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Angular Momentum/Rolling Multiple Choice

  1. Apr 3, 2013 #1
    1. The problem statement, all variables and given/known data

    1. A cylinder of radius R = 6.0 cm is on a rough horizontal surface. The coefficient of kinetic
    friction between the cylinder and the surface is 0.30 and the rotational inertia for rotation
    about the axis is given by MR2/2, where M is its mass. Initially it is not rotating but its
    center of mass has a speed of 7.0m/s. After 2.0 s the speed of its center of mass and its angular
    velocity about its center of mass, respectively, are:
    A. 1.1m/s, 0
    B. 1.1m/s, 19 rad/s
    C. 1.1m/s, 98 rad/s
    D. 1.1m/s, 200 rad/s
    E. 5.9m/s, 98 rad/s

    Answer is D.

    2. As a 2.0-kg block travels around a 0.50-m radius circle it has an angular speed of 12 rad/s.
    The circle is parallel to the xy plane and is centered on the z axis, 0.75m from the origin. The
    component in the xy plane of the angular momentum around the origin has a magnitude of:
    A. 0
    B. 6.0kg · m2/s
    C. 9.0kg · m2/s
    D. 11 kg · m2/s
    E. 14 kg · m2/s

    Answer is C.

    3. A uniform disk, a thin hoop, and a uniform sphere, all with the same mass and same outer
    radius, are each free to rotate about a fixed axis through its center. Assume the hoop is
    connected to the rotation axis by light spokes. With the objects starting from rest, identical
    forces are simultaneously applied to the rims. Rank the objects according to their
    angular momenta after a given time t, least to greatest.
    A. all tie
    B. disk, hoop, sphere
    C. hoop, disk, sphere
    D. hoop, sphere, disk
    E. hoop, disk, sphere

    Answer is A.

    2. Relevant equations
    1. Friction=Iα/R, thus α=2μg/r
    1/2m(vi)^2 = 3/4m(vf)^2
    ωf=ωi+αt

    2. L=Iω
    I=MR^2

    3. L=Iω
    I=MR^2


    3. The attempt at a solution

    1. I'm getting the angular velocity fine, but I just cannot for the life of me get the linear velocity correct. Using α=2μg/r, I get the angular acceleration as 98 rad/s^2. So the angular velocity at t=2 should be 196 rad/s, or about 200 rad/s. Since a=αr, the linear acceleration should be 98*0.06, which equals -5.88 m/s^2, since the angular acceleration causes the body to slow down. When plugging that into ωf=ωi+αt, I get a negative number for the final angular velocity, which would yield a negative linear velocity, which isn't right. When I try to do it in with the energy conservation formula I also get the wrong answer. What am I doing wrong in solving for the linear velocity?

    2. Shouldn't this be zero? If you evaluate r x V using the right hand rule, the vector points up, so the angular momentum is solely in the z-plane, right?

    3. Not quite sure how to approach this one. If I'd just guessed, I probably would have gone with A, simply because I would think that though the moments of inertia vary, a small moment of inertia would yield a larger angular velocity, so all three would end up with the same angular momentum, based off L=Iω. Is this reasoning correct?

    Thanks in advance!
     
  2. jcsd
  3. Apr 3, 2013 #2

    TSny

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    Gold Member

    Hello, and welcome to PF!

    Here are some things to consider:

    Does a=αr hold for rolling with slipping?

    Did you draw a diagram showing the vector r and the velocity vector V?

    You haven't shown anything quantitative here. There is a famous equation that relates the concepts of torque and angular momentum.
     
  4. Apr 3, 2013 #3
    Ah, you're right, but we can use energy conservation, right? So .5m(vi)^2 = .5m(vf)^2 + .5(.5MR^2)ω^2. However, evaluating that doesn't give me 1.1 m/s for vf.

    I made a quick drawing on paint:

    28whunk.jpg

    Is this reasoning right?

