- #1
LD_90
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I've been working on this for awhile. I think I've got it. Ok here's the problem:
A particle moves in a circle of radius r completing one cycle in time T. What is the magnitude of the average acceleration over time T and over time T/2?
It seems to me that over time T the magitude of the ave acceleration should be zero, but over time T/2, it has to be a non-zero number. Since the acceleration is constant, the instantaneous acceleration should be equal to the average acceleration. v(final)-v(initial)=change in v
the (change in v)/(change in time)= a ; so at time T, v(final) - v(initial)=0 and so acceleration has to equal zero. Over time T/2, v(final ) and v(initial) are equal in magnitude but opposite in direction, so the change in v
should equal (2)(magnitude of v)
v=(2(pi)r)/T, then delta v = (4(pi)r)/T, delta t=T/2, so a=((8)(pi)r)/T^2
This seems right to me. Am I missing something.
A particle moves in a circle of radius r completing one cycle in time T. What is the magnitude of the average acceleration over time T and over time T/2?
It seems to me that over time T the magitude of the ave acceleration should be zero, but over time T/2, it has to be a non-zero number. Since the acceleration is constant, the instantaneous acceleration should be equal to the average acceleration. v(final)-v(initial)=change in v
the (change in v)/(change in time)= a ; so at time T, v(final) - v(initial)=0 and so acceleration has to equal zero. Over time T/2, v(final ) and v(initial) are equal in magnitude but opposite in direction, so the change in v
should equal (2)(magnitude of v)
v=(2(pi)r)/T, then delta v = (4(pi)r)/T, delta t=T/2, so a=((8)(pi)r)/T^2
This seems right to me. Am I missing something.
