If the print is too small the question is this: Five situations of circular moti
ID: 1963003 • Letter: I
Question
If the print is too small the question is this:
Five situations of circular motion are described below with corresponding velocity and
acceleration vectors shown. Rank the magnitude of the acceleration in each case from
smallest to largest. Indicate if two particles have equal magnitudes of acceleration.
Case A: Constant velocity
Case B: Speeding up
Case C: Constant velocity
Case D: Constant velocity
Case E: Slowing down
As you see the answer is already there, I just don't know how to get that. If somebody could explain how to get that answer that would awsome.
Explanation / Answer
Just consider cases A C and D first. They say "constant velocity", but they really mean "constant speed" (the velocity is changing because the particle is changing direction, moving in a circle.) We know that the acc of circular motion is given by a = v^2 / r So it should be obvious that when comparing cases C and D, D will have greater acc (due to smaller radius, but same speed). Therefore, D>C Then compare A and D. Going from case A to case D, we cut the speed in half and the radius by a factor of 4. This means that v squared is reduced by a factor of 4 (i.e. one half squared). So going from A to D, there is a factor of 1/4 in the numerator (with v^2) and in the denominator (with r). Therefore, acc of case D is the same as case A. Now we have A=D>C For cases B and E, the acc associated with circular motion is directed toward the center of the circle, while the acc associated with changing speed is tangent to the circle. These two accelerations then are perpendicular to each other, and to get the total we would have to make a tright triangle of the two accelerations... and the total acc in each case is the hypotenuse of the triangle. The legs of the triangle we would make for case B and E are the same... so the total accelerations (i.e. the hypotenuse in each case) is also the same. Therefore, B=E Finally, the acceleration of case B or E is greater than that of A, because B and E have the same acc of circular motion, but the additional acc of changing speed. Putting all thsi together, we get B=E > A=D > C (I wrote in in reverse order of theirs... I wasnt looking when I wrote mine).