Uniform Circular Montion problem:?

Question

I'm more confused on part (a.) on how to answer the question more than anything else, here is the whole thing:

A car initially traveling eastward turns north by traveling in a circular path at uniform speed as in the figure below. The length of the arc ABC is 210 m, and the car completes the turn in 32.0 s.

(a) What is the acceleration when the car is at B located at an angle of 35.0°? Express your answer in terms of the unit vectors and .

(b) Determine the car's average speed.

(c) Determine its average acceleration during the 32.0 s interval.

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I've tried answering part B without answering part a. which I assumed that the ABC arc was just another type of radius and thus r=210m and that my angle would just become 35 degrees. Then I solved for Velocity with the F= m(v^2/r) got the velocity which due to shortage of characters my answer which was incorrect was 37.96 m/s, so I'm stumped currently.

Appreciate the help. :+)

Answer 1

uniform speed...therefore the only acceleration component is normal(towards the center) use an-normal acceleration. The Arc Length means the distance on the circumfrence of the circle covered by ABC, not radius. If we assume B is halfway along the circle at 35 degrees. Then we know that the whole ABC is 70 degrees. 70/360 is the amount of the total circumfrence covered by the 210 m. So the whole circumfrence is 1080 m. 2piR gives the radius as 540*pi. Because there is no change in Tangential Acceleration, Normal acceleration will not change either(circle doesn't change, if it was ovular motion it would). So the fact of where it is on the circle doesn't matter.
An = (4pi^2R)/(T^2)
to find the period, T, we do the same proportion thing as we did to find the total circumfrence...70/360 = 32 s/T s
T=164.57 s
An = 2.473 m/s^2
so your components of acceleration are
An=^
At=0
if you are supposed to find Ax and Ay(which is stupid for a circular motion problem) do the geometry with the 35deg angle.
b) car's average speed
210m/32s
6.563m/s
c)average acceleration is just the An again.

If i misinterpreted a part of the question...possible since i don't have the figure, i hope I atleast showed you some ways to solve the problem

Answer 2

The velocity of the car around the arc v = 210/32 m/sec. The acceleration of an object moving in a circular path is v^2/r. Assuming the arc is 1/4 of a circle (turning from east to north), and that 210m is the full length of the arc, the circumference of the circle is 4*210, so the radius r = (4*210)/(2*π).

Without your figure it is hard to be sure of the answer. It may be that the only problem with your answer is the radius, which is not 210m.

Using the numbers above, I get accel = 0.322 m/sec^2

Answer 3

Without seeing the diagram, I assume the car enters the turn going E and exits going N? And that the arc ABC is just the circular path the car follows as it makes this turn?
The car is moving at a constant speed so its only acceleration as it turns is the centripetal acceleration a(c)=v^2/r. Therefore the acceleration at B will be the same as at any other point of the arc. There is no tangential acceleration.
The radius can be found using r=d/theta. 'd' is 210. The car is going from E to N so theta is pi/2 radians.

Answer 4

the information is implicit. you in all likelihood know intuitively that in case you whirl it slower, it is going to commute at a decrease, greater vertical attitude. in case you whirl it swifter the string will upward thrust and develop into greater horizontal. in case you do no longer whirl it in any respect, that is going to dangle on the instant down. So in some way the particular attitude is telling you techniques concerning the value. The question is the thank you to translate that right into a actual concept. Why does it bypass up once you whirl it? (I might desire to contemplate this for a minute). Edit: you are able to describe it from the "centrifugal rigidity" point of view because of the fact the centrifugal rigidity pulling outward on the string. From the "centripetal rigidity" description that is that the centripetal rigidity (mv^2/r) comes from the horizontal area of the rigidity. The vertical area of the rigidity is often basically adequate to help 50 g. So set mv^2/r = horizontal area of hysteria.