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!

Nonuniform circular motion

  1. Sep 30, 2008 #1
    If an object is accelerating tangentially, then how would i go about calculating the radial acceleration since the speed is changing?
     
  2. jcsd
  3. Sep 30, 2008 #2

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Obviously the radial acceleration will be a function of time. If the tangential acceleration is constant then you can simply apply the kinematics equations for rotation (which are analagous to the linear equation). If the tangential acceleration is not constant, then the solution is somewhat more complex. Perhaps if you posted the problem in question we could help you out a little more.
     
  4. Sep 30, 2008 #3
    Bruce is on a bike. He wants to change he speed from 25 to 30km/hr on a curved road, but for safety, the magnitude of his acceleration must not exceed 0.2g. If the radius of the curved road is 5km. what is the minimum time Bruce can change his speed?

    Vi=25km/h, so [itex]\omega[/itex]i=50 rad/s
    Vf=30km/h, so [itex]\omega[/itex]f=60 rad/s


    |a|^2=|R[itex]\alpha[/tex]|^2+|R[itex]\omega[/itex]|^2
    (0.2g)^2=|R[itex]\alpha[/tex]|^2+|R[itex]\omega[/itex]|^2

    a(tang)=dv/dt=(300-250km/h)/t
    [itex]\alpha[/itex]=(60-50rad/h)/t

    then i don't know where to go on to.
     
  5. Sep 30, 2008 #4
  6. Oct 1, 2008 #5

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi subwaybusker! :smile:

    (Hootenanny is out for the evening)

    (have an omega: ω and a squared: ² :smile:)

    You don't need ω … centripetal acceleration, ω²r, is also v²/r. :wink:

    Hint: since v²/r is changing, the maximum dv/dt will also change. :smile:
     
  7. Oct 1, 2008 #6
    okay, i get since that the bike is increasing it's speed, radial ac will increase, but why is the tangential acc changing too? can't dv/dt be a constant? i'm thinking that the tangential is dv(speed)/dt and that you mean the maximum dvelocity/dt which is the total acceleration is changing (correct me if i'm wrong). also, if the speed is always increasing, what do i do with the v²/r since v is never constant?
     
    Last edited: Oct 1, 2008
  8. Oct 1, 2008 #7

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Bruce wants the shortest time, which means the greatest speed, but his acceleration is limited, which means that if the radial acceleration is increasing, then the tangential acceleration must decrease, so that he stays within the safe limit. :wink:
     
  9. Oct 1, 2008 #8
    how can tangential acceleration decrease when it's increasing it's speed from 25 to 30km/h?
     
  10. Oct 1, 2008 #9

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    'cos the acceleration is decreasing, but is still positive. :wink:
     
  11. Oct 1, 2008 #10
    does that mean i'm only find dv/dt and radial acceleration at an instant?
     
  12. Oct 1, 2008 #11

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    At each instant, yes. :smile:
     
  13. Oct 1, 2008 #12
    does that mean i can find the radial acceleration at 30km/hr, use that to find dv/dt and repeat that for 25km/hr and find dv/dt at that point? but somehow i have to relate this to minimum time?

    wait, that doesn't sound right. do i have to use some sort of calculus? sorry, but i really don't know how to approach this?
     
    Last edited: Oct 1, 2008
  14. Oct 2, 2008 #13

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi subwaybusker! :smile:

    Yes, you've got the basic idea …

    but, as I think you've spotted, you have to do it not only for 25 and 30 , but also for every speed in between :tongue2: …

    and then integrate :cry:

    (and before you ask, no, I'm sorry, there isn't any shorter way! :biggrin:)

    have a go! :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Nonuniform circular motion
Loading...