idku
  • idku
One question.... physics
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
idku
  • idku
A solid disk of mass \(\color{#339999}{\rm m_1=9.1~kg}\) and radius \(\color{#339999}{\rm R=0.25~m}\) is rotating with a constant angular velocity of \(\color{#339999}{\rm \omega=32~rad/s}\). A thin rectangular rod with mass \(\color{#339999}{\rm m_2=3.4~kg}\) and length \(\color{#339999}{\rm L = 2R = 0.5~m}\) begins at rest above the disk and is dropped on the disk where it begins to spin with the disk. I have found that: (And all of this is exactly correct I checked) [1] The initial angular momentum of the rod and disk system. 9.10 kg-m^2/s [2] The initial rotational energy of the rod and disk system. 145.6 J [3] The final angular velocity of the disk. 24.91 rad/s [4] The final angular momentum of the rod and disk system. 9.10 kg-m^2/s [5] The final rotational energy of the rod and disk system. 113.32 J
idku
  • idku
The only thing that I need help with, is question 6 (above). The rod took t = 6.2 s to accelerate to its final angular speed with the disk. What average torque was exerted on the rod by the disk?
amistre64
  • amistre64
how do we define torque?

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idku
  • idku
Torque = I * α = (1/12)*m2*L^2 * w\(_f\) / t am I right ?
amistre64
  • amistre64
im not uptodate on my physics equations, so id have to google to verify
amistre64
  • amistre64
if we had applied a constant torque, for 6.2 seconds, what amount of torque would have got us to where we ended up?
IrishBoy123
  • IrishBoy123
yes, \(\tau = I \; \alpha\) they should be opposite and equal, if you calculate them for both rod and disc.
idku
  • idku
the torque for rod and torque for disk are equivalent, you mean ?
IrishBoy123
  • IrishBoy123
equal and opposite, ie Newton's 3rd Law in a rotational context.
amistre64
  • amistre64
ill defer to the more qualified :)
idku
  • idku
I keep loosing it, sorry
idku
  • idku
\(\tau = I \; \alpha\) \(\tau = (1/12)M_2*L^2 *w_f/t\) am I right?
idku
  • idku
\(\tau = (1/12)3.4*L^2 *24.91/6.2\) and I have to calculate the L, based on: Li = Lf = I\(_{disk}\) * w\(_0\) = (1/2)(m1)(R^2)w0
idku
  • idku
L = 1/2 * 9.1 * 0.25^2 * 32 = 9.1 \(\tau = (1/12)3.4*9.1^2 *24.91/6.2\)
IrishBoy123
  • IrishBoy123
is it |dw:1447018730708:dw| or |dw:1447018754733:dw| i think you take it as the latter, hence the \(\dfrac{1}{12}\)
idku
  • idku
|dw:1447018849175:dw|
IrishBoy123
  • IrishBoy123
thought so so you can use all the analogous equations of motion eg \(v = u + at \rightarrow \omega_f = \omega_i + \alpha t\) so \(\alpha = \dfrac{\omega_f - \omega_i }{t}\) \(\tau = I \alpha = I \dfrac{\omega_f - \omega_i }{t}\) \(I = \dfrac{m L^2}{12} \)\\ \(\tau = \dfrac{m L^2}{12} . \dfrac{\omega_f }{t}\) which is what you seem to be doing...
IrishBoy123
  • IrishBoy123
oh, hang on " rod with mass m2=3.4 kg and length L=2R=0.5 m "
idku
  • idku
did I get the correct L ? L = \(\rm I_{disk}\) * \(\rm w_0\) = (1/2) * \(\rm m_1\) * R^2 * \(\rm w_0\) L = 1/2 * 9.1 * 0.25^2 * 32 = 9.1
idku
  • idku
Oh, dang sorry these stupid notations...
idku
  • idku
L length and L angular momentum, I hate it !
IrishBoy123
  • IrishBoy123
for the rod \(\tau = \dfrac{3.4 \times 0.5^2}{12} . \dfrac{24.91 }{6.2}\) assuming, as you say, all of your initial calculations are correct
idku
  • idku
http://www.wolframalpha.com/input/?i=%283.4%C3%970.5%5E2%29%2F12*24.91%2F6.2 0.2846
idku
  • idku
Yes, thanks !
IrishBoy123
  • IrishBoy123
for the disc \[\tau = I \alpha = I \left(\dfrac{\omega_f - \omega_i }{t} \right )\] \[I = \dfrac{mr^2}{2}\] \[\tau = I \alpha = \dfrac{mr^2}{2}. \dfrac{\omega_f - \omega_i }{t} \\ = \dfrac{(9.1)(0.25)^2}{2}.\dfrac{24.91-32}{6.2}\]
IrishBoy123
  • IrishBoy123
that should, absent an stupid error, give the equal and opposite result
idku
  • idku
for what you gave me just now, before: \(\tau = \dfrac{3.4 \times 0.5^2}{12} . \dfrac{24.91 }{6.2}\) I obtained my correct result:)
idku
  • idku
but this just now isn't the same value (?)
idku
  • idku
it worked tho ' , ty
IrishBoy123
  • IrishBoy123
http://www.wolframalpha.com/input/?i=solve+%28+%289.1%29%280.25%29%5E2%29%2F%282%29+*+%2832%29+%3D+%28+%289.1%29%280.25%29%5E2%29%2F%282%29+*+%28x%29+%2B+%28+%283.4%29%280.5%29%5E2%29%2F%2812%29+*+%28x%29 just wondering if the final angular velocity is right.....from the conservation of angular momentum law
IrishBoy123
  • IrishBoy123
1) is right http://www.wolframalpha.com/input/?i=solve+%28+%289.1%29%280.25%29%5E2%29%2F%282%29+*+%2832%29+
IrishBoy123
  • IrishBoy123
2) is right
idku
  • idku
everything is right for this homework the answers 1-5 to which I have posted.

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