will survive the fall? Explanations must include physics principle(s).

since $$F_{av}= \Delta p/ \Delta t$$ I would use the previous found Impulse and then find out if the force is greater or less

(e). Where should the assistant place the net from the dome to break the daredevil's fall?

I would find the distance in the x direction based on the time it takes to get to zero thus the floor
http://img440.imageshack.us/img440/3397/weeefq5.th.png

Homework Equations

stated above except I don't know about the first part..

The Attempt at a Solution

I'm not sure how to approach the first part A and find that angle or how they want me to state . Please help me

mass of skater: 80kg
$$r_{dome}= 8m$$

 

[~christina~]

Now thinking about it I would say that it's circular motion

$$a_c= v^2/ r$$

but would the centripital acceleration be due to gravity??

 

[hotcommodity]

The skater gets displaced some angle theta with respect to the vertical. We want to find out when his skates stop touching the dome. What would Newton's Second Law look like if we summed the forces acting on the skater at some angle theta?

 

[~christina~]

Well It would look like:

$$\sum F = ma_c = m(v^2/r)$$

however I'm not sure about the angle...

 

[hotcommodity]

Ok, this is the net force that acts on the skater, but what forces make up this net force?

Edit: It may help to draw a free-body diagram of the skater displaced over some angle, if you haven't already.

 

[rl.bhat]

As the daredevil starts from the top of the dome, i) his velocity increases ii) PE decreases, iii) component of his weight towards the center of the dome decreases and iv) centrifugalce force on him increases. He will be on the dome as long as iii) is > iV) If theta is the angle between the vertical and the line joining his position on the dome where he is detached from the dome, then Mgcos(theta) = Mv^2/R. If h is the vertical depth of his position, V^2 = 2gh, and cos(theta) = (R - h)/R. If you substitute these values we get h = R/3 or cos(theta) = 2/3. After he detaches from the dome his horizontal component of the velocity remains the same but vertical compomnent increases. I think this much information is suffucient to solve the problem.

 

[~christina~]

http://img111.imageshack.us/img111/2525/39764518ey4.png

Hm..I say the free body is wrong that I drew...

I seriously don't know how would I split up the forces since I'm looking at my book but It doesn't show that...just net force

I forgot to include the centripital accleration pointing into the circle from the skater

 

[hotcommodity]

You've got the weight force but you're missing one other force. Why doesn't the skater immediately accelerate through the dome, what other force is acting on him?

 

[~christina~]

Oh..the normal force.. added that.. to pic..

 

[hotcommodity]

Ok, so now you can re-write the net force equation using the normal and weight forces, and using a centripetal coordinate system (choose if you'd like outward or inward to be the positive direction). Once you have this equation, think what has to be zero when the skater flys off of the dome. The only thing that you'll have to solve for is the skaters speed at this point, and apply it to your force equation. We were told that friction and air resistance is negligible, so we can assume that mechanical energy is conserved. Use this equation to solve for the speed.

 

[~christina~]

hm...$$\sum F = ma_c = m(v^2/r)$$

$$Fy= N-mg cos theta= m(v^2/r)$$

is this alright?

not sure about the x force though

 

[hotcommodity]

This is the correct equation, but we're not necessarily dealing in components of x and y. The equation above would refer to the net centripetal force on the skater, because we have a centripetal acceleration in the net force equation. Does that make sense?

Now that you have the equation you need, think about which force goes to zero when the skater flys off the building. After this, you only need to solve for the speed of the skater at this position using the conservation of mechanical energy. Let me know if you get hung up.

 

[~christina~]

This is the correct equation, but we're not necessarily dealing in components of x and y. The equation above would refer to the net centripetal force on the skater, because we have a centripetal acceleration in the net force equation. Does that make sense?

Yes...but doesn't the velocity have x and y components? And I would have to relate it to the vertical not the horizontal (angle) so I think it would matter...

[/QUOTE]Now that you have the equation you need, think about which force goes to zero when the skater flys off the building. After this, you only need to solve for the speed of the skater at this position using the conservation of mechanical energy. Let me know if you get hung up.[/QUOTE]

The force that goes the zero is the centripital accelleration.

