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Problem about an sphere.

  1. Feb 29, 2012 #1
    1. The problem statement, all variables and given/known data

    An 7.60-cm-diameter, 400 sphere is released from rest at the top of a 1.90-m-long, 16.0 incline. It rolls, without slipping, to the bottom. What is the sphere's angular velocity at the bottom of the incline? What fraction of its kinetic energy is rotational?

    2. Relevant equations

    Can't really come up with any.

    3. The attempt at a solution

    I don't even know who to approach the problem. Any help? or at least any hints??
     
  2. jcsd
  3. Feb 29, 2012 #2
    Try using conservation of energy.
     
  4. Feb 29, 2012 #3
    OK this is what i did for the first part.

    Kf=1/2mv^2+1/2Iw^2

    The moment of inertia for a solid sphere is 2/5mR^2

    SUBSTITUTING values: Kf= 1/2mv^2+1/2(2/5mR^2)(v/r)2=1/2mv^2+1/5mv^2

    Using conservation of energy we get

    Ki+Ui=Kf+Uf

    0+mgy=1/2mv^2+1/5mv^2+0

    Solving for v

    v=sqrt mgy/1/2m+1/5m and that gave me 2.70 m/s

    but they are asking radians per second and in the second part of the question they are asking me for a fraction. Did I do anything wrong here?
     
  5. Feb 29, 2012 #4
    What's the relationship between angular velocity and tangential velocity?
     
  6. Feb 29, 2012 #5
    angular is the rate at which the wheel rotates and tangential is the velocity along the curve. So are you saying that i got tangential velocity instead of angular velocity and if so how do i find the angular velocity without a time?
     
  7. Feb 29, 2012 #6
    well never mind i took this 2.70m/s and divided that by the radius of the sphere. But my question is, did the process to solve the problem is the correct one and if so did you get the same tangential speed?
     
  8. Feb 29, 2012 #7
    in the equations you used, v stands for tangential velocity (which is the same thing as regular velocity, we just add "tangential" to make it clear from angular velocity). This velocity is just the velocity at which the entire object is moving through space.

    Angular velocity is represented with ω, it is the velocity at which the object's mass is rotating around its rotational axis. It has units of angles per second, angles are measured in degrees or radians.

    So when you solved for v, you solved for the velocity at which the sphere was moving through space at the bottom of the hill. When you get the angular velocity by dividing the tangential velocity by the radius of the sphere you get the velocity at which the sphere's mass was rotating around its rotational axis. In this case, its center, since it's a sphere.

    Kinetic energy can be due to rotation or due to traveling through space. The tangential kinetic energy is related to tangential velocity, can you guess what angular kinetic energy is related to? These two forms of energy simply add together if an object is both moving through space (has tangential velocity) and is rotating (has angular velocity).


    and yeah it looks right to me
     
  9. Feb 29, 2012 #8
    :( i just divided 2.70 m/s / 0.076 m according to w=v/r and got 35.55 rad/s and it wasn't the right answer... :****O what do i do??? Help??
     
  10. Feb 29, 2012 #9
    If I read your problem right, .076m is the diameter of the sphere ;)
     
  11. Feb 29, 2012 #10
    In case you want to know here is waht i did in my calculations v=sqrt (.400kg)(9.8m/s^2)sin(16)(1.9m)/1/2(.400kg)+1/5(.400kg)=2.70m/s is that right?
     
  12. Feb 29, 2012 #11
    YESSS!!!! SUCCESS!!!! SUCCESS!!!!!! THANKS MAN!! for your great wisdom my master!

    now how do i do the second part?
     
  13. Feb 29, 2012 #12
    Well there are two parts to its kinetic energy, right? Tangential kinetic energy and rotational kinetic energy, how do you determine either kind of those types of kinetic energy?
     
  14. Feb 29, 2012 #13
    let me see... tangential is 1/2mv^2 and rotational is.. 1/2Iw^2 but we have a sphere here so.... we take the moment of inertia and substitute into the KErot and we get 1/5V^2.
    so we have 1/2mv^2+1/5mv^2 are you saying is 1/5??
     
  15. Feb 29, 2012 #14
    well, a fraction is a part of something divided by the whole. Like, if I have two marbles that are red out of five marbles, then the fraction of marbles that are red is 2/5.

    So, what's the total amount of kinetic energy? What fraction of that is rotational kinetic energy?
     
  16. Feb 29, 2012 #15
    is not 3/5 though i now is 2/7 but i really don't know how to get that fraction, i'm trying but is not getting there.
     
  17. Feb 29, 2012 #16
    here, first calculate each of these things: rotational kinetic energy, tangential kinetic energy and the total kinetic energy.
     
  18. Feb 29, 2012 #17
    I=2/5mr^2=2.3x10^-4 Ke rot=1/2(2.3x10^-4)(71.05)^2=0.585 J than Ke trans=1/2(.400)(2.7)^2=1.458 0.585/1.458= 290/729 I still dont see the fraction did i do it right?
     
  19. Feb 29, 2012 #18
    a fraction is a part of a whole, right? So which part of what whole are we looking for?
     
  20. Feb 29, 2012 #19
    LOL good question. Total KE?? But what is the total kineteci eenrgy here is it KE rot+ KE trans?? so is 1/2mv^2/1/2mv^2+1/5mv^2?
     
  21. Feb 29, 2012 #20
    Ke rot /KEtotal??? Is it like this 1/2mv^2/1/2mv^2+1/5mv^2?
     
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