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Rotational Dynamics Practice Problems Help

  1. Dec 9, 2007 #1
    My AP Physics C class has a test tomorrow about Rotation Dynamics and my teacher gave us a practive exam over the weekend and i felt really good about it thinking i was going to do really well, but this practice exam is really difficult and im stuck on every problem. there are four problems but ill try to get help with them 1 at a time.

    1. The problem statement, all variables and given/known data
    A large sphere rolls without slipping(pure rolling) across a horizontal surface. The sphere has a constant translational velocity of 10 m/s, a mass of 25kg, and a radius of 0.2 m. The sphere approaches the 25 degree incline of height 3 m as shown in the attached file and rolls up the incline without slipping. (entire problem and questions attached)

    Vo = 10 m/s
    m = 25 kg
    r = 0.2 m
    ϑ = 25 degrees
    h = 3 m
    (also i'm using g = 10 m/s^2)


    2. Relevant equations

    KE = 1/2 mv^2 + 1/2 Iw^2
    I = 2/5 MR^2 (solid sphere)
    TEa = TEb


    3. The attempt at a solution
    ok i did parts a and b but im stuck at c, and i feel really stupid for not being able to figure it out but heres what i have so far.

    a.) to find the total kinetic energy, i used
    KE = 1/2 mv^2 + 1/2 Iw^2 and plugged in the inertia and omega, and solved it down to
    KE = 1750 joules

    b.) to find the velocity, i used
    TEa = TEb for TEa i plugged in 1750 and for TEb i plugged in the kinetich and potential energy.
    1750 = 1/2 mv^2 + 1/2 Iw^2 + mgh plugged in everything and solved it down to
    v = 7.56 m/s

    now part c asks for how far in the x direction it travels after flying off the plane. i know that it will leave the inclined plane in a case 1 trajectory but i have forgotten a lot of the trajectory motion stuff we did at the begining of the year and i have completely forgotten how to find the time. once i have the time i would plug it into Sx=Vo(sin ϑ)t. i have an equation for the time of a case 1 trajectory but the sphere falls another 3 meters after it is level with the point it left from.

    if anyone could help me find the time for this spot i would much appreciate it.


    -EDIT-

    ok i figured out #1 by myself. to find the time you set Sy = -3 and use the equation

    Sy = Vo(Sin ϑ)t - 1/2gt^2 plug in and move it around to get the quadratic
    0 = -5t^2 = 3.18t - 3 use the quadractic equation and use the positive result as the time
    t = 1.16 s

    then plug that into
    Sx = Vo(Cos ϑ)t plug in all variables and solve
    Sx = 7.98 m


    I also solved my problems with #2 so i wont post that. but i have some issues on #3 and 4 as well so ill post them next.
     

    Attached Files:

    Last edited: Dec 10, 2007
  2. jcsd
  3. Dec 9, 2007 #2
    I must say that you have indeed,beautifully described the problem. It is really a good thread to start with.

    The c part is much easier than the rest.
    You know the velocity at the highest point now we know that

    [tex] H = \frac{v^{2}sin^2\theta}{2g}[/tex]

    [tex] T = \frac{2vsin\theta}{g}[/tex]

    solving them we get that

    [tex] T = \sqrt{\frac{2H}{g}}[/tex]

    We want H and that we have it already.
     
  4. Dec 10, 2007 #3
    1. The problem statement, all variables and given/known data

    Hard to explain. See the attachment for the problem statement.
    [​IMG]

    each sphere has
    - mass = m
    - small r
    the rod has
    - length = l
    - negligable mass.
    the clay lump has
    - mass = m
    - initial velocity = Vo

    all answers should be in terms of m, Vo, l, and fundamental constants.


    2. Relevant equations

    KE = 1/2 mv^2
    W = change in KE (KEf - KEo)
    Xcm = (m1x1 + m2x2)/(m1+m2)
    Pi = Pf
    Li = Lf


    3. The attempt at a solution

    for part a, you need to find the velocity after the collision using
    m1v1 + m2v2 = (m1 + m2)v'
    mVo + 0 = (m + 2m) v'
    v' = Vo / 3

    then for ai.) use
    KE = 1/2 mv^2 plug in Vo / 3 fo V and solve down to
    KE = (mVo^2) / 6

    for aii.) use
    W = change in KE
    W = KEf - KEo
    W = 1/6 mv^2 - 1/2 mv^2
    W = 1/6 mv^2 - 3/6 mv^2
    W = -(mVo^2)/3


    for part b.) you know that once the clay lump hits the other object, the new object is going to rotate about the center of mass, and the center of mass is going to continue to move in a linear path. so first you need to find the center of mass.

    bi.)
    Xcm = (m1x1 + m2x2)/(m1 + m2) using the left sphere as the 0 point,
    Xcm = 0 + 2ml / 3m
    Xcm = 2/3 L

    bii.)
    as i said before, the CoM continues to move in a linear path so the direction would be straight upward on the paper.

    biii.)
    we know that both angular and linear momentum are conserved so the linear speed will be the same as if it was hit in the center. so
    V' = Vo/3

    heres where im stuck though. im not sure how to go about finding the angular momentum and thus, the angular speed. of the object. and im also not sure what to do for bv. either. would it be the same as aii.) since the V' is the same?
     

    Attached Files:

    Last edited: Dec 10, 2007
  5. Dec 10, 2007 #4
    I am not able to see anything.

    So please write down the problem.
     
  6. Dec 10, 2007 #5
    i hosted the image on an outside hosting site and edited it into the Problem statement section.
     
  7. Dec 10, 2007 #6
    i have a feeling that to find the angular velocity im going to need to use

    v = ωR

    just im not sure how i would go about implementing this formula. i have the linear velocity of the center of mass, but it isnt at the center. it is 2/3 the length from the left side of the system. and if i use the equation at the center of mass, the R is 0 so the whole thing cancels out to 0 so im not sure what to do here.
     
  8. Dec 10, 2007 #7
    Now we know that

    [tex]Impulse(J) = mv[/tex]

    Also

    [tex] Jx = I\omega[/tex]

    Where x is the distance of impulse from the point of rotation.Solving them we can get relation between v and omega.

    v is not equal to omega(r) as it is not pure rolling.
     
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