The conservation of mechanical E

$$KE_i + PE_i = KE_f + PE_f$$

$$1/2 mv_i^2 + mgy = 1/2 mv_f + mgy$$

well at the beginning the person's velocity is negligible according to the problem...

since I have to find the final height...not sure bout this

m= 80kg
initial y= 8m + 2(8m)= 24m
final y=


1/2 (80kg)(0)^2 + 80kg (9.8m/s^2)(24m)= 1/2mvf + mgy ==> stuck

$$Fy= N-mg cos theta= m(v^2/r)$$

you said that I need only to find the velocity but don't I need the angle that the person makes relative to the vertical?? (problem asks for the angle thus telling me how much of dome has paint)

How would I incorperate the F equation into the conservation of momentum equation?

 

[hotcommodity]

Yes...but doesn't the velocity have x and y components? And I would have to relate it to the vertical not the horizontal (angle) so I think it would matter...

The force that goes the zero is the centripital accelleration.

Velocity does have x and y components but the wonderful thing about centripetal acceleration (and kinetic energy for that matter) is that it deals with speed, so we don't have to worry about direction.

This is where terminology can get tricky. There's really no such thing as a centripetal force, it's simply a concept that's used to describe an objects circular motion. So between the weight force and the normal force, which force will become zero when the skater flys off of the dome?

The conservation of mechanical E

$$KE_i + PE_i = KE_f + PE_f$$

$$1/2 mv_i^2 + mgy = 1/2 mv_f + mgy$$

well at the beginning the person's velocity is negligible according to the problem...

since I have to find the final height...not sure bout this

m= 80kg
initial y= 8m + 2(8m)= 24m
final y=


1/2 (80kg)(0)^2 + 80kg (9.8m/s^2)(24m)= 1/2mvf + mgy ==> stuck

$$Fy= N-mg cos theta= m(v^2/r)$$

I'm not exactly sure where you're getting 24m. It may help you to think of the center of the circle as the point of zero potential energy. Your final speed will have this form:

$$v_f = \sqrt{ \frac{2[mgh_0 -mgh_f]}{m}}$$

Does that make sense?

 

[~christina~]

Velocity does have x and y components but the wonderful thing about centripetal acceleration (and kinetic energy for that matter) is that it deals with speed, so we don't have to worry about direction.

This is where terminology can get tricky. There's really no such thing as a centripetal force, it's simply a concept that's used to describe an objects circular motion. So between the weight force and the normal force, which force will become zero when the skater flys off of the dome?

technically the normal force would disappear since there wouldn't be any surface to push up on the skater.


$$Fy= N-mg cos theta= m(v^2/r)$$

Fy (when the skater falls off) = -mg cos theta = m(v^2/r)

I'm not exactly sure where you're getting 24m. It may help you to think of the center of the circle as the point of zero potential energy. Your final speed will have this form:

$$v_f = \sqrt{ \frac{2[mgh_0 -mgh_f]}{m}}$$

Does that make sense?

Fy (when the skater falls off) = -mg cos theta = m(v^2/r)

How would this fit? where would I put this eqzn?


It does make sense what you did but how would I find the angle in relation to the original point ?
and I don't have the final height...or angle

 

[hotcommodity]

technically the normal force would disappear since there wouldn't be any surface to push up on the skater.


$$Fy= N-mg cos theta= m(v^2/r)$$

Fy (when the skater falls off) = -mg cos theta = m(v^2/r)

Correct, the normal force will become zero. Notice that since you took the positive direction to be centripetally outward, both the weight and centripetal forces will have a negative sign. So now each side of your equation will be positive. Sorry I didn't catch the sign error earlier.

Fy (when the skater falls off) = -mg cos theta = m(v^2/r)

How would this fit? where would I put this eqzn?


It does make sense what you did but how would I find the angle in relation to the original point ?
and I don't have the final height...or angle

Let's say we take the center of the circle to have zero potential energy, this is our reference point for the height. The initial height of the skater will simply be the radius of the dome, and the final height will be the radius times the cosine of theta, right? You can draw a right triangle for the angle theta to see this. Now you can solve for v_f, and plug that into your force equation, and subsequantly solve for theta.

