Question

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 0.251 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 0.0101 rad/s². What is the shortest time it takes for the child to catch up with the horse?

Answer 1

The total angular displacement is,

        d(theta) = (2*pi / 4) + 0.5(alpha)(t²)

              wt = (pi / 2) + 0.5(alpha)(t²)

Write the equation in the form of quadratic relation,

     0.5*(alpha)*(t²) - wt + (pi / 2) = 0

     0.5*(0.0101 rad/s²)(t²) - (0.251 rad/s)t + (pi/2) = 0

Then, the solutions of the above equation are, t = 7.34 s and 42.36 s.

Hence, the shortest time is, 7.34 s