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Dynamics of Rotational Motion

  1. Dec 4, 2007 #1
    [Solved] Dynamics of Rotational Motion

    1. The problem statement, all variables and given/known data
    A uniform disk has a mass of 3.4kg and a radius of 0.53m. The disk is mounted on frictionless bearings and is used as a turntable. The turntable is initially rotating at 50rpm. A thin-walled hollow cylinder has the same mass and radius as the disk. It is released from rest, just above the turntable, and on the same vertical axis. The hollow cylinder slips on the turntable for 0.20s until it acquires the same final angular velocity as the turntable. The average torque exerted on the hollow cylinder during the 0.20s time interval in which slipping occurs is closest to:
    A) 13 N m
    B) 16 N m
    C) 6.3 N m
    D) 8.3 N m
    E) 17 N m

    2. Relevant equations
    Moment of inertia of a solid cylinder: [tex]\frac{1}{2}MR^2[/tex]
    Moment of inertia of a thin-walled cylinder: [tex]MR^2[/tex]

    3. The attempt at a solution
    [tex]50 rpm * \frac{2 \pi}{60} = 5.236 rad/s[/tex]
    [tex]K_{1} = 0[/tex]
    [tex]K_{2} = \frac{1}{2}I_{c}\omega^2[/tex]
    [tex]K_{1} + W_{other} = K_{2}[/tex]
    [tex]W = \frac{1}{2}I_{c}\omega^2[/tex]
    [tex]W = \frac{1}{2}(MR^2)(5.236)^2[/tex]
    [tex]W = \frac{1}{2}(3.4)(0.53^2)(5.236)^2 = 13.09 J[/tex]
    I'm not sure how to relate this to average torque. I thought maybe W/time = torque, but that doesn't really work at all or make sense. W = t * d theta, but then I need theta for a slipping disk which seems hard to find.

    [tex]\alpha = \frac{\Delta\omega}{\Delta t}[/tex]
    [tex]\alpha = \frac{5.236}{0.20s} = 26.18 rad/s^2[/tex]
    [tex]\tau = I\alpha[/tex]
    [tex]\tau = (MR^2)(\alpha) = (3.4 * 0.53^2)(26.18)[/tex]
    [tex]\tau = 25[/tex]

    Which is nothing near any of the choices. I think I can't use the [tex]\tau = I\alpha[/tex]. :confused:
    Last edited: Dec 4, 2007
  2. jcsd
  3. Dec 4, 2007 #2

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    Initial ang mom = I1w1 = I1w2 + I2w2 = final ang mom, where I2 is MI of cylinder, and w2 is angular velo of both.

    Avg Torque = I2*(change in ang velo of cylinder/delta-t).

    Find w2 and plug in.
  4. Dec 4, 2007 #3
    [tex]I_{c} = MR^2 = (3.4)(.53)^2 = 0.95506[/tex]
    [tex]I_{d} = \frac{1}{2}MR^2 = \frac{1}{2}(3.4)(.53)^2 = 0.47753[/tex]

    From what you said,
    [tex]I_{d}\omega_{1} = I_{d}\omega_{2} + I_{c}\omega_{2}[/tex]

    [tex]\omega_{2} = \frac{I_{d}\omega_{1}}{I_{d} + I_{c}}[/tex]

    [tex]\omega_{2} = \frac{(0.47753)(5.236)}{0.47753 + 0.95506} = 1.745[/tex]

    [tex]\tau = I_{c}\frac{\Delta\omega}{\Delta t}[/tex]
    [tex]\tau = (0.95506)\frac{5.236 - 1.745}{0.20s} = 16.67[/tex]

    Wow, thanks! I was beginning to lose hope too :)

    So, if I understand this right, the one is spinning on a frictionless surface without being forced to spin, just spinning on its own at a given angular velocity. Then the cylinder falls on, and takes some of the energy of the disk and they both move at a new slower angular velocity of 1.745 rad/s?
  5. Dec 4, 2007 #4

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    Right. The total ang mom is conserved, but there is loss in toatl KE of the system, being dissipated into heat etc.
  6. Dec 7, 2007 #5
    I got this one wrong. The correct answer is 8.3 Nm. I was wondering, could anyone explain to me where I messed up?
  7. Dec 7, 2007 #6

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    I'm getting 8.3 Nm, using the method I'd told you. Check your calc.
  8. Dec 7, 2007 #7

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    You guys do such messy calculations that I wanted to waste some time in just telling you please do not to plug in the numbers right from the start. Operate with symbols and plug in the numbers at the end: e.g., as I’m showing how in this particular problem.

    I1w1 = I1w2 + I2w2 => w2 = I1w1/(I1 + I2). ..(1)

    Noting that I1 = mr^2/2 and I2 = mr^2, we get, w2 = (1/2)w1/(3/2) = (w1)/3 (from 1).

    Avg torque = N
    = I2(w2-0)/delta_t
    = I2*(w1/3)/0.2
    = m(r^2)*(2*pi*(50/60)/3)/0.2
    = 3.4*(0.53^2)(2*pi*50/60)/(3*0.2)
    = 8.33.
  9. Dec 8, 2007 #8
    Ah, I see where I went wrong. I let the cylinder start at 5.236 and go to 1.745 instead of start at 0 and go to 1.745.
    The very last line should read:
    [tex]\tau = (0.95506)\frac{1.745 - 0}{0.20s} = 8.3[/tex]

    Is this better?:
    [tex]I_{d}\omega_{1} = I_{d}\omega_{2} + I_{c}\omega_{2}[/tex]
    [tex]\omega_{2} = \omega_{1}\frac{I_{d}}{I_{d} + I_{c}} = \frac{\omega_{1}}{3}[/tex]
    [tex]\tau = I_{c}\frac{\Delta\omega}{\Delta t} = MR^2\frac{\omega_{1}}{3t} = (3.4)(.53)^2\frac{5.236}{3(0.2)} = 8.3[/tex]
    Last edited: Dec 8, 2007
  10. Dec 8, 2007 #9

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    Much, much better!
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