 

[~christina~]

Correct, the normal force will become zero. Notice that since you took the positive direction to be centripetally outward, both the weight and centripetal forces will have a negative sign. So now each side of your equation will be positive. Sorry I didn't catch the sign error earlier.

Fy (when the skater falls off) = -mg cos theta = -m(v^2/r)

so...mgcos theta = m(v^2/r)

-Let's say we take the center of the circle to have zero potential energy, this is our reference point for the height. The initial height of the skater will simply be the radius of the dome, and the final height will be the radius times the cosine of theta, right? You can draw a right triangle for the angle theta to see this. Now you can solve for v_f, and plug that into your force equation, and subsequantly solve for theta.

I've got to sleep but I'll get back to this tommorow after school and if your around I'd appreciate it if you just checked and see what put up.

Thanks for your help

 

[azatkgz]

It's frequent physics problem.It exists in many problem books.

 

[~christina~]

Let's say we take the center of the circle to have zero potential energy, this is our reference point for the height. The initial height of the skater will simply be the radius of the dome, and the final height will be the radius times the cosine of theta, right? You can draw a right triangle for the angle theta to see this. Now you can solve for $$v_f$$, and plug that into your force equation, and subsequantly solve for theta.

Okay..finally I get back to this problem..

here's what I did.

$$mgcos\theta= m(v^2/r)$$

$$v_f= \sqrt{ 2gm(h_o- h) / m}$$

$$v_f= \sqrt{2g(h_o- h)}$$

I then took this and plugged it into the force eqzn.

$$mgcos\theta= m(v^2/r)$$

$$rg cos\theta = v^2$$

plugging in for v

$$rg cos\theta= (\sqrt{ 2g(8m- 8m*cos\theta})^2$$

$$rg cos\theta= 2g(8m- 8m*cos\theta)$$

$$(8m*cos\theta)/2 = 8m- 8m*cos\theta$$

$$(8m*cos\theta)/2 + 8m*cos\theta = 8m$$

$$(8m*cos\theta)/2 + 2(8m*cos\theta)/2 = 8m$$

$$(24m*cos\theta)/2 = 8m$$

$$24m*cos\theta = 16m$$

$$cos \theta = 2/3$$

$$\theta = 48.19^o$$ ==> I think that is the angle with respect to the horizontal not vertical thus...


The final height that the person is ..

$$h= 8m cos (48.19^o) = 5.333m$$


I need the vf for the next parts

$$v_f = \sqrt{2gh}$$

$$v_f = \sqrt{2(9.81m/s^2)(5.333m)}$$

$$v_f= 10.23m/s$$

for part b) what velocity would the painter have when he hit the ground if the net was removed?

I know I need the angle below the horizontal when the painter flies off the dome and I have the [texv_f[/tex] but I don't have the angle so how can I find the angle or is it the one I found before?

I wanted to plug into this eqzn though.

$$v_f^2 = v_i^2 + 2a(y_f - y_i)$$

choosing the down direction as negative..

y= 5.333m + 2R= 2.667m + 16m = -18.667m

$$v_{yi}= Vcos\theta$$ ==> PROBLEM is I don't have theta...

$$v_{yi} = 10m/s$$

a= -9.81m/s


Part c) Impulse the ground delivers to the skater.

$$I= \Delta p = p_f-p_i$$

do know : pf = 0
pi= - ?


and also the x and y components of the velocity for finding the impulse?



Help please

 

[azatkgz]

When he loses contact with the doom,he is at the height $$h=2R+R\cos\theta=\frac{8}{3}R$$,

and his component of velocity in the vertical direction is

$$v_y=\sqrt{2gR(1-\cos\theta)}\sin\theta=\frac{\sqrt{5}}{3}\sqrt{\frac{2}{3}gR}$$

now we can find the vertical component of the velocity,when he hits the ground from this equation

$$h=\frac{v_f^2-v_o^2}{2g}$$

Impulse is just $$I=mv_f$$

 

[hotcommodity]

Okay..finally I get back to this problem..

here's what I did.

$$mgcos\theta= m(v^2/r)$$

$$h_{final height}= r- rcos\theta$$

Let's be careful about what you're saying here. Your equation for the final height is actually the equation for the change in height. The initial height is r, and the final height is $$r cos \theta$$.

$$cos \theta = 2/3$$

$$\theta = .841$$ ==> I think that is the angle with respect to the horizontal not vertical thus...

