A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

ganeshie8

  • one year ago

Consider a disk under uniform circular motion. The acceleration of a particle on disk \(r\) units away from center is given by : \(a_c = \dfrac{v^2}{r}\) does that mean the acceleeration is inversely proportional to the radius ? why or why not ?

  • This Question is Closed
  1. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @ParthKohli

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

    If it's a rigid disk, then all particles have the same angular acceleration \(\alpha\), but their acceleration \(a = r \alpha\) so acceleration is actually directly proportional to radius here's the flaw with your argument: v is not constant. it also depends on the radius.

  3. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    How do you explain it more lucidly to a 11th grader ?

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

    You can't say that time is inversely proportional to time just because \(t = \dfrac{t^2}{t}\) The top (numerator) of the fraction is a function of radius too so you can't say acceleration is inversely proportional to radius.

  5. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I could show him that \(a_c = \omega^2 r\), but how to really make him see that \(r\) is somehow more important than \(v\) ?

  6. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that is nice so what you saying is that just because something is there in the denominator wont make the quantity vary inversely

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

    well you could either use that example about time the thing he'd need to understand is that the numerator is not constant, so you'd need to get something constant in the way. direct and inverse proportion always means that the rest is constant right?

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

    yeah exactly

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

    there was another question about circuits which @shrutipande9 asked in which she was confused about a similar thing: is power due to resistance \(\propto V^2\) or \(\propto V\) cuz \(P = VI = V^2/R\)

  10. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wow yeah same thing \[P=VI = \dfrac{V^2}{R} = I^2R\]

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

    varying voltage does affect current but not resistance

  12. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    who is depending on whom

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

    yes! so it is proportional to \(V^2\) and since \(V \propto I\) we know \(I^2\) is also a choice

  14. 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.