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Painting a Dome with rollerskates

  1. Nov 20, 2007 #1


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    [SOLVED] Painting a Dome with rollerskates

    1. The problem statement, all variables and given/known data

    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[tex] p_f= p_i thus I= \Delta p= p_f-p_i[/tex]

    (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 [tex]F_{av}= \Delta p/ \Delta t[/tex] 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

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

    3. The attempt at a solution

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

    mass of skater: 80kg
    [tex] r_{dome}= 8m [/tex]
    Last edited: Nov 20, 2007
  2. jcsd
  3. Nov 20, 2007 #2


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

    [tex]a_c= v^2/ r [/tex]

    but would the centripital acceleration be due to gravity??
  4. Nov 20, 2007 #3
    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?
  5. Nov 20, 2007 #4


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    Well It would look like:

    [tex]\sum F = ma_c = m(v^2/r)[/tex]

    however I'm not sure about the angle...
  6. Nov 20, 2007 #5
    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 havent already.
  7. Nov 20, 2007 #6


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    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.
  8. Nov 20, 2007 #7


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    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
    Last edited: Nov 20, 2007
  9. Nov 20, 2007 #8
    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?
  10. Nov 20, 2007 #9


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    Oh..the normal force.. added that.. to pic..
  11. Nov 20, 2007 #10
    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.
  12. Nov 20, 2007 #11


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    hm...[tex]\sum F = ma_c = m(v^2/r)[/tex]

    [tex]Fy= N-mg cos theta= m(v^2/r)[/tex]

    is this alright?

    not sure about the x force though
  13. Nov 20, 2007 #12
    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.
  14. Nov 20, 2007 #13


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    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

    [tex] KE_i + PE_i = KE_f + PE_f [/tex]

    [tex]1/2 mv_i^2 + mgy = 1/2 mv_f + mgy[/tex]

    well at the begining 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

    [tex]Fy= N-mg cos theta= m(v^2/r)[/tex]

    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?
  15. Nov 20, 2007 #14
    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?

    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:

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

    Does that make sense?
  16. Nov 21, 2007 #15


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

    [tex]Fy= N-mg cos theta= m(v^2/r)[/tex]

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

    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
  17. Nov 21, 2007 #16
    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.

    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 vf, and plug that into your force equation, and subsequantly solve for theta.
  18. Nov 21, 2007 #17


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

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

    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 :smile:
  19. Nov 21, 2007 #18
    It's frequent physics problem.It exists in many problem books.
    Last edited: Nov 21, 2007
  20. Nov 24, 2007 #19


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

    here's what I did.

    [tex]mgcos\theta= m(v^2/r) [/tex]

    [tex]v_f= \sqrt{ 2gm(h_o- h) / m} [/tex]

    [tex] v_f= \sqrt{2g(h_o- h)}[/tex]

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

    [tex]mgcos\theta= m(v^2/r) [/tex]

    [tex] rg cos\theta = v^2[/tex]

    plugging in for v

    [tex]rg cos\theta= (\sqrt{ 2g(8m- 8m*cos\theta})^2[/tex]

    [tex]rg cos\theta= 2g(8m- 8m*cos\theta)[/tex]

    [tex](8m*cos\theta)/2 = 8m- 8m*cos\theta[/tex]

    [tex](8m*cos\theta)/2 + 8m*cos\theta = 8m[/tex]

    [tex] (8m*cos\theta)/2 + 2(8m*cos\theta)/2 = 8m[/tex]

    [tex](24m*cos\theta)/2 = 8m [/tex]

    [tex] 24m*cos\theta = 16m [/tex]

    [tex] cos \theta = 2/3 [/tex]

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

    The final height that the person is ..

    [tex]h= 8m cos (48.19^o) = 5.333m [/tex]

    I need the vf for the next parts

    [tex] v_f = \sqrt{2gh}[/tex]

    [tex] v_f = \sqrt{2(9.81m/s^2)(5.333m)}[/tex]

    [tex]v_f= 10.23m/s[/tex]

    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.

    [tex]v_f^2 = v_i^2 + 2a(y_f - y_i)[/tex]

    choosing the down direction as negative..

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

    [tex] v_{yi}= Vcos\theta [/tex] ==> PROBLEM is I don't have theta....

    [tex] v_{yi} = 10m/s[/tex]

    a= -9.81m/s

    Part c) Impulse the ground delivers to the skater.

    [tex]I= \Delta p = p_f-p_i [/tex]

    do know : pf = 0
    pi= - ?

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

    Help please:frown:
    Last edited: Nov 24, 2007
  21. Nov 24, 2007 #20
    When he loses contact with the doom,he is at the height [tex]h=2R+R\cos\theta=\frac{8}{3}R[/tex],

    and his component of velocity in the vertical direction is


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


    Impulse is just [tex]I=mv_f[/tex]
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