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Homework Help: Tangential/radial/total acceleration & angle

  1. Nov 2, 2003 #1
    Question: A race car starts from rest on a circular track. The car increases its speed at a constant rate at as it goes once around the track. Find the angle that the total acceleration of the car makes-with the radius connecting the center of the track and the car-at the moment the car completes the circle.

    OK now it just so happens this question is odd numbered and the answer is in the back of the book:

    Answer: Theta = Tan^(-1)(1/(4Pi)) = 4.55 degrees.

    Basically there seems to be zero information in the question so I don't understand how the answer is a precise angle.. I don't even know how to approach this question.

    This second one I've done but I dont understand what I'm doing very well (specifically on part c), I just tried to follow the notes.

    Second question:

    An automobile whose speed is increasing at a rate of 0.600 m/s^2 travels along a circular road of radius 20.0m. When the instantaneous speed of the automobile is 4.00 m/s, find (a) the tangential acceleration component, (b) the centripetal acceleration component and (c) the magnitude and direction of the total acceleration.

    OK so (a) we want at (tangential acceleration):

    ok so the speed is tangential, so at is just the magnitude of the rate of change of the speed, so at=0.6m/s^2

    b) radial acceleration, ar:

    so 4^2/20 = 0.8 m/s^2

    c) lastly magnitude and direction of total acceleration, a:

    a = sqrt(at^2 + ar^2) = 1 m/s^2
    (assuming it starts at rest even tho not stated?)
    er tangential acceleration = ai now, t doesn't work well.
    v(t) = v(0) + ai*t = 0 + ai*t, v(t) = 4 m/s so t = 4/1
    = 4 s

    heres where I get really lost, and the notes and text book examples differ though im not sure why. The lecture notes posted are very wrong (mistakes in formulas) so i cant double check my notes.. anyway

    c(t) = c(0) + v(o)t + (a/2)*t^2 = so (1/2)*4^2=8 m

    so 8 meters traveled at time instaneous speed of 4 m/s is reached

    the books example now just solves theta as theta= tan^(-1)(ar/ai)
    so just inverse tan of radial acceleration over tangential acceleration..

    My notes for a very similar question with different numbers are something like this:

    angle = dist/circum = (8m * 2 * Pi) / (2 * Pi * 20m)

    then says angle = Pi - 2/5. Now I'm not sure how he arrived at the conclusion the angle needed to be subtracted from Pi in the first question so I dont know if the same needs to be done here or not. However the angle for the similar question is 5/2 so im guessing that subtracting the angle from Pi was only neccesary because it was >90 and hence is not neccesary in this question?

    lastly tangential and radial accelerations are reduced to vector components

    aix = -ai*sin(theta)
    aiy = -ai*cos(theta)

    arx = ar*cos(th)
    ary = -ar*sin(th)

    then a vector = ai vector + ar vector so just
    ax =-ai*sin(theta)+ar*cos(th)
    ay = -ai*cos(theta)+-ar*sin(th)

    I thought we already had the magnitude of the total acceleration so im not really sure what the last part w/ vectors is even for..
    Any help would be appreciated, I'm off to bed heh.

    Edit: Oh ya the problem I used as a template to do the second problem was worded about the same numbers were as follows: Radius 20m (same), instanteous speed, 10 m/s, constant change in speed of magnitude 1 m/s^2. Just incase there is any confusion as to methods that that might help to alleviate. Going back to bed now.

    Last edited: Nov 2, 2003
  2. jcsd
  3. Nov 2, 2003 #2


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    ok, for the first qn:

    the 2 components of acceleration acting on the car during its motion are tangential and centripetal. the tangential component is a constant - let's call it A. the centripetal component is v^2/r, where v is the velocity of the car as it completes its first round, and r is the radius of the circular path.

    since the 2 components are mutually perpendicular, they can actually be arranged head-to-tail to form the 2 shorter legs of a right-angled triangle. the question requires the angle that the hypotenuse makes with the centripetal component. in other words, we have to find angle x where

    x = tan^(-1) [ A / (v^2/r) ]

    but notice the 2 relationships:

    v = AT ----- (1)


    (1/2)vT = 2.pi.r ----- (2)

    where T is the time taken for the complete circle.

    simplifying the expression for x, we have

    x = tan^(-1) [ A / (v^2/r) ]
    x = tan^(-1) [ Ar / v^2 ]

    using eqn (1),

    x = tan^(-1) [ Ar / A^2.T^2 ]
    x = tan^(-1) [ r / A.T^2 ]

    using eqn (1) again,

    x = tan^(-1) [ r / vT ]

    using eqn (2).

    x = tan^(-1) [ r / 4.pi.r ]
    x = tan^(-1) [ 1 / 4.pi ]

    and there's your answer. so it's actually just a spot of symbol manipulation. off to bed now.
  4. Mar 15, 2004 #3

    Perhaps I missed something, but shouldn't that be vT = 2 * pi * r ?

    I'm not really sure where the (1/2) came from on the LHS, which is possibly why I keep on coming up with x = arctan 1 / ( 2 * pi) instead of x = arctan 1 / (4 * pi).

    Could someone please elaborate?
    Last edited: Mar 15, 2004
  5. Mar 16, 2004 #4


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    (1/2)vT = 2.pi.r -----(2)

    is an equation concerning distance travelled. the distance travelled is equal to the area under the speed-time graph, and is therefore (1/2)vT since acceleration is constant. the RHS is the circumference of the circle, i.e. the distance around the track. hope this clears things up.
  6. Mar 16, 2004 #5

    Thank you very much, that cleared up everything. For some really odd reason I didn't use an equation which factored in acceleration, and instead used an equation which assumed constant velocity (t = 2piR / v). Thanks again for clearing things up.
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