Posted by Kyle on Tuesday, August 16, 2011 at 7:38pm.
The centripetal acceleration is constant:
v^2/r = 5.80^2/2.3 = 14.6 m/s^2
But is centripetal acceleration the same as average acceleration?
What's the point of the problem giving information about time if it's not even used in the process?
Now if you wanted to approximate that in x y coordinates you could do the following:
w = angular frequency = 2pi/period
period = 2 pi r/v
so w = v/r = 5.8/2.3 = 2.52 radians/second
Vx = -5.8 sin wt = 0 at t = 0
Vy = 5.8 cos w t = 5.8 at t = 0
for going in a circle starting at (5.8 , 0)
after .5 s
Vx = 5.8sin1.26radians = -5.52
Vy = 5.8cos1.26 = 1.77
change in Vx = -5.52
change in Vy = 1.77 - 5.8 = -4.03
for acceleration A divide by .5 sec
Ax = -11.04
Ay = -8.06
magnitude of A = sqrt(121+64) = 13.6 etc
but what a waste of time :)
I suspect the 1/2 second and average was just given to confuse you but you could do it the way I did calculating the change in velocity over he half second and dividing by the time. I suspect that they did not expect you to do all that.
The point is that although the centripetal acceleration is constant, you can approximate it by using
average Acceleration = change in velocity / change in time
By the way that averaging came out a lot closer to the real answer than I thought it would. After all we went almost 1/4 of the way around the circle during that half second.
Oh okay. I think this makes sense. Thanks for the explanation!
For the first Vx, how come the 5.8 is negative?
Okay I understand the problem now and I got the answer right doing it your way.
Thanks a lot for the explanation. It really helped!
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