ganeshie8
  • ganeshie8
Another classic problem. A rod of uniform mass 1kg and length 1.5m was acted upon by a force of magnitude 10N for 1 second. The force was applied 0.5m away from the center of mass of rod. Find the velocity of center of mass of the rod and the angular velocity of the rod about the center of mass.
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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ganeshie8
  • ganeshie8
|dw:1443348109250:dw|
ganeshie8
  • ganeshie8
Assume the rod is in space, no gravity.
ParthKohli
  • ParthKohli
That's it?\[\int \tau dt = \Delta L\]

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ParthKohli
  • ParthKohli
\[\int F dt = \Delta p \]for the velocity of the center of mass
ParthKohli
  • ParthKohli
What I wrote above was just impulse.
ganeshie8
  • ganeshie8
yeah, nice, that should give velocity of center of mass
ParthKohli
  • ParthKohli
Here, \(F\) (and \(\tau\)) are constant so you can just say \(F \cdot t = m \Delta v_{CM}\)
ParthKohli
  • ParthKohli
\[\tau = (10 N)(0.5 m) = 5 N\cdot m\]\[t = 1 ~sec\]\[\rm Angular ~ impulse = \Delta L = 5 N\cdot m \cdot sec = L_f - L_i = L_f\]
ParthKohli
  • ParthKohli
\[L_f = I\omega = \frac{1}{12}M\ell ^2\omega = 5 ~SI ~units \]And \(M = 1 kg, \ell = 1.5 m\)
ganeshie8
  • ganeshie8
Newton's second law : \(F=Ma_{CM} \) multiply both sides by \(\Delta t\) : \(F*\Delta t = Ma_{CM}\Delta t = Mv_{CM} \\\implies v_{CM}=\dfrac{F*\Delta t}{M}\)
ParthKohli
  • ParthKohli
Yessss, exactly.
ganeshie8
  • ganeshie8
Are you taking center of mass as reference point for your \(\tau \)?
IrishBoy123
  • IrishBoy123
|dw:1443352163127:dw| i do not believe that this is so trivial, though i would love to be proven wrong QU: after the rod starts rotating, does the force act as per blue or green arrow? because you either have a variable linear force or torque and from what i can see a DE that can't be solved without linearisation eg if it is green then you must have something \(\tau = F (0.5) \cos \theta = I \ddot \theta\)
ParthKohli
  • ParthKohli
That's a great observation. Problems are rarely as complicated as that. Maybe we can try that too?

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