1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: A 0.300 kg hollow sphere is rolling along a horizontal floor

  1. Dec 9, 2009 #1
    1. The problem statement, all variables and given/known data

    A 0.300 kg hollow sphere is rolling along a horizontal floor at 5.0 m/s when it comes to a 30 degree incline. How far up along the incline does it roll before reversing direction?

    2. Relevant equations

    Kinematics equations?


    3. The attempt at a solution

    I didn't even know where to begin...
     
  2. jcsd
  3. Dec 9, 2009 #2
    Since you're not given the radius of the sphere or any further angular information on it i'd simply use the conservation of energy;

    all of its' kinetic energy will be converted into gravitational potential, should be easy from there.
     
  4. Dec 9, 2009 #3
    Any way you could walk me through it, I'm really struggling in this course and my understanding of things is relatively low...
     
  5. Dec 9, 2009 #4
    Of course, sorry;

    you know that you need to find the distance at which it stops momentarily (reaches its maximum height) just before it rolls down again.

    The kinetic energy it has before encountering the slope is [tex]\frac{1}{2}mv^{2}[/tex], right? Now you know by the time it stops it has lost all of this kinetic energy-> v = 0, so all of the energy (for the sake of this problem you must assume there is no friction) must have been converted into gravitational potential enegy, you rightly quoted mgh;

    however be wary, you need to know the distance it travels up the plane, but remember all you're interested in is the vertical distance (for h), you know that the angle of the plane is 30 degrees, so given;

    [tex]\frac{1}{2}mv^{2} = mgh [/tex] , can you think of a way of having h, the vertical height travelled, in terms of x, the distance along the plane travelled? (hint: sin)
     
  6. Dec 9, 2009 #5
    Ok so KE before the hill is 1/2(0.3)(5)^2 = 3.75 J.

    From there I can set that to m*g*sin(30)?...How am I finding the height?
     
  7. Dec 9, 2009 #6
    the sin(30) is correct but think about it, you need the vertical component, if you think of it like a triangle;

    x is the hypotenuse, the opposite side is the height (h);

    [tex] sin(30) = \frac{Opposite}{Hypotenuse}[/tex]

    do you see why?
     
  8. Dec 9, 2009 #7
    Ok I get that.

    So set 3.75 = m*g*h. Solve for h. Once I have that I can find x using sin(30) = h/the hypotenuse. But how do I find that?
     
  9. Dec 9, 2009 #8
    Yep you're right

    sin(30) = h / hypotenuse, the hypotenuse, remember is the distance travelled, from there i'd use your calculator, what you're doing is like;

    10 = 5 / x, solve for x, but instead obviously you have sin(30) and the h which you found.
     
  10. Dec 9, 2009 #9
    Ok so,

    3.75 J = (0.3)(9.8)h, h = 1.28 m

    sin(30) = 1.28/ hypotenuse ---> hypot = 1.28sin(30) = 0.64 m?

    Did I do that properly?
     
  11. Dec 9, 2009 #10
    that's all correct apart from the last bit...

    sin(30) = 1.28 / hyp so if you times both sides by the hyp;

    hyp sin(30) = 1.28, and now divide by sin(30)

    hyp = 1.28/sin(30)

    do you see that?
     
  12. Dec 9, 2009 #11
    So once I have the hypotenuse that's my answer?

    I see what you did, and I apologize for all the stupid mistakes I'm making. Things are so much simpler when someone holds your hand the entire way...

    1.28/sin(30) = 2.56 m
     
  13. Dec 9, 2009 #12
    Correct
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook