Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

geerky42

  • 3 years ago

Help Needed... (Rotation)

  • This Question is Closed
  1. geerky42
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

  2. Diwakar
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Angular acc = torque / Moment of inertia about the axis of rotation I = ML^2/3 since axis of rotation passes through the end point. Now since the rod is of uniform density , the centre of gravity is at the mid point of rod. torque equals : force * (perpendicular distance between point of suspension and the line of force ) which is = mg * (Lcos68)/2

  3. JamesJ
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Does what Diwakar wrote make sense? He has laid it out for you. In notation, \( \tau = I\alpha \) and hence \( \alpha = \tau/I \). Here, \( I = \frac{1}{4}mL^2 \). Now \( \tau = \frac{1}{2} mgL\cos(68) \) and you can therefore calculate \( \alpha \).

  4. geerky42
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Ah, that makes sense. Thanks.

  5. 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