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.

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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

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)

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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joc
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)

and

(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.

Hmm

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

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joc
(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.

Excellent

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.