You have the correct equation, but this will be the angle with respect to the vertical. The reason you're getting a funny answer is because you're mixing radians with degrees. If someone else doesn't help you out I'll come back and look at the rest of your work, but I have to run out, good luck :)

 

[~christina~]

if you check it out I'd appreciate it.

I fixed what you said was incorrect and I'm working on the next parts.

The only thing really confusing me is vy and vx components of velocity and would I need both to find impulse.

Thanks

 

[hotcommodity]

I'm taking a look at it. Can you check the question one more time and see if its asking for the skaters "speed" or "velocity" just before he hits the ground. If they're asking for speed it will simplify things greatly.

 

[~christina~]

it says velocity...just to complicate it ...

I was looking at the other person's post of how to get the vy (same pg)

but I don't know how he/she got the equation which wouldn't help me on a test unfortunately.

 

[hotcommodity]

He/she got it by using the equations for projectile motion; $$v_{0x} = v_0 cos \theta , v_{0y} = v_0 sin \theta$$. I'm not sure how that helps either, but I'll post if I come up with something...

 

[~christina~]

ok then.

technically I would need the vy and vx though for the momentum then to find the impulse I guess since the skater has a x and y component.

 

[BlackWyvern]

N = mgcos0
(normal force is given by mgcos0, derive it yourself, or just remember this)

mgcos0 = mv^2/r
(the centripetal acceleration is provided by the normal force)

mgr = mg(r - h) + (0.5)mv^2
2gh = v^2
(the energy at the top, equals the energy at the point where the object has fallen distance h, and has a speed)

cos0 = ((r-h) / r)
(using parallel lines to get this ratio, look closely at your diagram)

mgcos0 = mv^2/r
(let's sub stuff into this)

mgcos0 = m(2gh)/r
(mgr-mgh) / r = 2mgh/r
(turning it all into h's and r's and mg's through subbing in stuff)

r/3= h
solved.

:)

This is the case for all circles. The object will fall off a distance 1 third of the radius down. Use this to help you finish the problem.

 

[rl.bhat]

When the skater leaves the dome loss of PE = gain in KE. Hence mgh = 0.5mvf^2* and vf = sqrt(2g*R/3). When the skater leaves the dome his velocity is tangential to the dome. Its vertical component is vfsin(theta) = sqrt.5vf/3. And its horizontal component is vfcos(theta) = 2vf/3 = vx, and it remains constant till skater hits the groung, because acceleration does not acts in that direction. Now vy^2 = [vfsin(theta)]^2 +2g*total height =[vfsin(theta)]^2 + (16 + 5.333)*2g. Now you can complete the problem.

 

[~christina~]

When the skater leaves the dome loss of PE = gain in KE. Hence mgh = 0.5mvf^2* and vf = sqrt(2g*R/3). When the skater leaves the dome his velocity is tangential to the dome. Its vertical component is vfsin(theta) = sqrt.5vf/3. And its horizontal component is vfcos(theta) = 2vf/3 = vx, and it remains constant till skater hits the groung, because acceleration does not acts in that direction. Now vy^2 = [vfsin(theta)]^2 +2g*total height =[vfsin(theta)]^2 + (16 + 5.333)*2g. Now you can complete the problem.

but for theta is it the angle I found with respect to the vertical?

$$(48.19^o)$$

 

[rl.bhat]

Draw the the figure. You can see that same angle appeares for vertical and horizontal component of vf.

 

[~christina~]

http://img119.imageshack.us/img119/1756/72948630sr0.png

Your saying that these 2 angles are the same? (yellow line)

I was referring to the initial velocity after leaving dome

 

[rl.bhat]

Yes. They are the same. Tangent to the dome is the initial velocity v. Later on it acts like a projectile and falls freely under gravity.
The final vertical height that the person leaves the dome is ..h = R/3

 

[~christina~]

Ok then. That was my only confusion I think.

Thank You