A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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
  1. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1443348109250:dw|

  2. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Assume the rod is in space, no gravity.

  3. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    That's it?\[\int \tau dt = \Delta L\]

  4. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\int F dt = \Delta p \]for the velocity of the center of mass

  5. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    What I wrote above was just impulse.

  6. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yeah, nice, that should give velocity of center of mass

  7. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Here, \(F\) (and \(\tau\)) are constant so you can just say \(F \cdot t = m \Delta v_{CM}\)

  8. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\tau = (10 N)(0.5 m) = 5 N\cdot m\]\[t = 1 ~sec\]\[\rm Angular ~ impulse = \Delta L = 5 N\cdot m \cdot sec = L_f - L_i = L_f\]

  9. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[L_f = I\omega = \frac{1}{12}M\ell ^2\omega = 5 ~SI ~units \]And \(M = 1 kg, \ell = 1.5 m\)

  10. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Newton's second law : \(F=Ma_{CM} \) multiply both sides by \(\Delta t\) : \(F*\Delta t = Ma_{CM}\Delta t = Mv_{CM} \\\implies v_{CM}=\dfrac{F*\Delta t}{M}\)

  11. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yessss, exactly.

  12. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Are you taking center of mass as reference point for your \(\tau \)?

  13. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |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 \(\tau = F (0.5) \cos \theta = I \ddot \theta\)

  14. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    That's a great observation. Problems are rarely as complicated as that. Maybe we can try that too?

  15. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.