A small object with mass 3.90 kg moves counterclockwise with constant speed 1.55 rad/s in a circle of radius 3.20 m centered at the origin. It starts at the point with position vector 3.20i hat m. Then it undergoes an angular displacement of 9.15 rad.

Question

anonymous · Answer

(a) In what quadrant is the particle located and what angle does its position vector make with the positive x-axis? 

(b) What is its velocity? (m/s)

(c) In what direction is it moving? (degrees, from the +x-direction)

(d) What is its acceleration? (m/s^2)

(e) What total force is exerted on the object? (N)

anonymous · Answer

Initially the object is at 3.20 i which means at x-axis.
Now 9.15 rad = 2pi + 2.87 rad
2pi means i full round, so the object is again at 3.20i. Now, pi/2 < 2.87 < pi, that is the object is in second quadrant and the angle is 2.87 radians.
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