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Rotation Question

  1. Oct 27, 2006 #1
    if eight 60 kg skaters join hands and skate in a line thats 12m wide at 4.6m/s, and one skater at the end stops abruptly, and the whole line rotates about that skater rigidly, what is the angular speed of the outermost skater?

    ... would the momentum be conserved?
     
    Last edited: Oct 27, 2006
  2. jcsd
  3. Oct 27, 2006 #2

    OlderDan

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    There would be an external force required to stop the one skater, but there is momentum, and there is __________ momentum. One is conserved if there is no external force, and one is conserved if there is no external torque.
     
  4. Oct 27, 2006 #3

    radou

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    Are you sure you need to find the angular speed of a skater? When one skater at the end stops, the rest rotate around him in a line, and this line has an angular speed. Now, every skater has his translatoral speed, except for the one standing, who has zero speed, and translatoral speed is proportional to angular speed, i.e. [tex]v = \omega \cdor r[/tex], where r is the distance from the center of rotation to the point (i.e. skater) which translatoral speed we are looking for.
     
  5. Oct 27, 2006 #4
    Oops I forgot to tell you, the line of skaters were originally skating at 4.6m/s until the one guy at the end stopped.

    the question has 3 parts.
    a) what is the angular speed of outermost skater
    b) what is the linear speed of outermost skater
    c) what is the force exerted on the outermost skater.
     
  6. Oct 27, 2006 #5
    By conservation of momentum I meant...

    L = L'
    r x p = Iw ... then taking middle of the line as centre of mass
    (12/2) (8*60) (4.6) sin 90 = 1/3 (8*60)(12)^2 w .... inertia approximated by a rod rotating about one end with rotational inertia I=1/3ML^2.
    then solve for w....

    However, is that valid, is momentum conserved?

    another way i thought of is..
    Etotal = Etotal'
    Ek = 1/2 I w^2 ... then since 1 of 8 skaters stopped, 7 are still moving, the energy of the system available for rotating is 7*Ek
    7*1/2(60)(4.6)^2 = 1/2 (1/3)(8*60)(12)^2 w^2 ....again, inertia approximated by a rod rotating about one end with rotational inertia I=1/3ML^2.

    then solve for w...
     
    Last edited: Oct 27, 2006
  7. Oct 27, 2006 #6

    OlderDan

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    It would really be a lot easier to follow what you are doing if you used variables like v for velocity and L for length of line, etc. It's harder to follow when you throw in all the numbers, especially when you leave out all the units.

    Your energy approach is not valid. It might be interesting to compare initial and final energy, but your assumption is not valid.

    You can calculate angular momentum about any point you choose. It does not even have to be touching the object. The initial angular momentum about the CM is zero, but there is going to be a torque applied that will change this. You might want to calculate the angular momentum about a point where there will be no applied torque so that angular momentum is conserved.
     
    Last edited: Oct 28, 2006
  8. Oct 29, 2006 #7
    Here are the two methods I have (hopefully easier to read this time)... Momentum doesnt seem to be conserved though, doesn't an external force stop the skater at the end, so net external force is nonzero? I can't see how the whole line can start rotating without an external force applied. The energy approach to me seems to make more sense since it is a crazy situation where the skater stops instantly and loses all his energy...

    [tex]
    \noindent \underline{{\bf Momentum}}:

    $ L = L ' $ \\

    $ { \left| \overrightarrow{r} \times \overrightarrow{p} \right| } = I\mathrm{ \omega } $ where the line of skaters is approximated as a rod rotating about one end ($ I = \frac {1}{3} \mathrm{ML}^{2} $).\\

    $ r\hspace{0.222222em}p\hspace{0.222222em}\mathrm{sin}\hspace{0.222222em}\mathrm{ \theta }\hspace{0.222222em} = ( \frac {1}{3} \mathrm{ML}^{2} ) \mathrm{ \omega } $ \\

    $ r\hspace{0.222222em}m\hspace{0.222222em}v\hspace{0.222222em}\mathrm{sin}\hspace{0.222222em}\mathrm{ \theta }\hspace{0.222222em} = ( \frac {1}{3} \mathrm{ML}^{2} ) \mathrm{ \omega }\hspace{0.222222em} $ \\

    Then taking the centre of mass of this 'rod' as the middle, and the rotational axis at one end of the line, r = 6m\\

    $ ( 6m ) ( 8 \times 60\mathrm{kg} ) ( 4.6m / s ) \hspace{0.222222em}\mathrm{sin}\hspace{0.222222em} {90\hspace{0.222222em} = \hspace{0.222222em} [ \frac {1}{3} \times 8 \times 60\mathrm{kg} \times ( 12m ) ^{2} ] \mathrm{ \omega }}^{} $ \\

    Then solve for $ \mathrm{ \omega } $\\
    [/tex]

    [tex]

    \underline{{\bf Energy}:}\\

    $ E_{\mathrm{total}}^{} = E_{\mathrm{total}} ' $ \\

    $ \mathrm{ \Sigma } E_{\mathrm{kinetic}} = E_{\mathrm{rotation}} $\\

    $ ( 7 \times \frac {1}{2} \mathrm{mv}^{2} + \frac {1}{2} \mathrm{mv}^{2} ) = \frac {1}{2}I \mathrm{ \omega }^{2} $\\

    Then the line of skaters is approximated by a rod rotating about one end ($ I = \frac {1}{3} \mathrm{ML}^{2} $), and since at the instant where the last skater abruptly stop, he loses all his kinetic energy.\\

    $ ( 7 \times \frac {1}{2} { ( 60kg ) ( 4.6m / s ) }^{2} + \frac {1}{2} { ( 60\mathrm{kg} ) ( 0 ) }^{2} ) = \frac {1}{2} [ \frac {1}{3} \times 8 \times 60\mathrm{kg} \times ( 12m ) ^{2} ] \mathrm{ \omega }^{2} $ \\

    Then solve for $ \mathrm{ \omega } $\\
    [/tex]

    Those are the two methods I have... Momentum doesnt seem to be conserved though, doesn't an external force stop the skater at the end, so net external force is nonzero? I can't see how the whole line can start rotating without an external force applied. The energy approach to me seems to make more sense since it is a crazy situation where the skater stops instantly and loses all his energy...
     
    Last edited: Oct 29, 2006
  9. Oct 29, 2006 #8

    OlderDan

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    There is an external force that stops the one skater. And that skater exerts a force (internal to the line) on the next skater and so on and so on. But if the rotation is around that one end of the line, at the point of rotation there is no torque from that external force around that point.

    The initial angular mometnum about the point of rotation is the rpsinθ of the CM of the line (which, by the way, is the sum of the indiviual rpsinθ terms for each skater). Wherever the line is as it approaches the rotation point, rsinθ for the CM is half the length of the line or 6m. You need to compute this initial angular momentum. Then use your Iω expression to find the resulting ω using angular momentum conservation. You do not know the speed of the center of the line once the rotation starts. The (unrealistic) force in this problem is huge. It stops the end skater instanly, and instantly reduces the speed of other skaters. You will find out how fast they are going by first doing a correct calculation of ω.

    Actually, the skater at the end of the line does not have to stop instantly. As long as the external force is acting at the point about which the line is rotating, the angular momentum about that point will be conserved.
     
    Last edited: Oct 29, 2006
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