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.(adsbygoogle = window.adsbygoogle || []).push({});

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.

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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:

ar=-v^2/r

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)

=2/5

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

tangential:

aix = -ai*sin(theta)

aiy = -ai*cos(theta)

radial:

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.

-Klion

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# Tangential/radial/total acceleration & angle

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