    Oh dear, I completely forgot about T=dl/dt. Makes complete sense now - if torque is the same for the objects, the angular momentum will be also.
     
  5. Apr 3, 2013 #4

    TSny

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    Slipping will generate heat energy. That makes energy conservation messy.

    There's a fundamental theorem of mechanics that states that the net force acting on a body equals the mass of the body times the acceleration of the CM of the body.

    You want to find the angular momentum about the origin (of the coordinate system). So, where should vector r start from?
     
    Last edited: Apr 3, 2013
  6. Apr 3, 2013 #5
    Completely slipped my mind that it's the frictional force doing the deceleration. I get the right answer that way. Thanks!

    Oops, the vector r points from the origin to (0.5,0.75,0). Then to find the component of the angular momentum in the xy plane, I'll need to use the vertical component of vector r. That yields 9 using L=mvr.

    All understood. Thanks for the help!
     
  7. Dec 9, 2013 #6
    How exactly was α = 2μg/r derived? I keep doing

    mgμ = ma
    αr = gμ
    α = gμ/r and then ωf = αt = tgμ/r....i understand that t = 2s but where does the original two come from when angular acceleration is solved for? Using Newton's laws I was able to derive the linear velocity, but came up with the wrong answer for the angular velocity. Also, when trying a different method, I did:

    fr = Iα
    mgμr = (1/2)mr2α
    2gμ/r = α

    Now, I understand that when I do it through this method, I get the correct answer. I just don't seem to understand why it does not work using both methods...any clarification would be great!
     
  8. Dec 9, 2013 #7

    TSny

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    Hello.

    Looks like you've assumed that ##a = \alpha r##. But that is not true for rolling with slipping.
     
  9. Dec 9, 2013 #8
    Hmmm...okay, thanks! As for the second question, I tried to set up the system and did this:

    Δθ = ωΔt
    v = ωr = 6 m/s
    r = (0.75 + 0.5)1/2 m (distance from the origin to the centre)

    L = mrv = 2kg (0.9 m)(6 m/s) = 10.8 kg m2/s

    Is this the correct way since I found linear speed, and the actual distance can be found using r2 = x2 + y2

    Now, I don't seem to calculate the right component in the z-direction though for the angular momentum...(this is a separate question)

    I did:

    [ (0.5, 0.75, 0)m x (0, 6, 0) m/s ](2 kg) and I get 9 kgm^2/s....the answer is 6 kgm^2/s
     
    Last edited: Dec 9, 2013
  10. Dec 9, 2013 #9

    TSny

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    Yes, I think this is correct for the magnitude of the total angular momentum vector.

    You have 0.75 as the y-component of the position vector. I don't think that's correct.
     
  11. Apr 1, 2014 #10
    #1 Cylinder with r=.06m question

    Answer D is what you get when you use t=2 secs. in the equations calculated for translational velocity (Vcom = 7 - 2.94t) and for angular velocity (ω = 98t). However the cylinder should not be slipping for that long of a time. The cylinder stops slipping when the velocity of the center of mass equals the tangential velocity of the cylinder (Vcom = Vt). Vt = rω = .06(98)t = 5.88t. Set Vcom = Vt and solve for time. (7 - 2.94t) = 5.88t , t = .794 sec

    After .794 seconds, the Vcom has slowed from 7 m/s to 4.67 m/s
    Vcom = 7 - 2.94(.794) = 4.67
    Vt has increased from 0 m/s to 4.67 m/s
    Vt = 5.88(.794) = 4.67
    the angular velocity (ω) has increased from 0 rad/s to 77.8 rad/s
    ω = 98(.794) = 77.8
    and a cylinder of radius .06 m and angular velocity of 77.8 rad/s has a Vt =rω
    Vt = .06(77.8) = 4.67 m/s
    so it all checks out.

    Once it stops slipping the kinetic frictional force disappears, and the cylinder will continue to keep rolling at those speeds (in equilibrium) until t = 2 seconds.
     
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