A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

A cubical block of side "a" is moving with velocity v on a horizontal smooth plane. It hits a ridge at point O. Find the angular speed of the block after it hits the point O. Please view the attached image for more details

  • This Question is Closed
  1. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    This is the attached diagram

    1 Attachment
  2. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So this thing is just heading towards O which then acts as a point of rotation?

  3. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes

  4. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay...I just to do something, then I'll take a look.

  5. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    *just go to do something*

  6. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay, I will be waiting

  7. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Still waiting?

  8. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes

  9. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm thinking you can attack the problem by considering conservation of energy, assuming only internal, conservative forces are acting.

  10. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You have two regimes - translational (before the ridge), then rotational.

  11. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The kinetic energy of the block before hand is \[K_i=\frac{1}{2}Mv^2\]

  12. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The kinetic energy of the block just after impact, as it enters the rotational regime, is\[K_f=\frac{1}{2}I \omega^2\]

  13. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Is this making sense so far?

  14. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes

  15. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    If I am not mistaken then the angular speed is coming to be (v/a)(sqrt6)

  16. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Now, your problem is now about finding the moment of inertia for a block whose axis of rotation is about one of its edges. A solid, rectangular parallelpiped is given by...

  17. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I get something close, but different.

  18. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh I am sorry, I calculated the I around the Cm

  19. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Which is wrong

  20. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes

  21. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So how do you calculate it around that corner?

  22. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Parallel Axis Theorem

  23. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh yes, you are right

  24. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The new moment of inertia will be\[I=I_{cm}+Mh^2\] where h is the distance from the CM axis to the new, parallel axis.

  25. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[I_{cm}=\frac{Ma^2}{6}\]

  26. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[Mh^2=M \left( \frac{a}{2} \right)^2=M \frac{a^2}{4}\]

  27. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I think it is \[1/6(Ma^2)+2M\]

  28. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So \[I=\frac{5Ma^2}{12}\]

  29. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You see the corner wise distance will be sqrt 2

  30. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I mean sqrt 2 times a

  31. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[w^2=\frac{12v^2}{5a^2}\]

  32. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes, you're right there.

  33. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So change h to a.sqrt(2)

  34. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    a/sqrt(2)

  35. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[h=\frac{a}{\sqrt{2}}\]

  36. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So I am supposed to solve the equation \[0.5 Mv^2=0.5[(1/6)Ma^2+2M]\omega\] Am I right

  37. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    am i right?

  38. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I think the moment of inertia should be\[I=\frac{Ma^2}{6}+Mh^2=\frac{Ma^2}{6}+M \left( \frac{a}{\sqrt{2}} \right)^2=\frac{Ma^2}{6}+\frac{Ma^2}{2}\]

  39. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[=\frac{4Ma^2}{6}=\frac{2Ma^2}{3}\]

  40. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Then\[\frac{1}{2}Mv^2=\frac{1}{2}\frac{2Ma^2}{3} \omega ^2\]

  41. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i.e.\[v^2=\frac{2a^2}{3} \omega ^2 \rightarrow \omega ^2 = \frac{3v^2}{2a^2}\]

  42. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ok I understand that

  43. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Are you agreeing? I'm doing this on-the-fly, so if you pick up anything, interrupt.

  44. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Any how I have got the thing

  45. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Did your answer agree?

  46. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Lokisan, I have got what I expected, I can handle the rest

  47. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    There was some problem with the site, and some of our converstation got split up

  48. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay. Happy physics.

  49. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lol, thanks for the genius comment. we'll just see how this question pans out.

  50. 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.