1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Desperate Circular motion

  1. Nov 18, 2008 #1
    Desperate !!! Circular motion

    1. The problem statement, all variables and given/known data

    Particles travels counterclockwise in circles of radius 5m. The accleration vectors are indicated at three specific times. Find the values of v and dv/dt at each time.

    a. a=20 m/s^2 . vector a points toward the center of the circle, vector v tangent to the circle

    b. a=30 m/s^2. vector a points toward the center of the circle at a 30 degree angle in the direction of motion ( so imagine the case a, vector a point directly to the center, this case vector a and the vector a in part a creates a 30 degree)

    c. a= 50 m/s^2. vector a creates a 45 degree angle clockwise with the radius.

    2. Relevant equations

    a=mv^2/s

    3. The attempt at a solution

    I know that it's confusing. Please give some input. if you guys are unclear I can explain more.
     
  2. jcsd
  3. Nov 19, 2008 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi nns91! :smile:

    (have a theta: θ :wink:)

    I assume you know how to do a., using centripetal acceleration.

    ok, for the others, you have the usual centripetal acceleration formula, and you also need to calculate the tangential acceleration (for constant r, that's at = rθ'' = v').

    Acceleration is a vector, so you add accelerations just like velocities … add the centripetal and tangential accelerations, to get the required angle. :wink:
     
  4. Nov 19, 2008 #3
    Re: Desperate !!! Circular motion

    Thanks.

    I know acentripetal=mv2/2

    The tangential part confuses me the most. How can I calculate it from that formula at=r[tex]\theta[/tex]'=v' ??

    What do you mean by to get the require angle ?
     
  5. Nov 19, 2008 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi nns91! :smile:

    (why didn't you use that θ i gave you? :rolleyes:)

    at = dv/dt, which you get from the question.

    And if you want the total acceleration to be at 45º, for example, then obviously you need the centipetal acceleration to be equal to the tangential acceleration … just like adding velocities! :wink:
     
  6. Nov 19, 2008 #5
    Re: Desperate !!! Circular motion

    But I don't have time ? so how can I get dv/dt ?
     
  7. Nov 19, 2008 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Well, for 45º, for example, you need v2/r = dv/dt, so v = … ? :smile:
     
  8. Nov 19, 2008 #7
    Re: Desperate !!! Circular motion

    How about a 30 degree angle ?
     
  9. Nov 19, 2008 #8

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Oi! Do it yourself!
     
  10. Nov 19, 2008 #9
    Re: Desperate !!! Circular motion

    I mean I still have not fully understand.

    So they give a=20 m/s^2 in part a.

    Thus: v=sqrt(ar)= sqrt(20*5)= 10 m/s

    Is that right ??

    I still don't understand the calculation of acceleration
     
  11. Nov 19, 2008 #10

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    That's right. :smile:

    Now try c.
     
  12. Nov 19, 2008 #11
    Re: Desperate !!! Circular motion

    c.

    v= sqrt(a*r)=sqrt(50.5)= 15.8 m/s ??

    So I got v.

    How then can I calculate dv/dt ?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Desperate Circular motion
  1. Circular Motion (Replies: 4)

  2. Circular motion (Replies: 8)

Loading...