Posted by Anonymous on .
A child, hunting for his favorite wooden horse, is running on the ground around the edge of a stationary merry-go-round. The angular speed of the child has a constant value of 0.264 rad/s. At the instant the child spots the horse, one-quarter of a turn away, the merry-go-round begins to move (in the direction the child is running) with a constant angular acceleration of 0.00500 rad/s2. What is the shortest time it takes for the child to catch up with the horse?
The child reaches the horse when the angular positions of the child and horse, measured from initial angular position of the child, are the same.
0.264 t = (pi/2) + (1/2)(0.005)t^2
Solve for t in seconds.
The pi/2 term represents the quarter turn lead of the horse.