1. Limited time only! Sign up for a free 30min personal 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!

Finding the total acceleration

  1. Jan 6, 2013 #1
    1. The problem statement, all variables and given/known data
    Gaetan Boucher started from rest and skated around a circular ice track of radius 100m. He maintained a constant rate of increase of his speed and finished one complete lap in 30s. Calculate the magnitude of his total acceleration as he passed a point 1/4 of the distance around the track from the start.

    3. The attempt at a solution
    v1 = 0m/s
    r = 100m
    d = 1/4d
    a = ?
    t = 30s

    c = 1/4 x 2rπ
    c = 157 m

    t = 1/4 (30s)
    t = 7.5s
    157 m = (0m/s + v2)/2 (7.5s)
    41.866 m/s = v2

    157 m = 1/2a(t)2
    5.58 m/s^2 = a
     
  2. jcsd
  3. Jan 6, 2013 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    There are two accelerations to be solved for here: tangential and radial.

    First get tangential accel: this should be easy from the fact that this accel is constant.

    Then you need to solve for speed v at the 1/4 point. You can do that by solving for the time it takes him to go to the 1/4 point (again using the tangential accel.) , then use your value of tangential accel one more time to get v. Finally, you compute the other component of accel (hint: it's called centripetal) to get the total accel.

    The total accel is not the numerical sum of the two. The two accels are separate vectors, one pointing tangential to his course and the other towards the center of the ring. If you don't know vectors just leave the answer as two separate accelerations.
     
  4. Jan 6, 2013 #3
    Okay i am a bit confused.
    you said tangential acceleration,
    a = (rw) / t

    right?
    i have not learned the tangential acceleration thoroughly so.
     
  5. Jan 6, 2013 #4

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    What is the angular acceleration ?
     
  6. Jan 6, 2013 #5
    using this formula ω = θ / t

    but i do not know the angle .. not wait i do.
    1/4 x 360 = 90 deg

    ω = θ / t
    ω = n/2 / 7.5s
     
  7. Jan 6, 2013 #6

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    That's the average angular velocity for the first 1/4 of a lap.

    The skater starts with ωi = 0, & averages π/(2(7.5)) rad/s ?

    What is ωf at 1/4 lap ?

    (Just as in the case of constant linear velocity, for constant angular velocity, ωAverage = (ωi + ωf)/2 .)
     
  8. Jan 6, 2013 #7
    π/(2(7.5)) rad/s = (0 +ωf / 2)
    π/7.5 rad/s = ωf
     
  9. Jan 6, 2013 #8

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Yes.

    From ωf & ωi , you should be able to find the angular acceleration, α .
     
  10. Jan 6, 2013 #9
    π/7.5 rad/s = at
    π = a
     
  11. Jan 6, 2013 #10

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    No. Bad algebra.

    t = 7.5 s ,

    so a = (π/7.5)/7.5 rad/s2

    = ?
     
  12. Jan 6, 2013 #11
    a= π/56.25 rad/s^2
     
  13. Jan 6, 2013 #12

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    The tangential component of acceleration is [itex]\ \ a_t=\alpha R\ .[/itex]

    The radial component of acceleration is simply the centripetal acceleration.
     
  14. Jan 6, 2013 #13
    like this?

    π/56.25 rad/s^2 x 100 m = a
    5.58 radm/s^2 = a
     
  15. Jan 6, 2013 #14

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    That's only the tangential component.
     
  16. Jan 6, 2013 #15
    a = 5.58 radm/m^s2
    v at the 1/4 point. You can do that by solving for the time it takes him to go to the 1/4 point
    d = 200 m
    C = 200 m x 3.14
    C = 628 m
    D = 1/4 628 m = 157 m

    157 m = 1/2 (5.58 radm/m^s2)(t)^2
    t = 7.5 s

    157 m x 2 / 7.5 s =v2
    41.866 m/s = v2

    a = 41.866 m/s /100m
    a = 0.041866 m/s^2
     
  17. Jan 6, 2013 #16

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    You went through some needless steps there.

    You already have ω at t = 7.5s, which you already know is the time at the 1/4 lap mark.

    v = ωR .

    Centripetal acceleration is ac = v2/R .

    (It's also equal to 2R)

    I get a value different from yours.

    By the way, rad∙m/s2 is merely m/s2 .
     
  18. Jan 6, 2013 #17
    157 m x 2 / 7.5 s =v2
    41.866 m/s = v2



    a = v^2 / R
    a = 41.866 m/s^2 / 100m
    a = 17.52 m/s^2
     
  19. Jan 6, 2013 #18

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    That's about right.

    Tangential acceleration is in the direction of motion. Centripetal acceleration is towards the center of the circle. Therefore, they're perpendicular to each other.

    Find the magnitude of the acceleration using your two results.
     
  20. Jan 7, 2013 #19
    a = 17.52 m/s^2[W] + 5.58 m/s^2
    like this?
     
  21. Jan 7, 2013 #20

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member


    No.

    Acceleration is a vector quantity.

    Tangential component is perpendicular to radial.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook