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ganeshie8
 one year ago
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.
ganeshie8
 one year ago
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.

This Question is Closed

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1443348109250:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Assume the rod is in space, no gravity.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1That's it?\[\int \tau dt = \Delta L\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1\[\int F dt = \Delta p \]for the velocity of the center of mass

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1What I wrote above was just impulse.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1yeah, nice, that should give velocity of center of mass

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Here, \(F\) (and \(\tau\)) are constant so you can just say \(F \cdot t = m \Delta v_{CM}\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1\[\tau = (10 N)(0.5 m) = 5 N\cdot m\]\[t = 1 ~sec\]\[\rm Angular ~ impulse = \Delta L = 5 N\cdot m \cdot sec = L_f  L_i = L_f\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1\[L_f = I\omega = \frac{1}{12}M\ell ^2\omega = 5 ~SI ~units \]And \(M = 1 kg, \ell = 1.5 m\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Newton's second law : \(F=Ma_{CM} \) multiply both sides by \(\Delta t\) : \(F*\Delta t = Ma_{CM}\Delta t = Mv_{CM} \\\implies v_{CM}=\dfrac{F*\Delta t}{M}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Are you taking center of mass as reference point for your \(\tau \)?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0dw: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
 one year ago
Best ResponseYou've already chosen the best response.1That's a great observation. Problems are rarely as complicated as that. Maybe we can try that too?
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