Posted by **Anonymous** on Friday, March 11, 2011 at 11:28am.

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?

- physics -
**drwls**, Friday, March 11, 2011 at 12:42pm
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.

## Answer this Question

## Related Questions

- Physics - In a playground, there is a small merry-go-round of radius 1.20 m and ...
- Physics - A child (48 kg) stands on the edge of a stationary merry-go-round (...
- Physics - In a playground, there is a small merry-go-round of radius 1.20 m and ...
- Physics - A merry-go-round is at rest before a child pushes it so that it ...
- Physics - A child pushes a merry go-round from rest to a final angular speed of ...
- Physics - In a playground, there is a small merry-go-round of radius 1.20 m and ...
- Physics - A child of mass m is playing on a merry-go-round which has mass 6m and...
- physics - A merry-go-round is stationary. A dog is running on the ground just ...
- physics - A merry go round in the park has a radius of 1.8 m and a rotational ...
- Physics - A merry-go-round is rotating at 4.775 revs/min with a 45 kg child on ...

More Related Questions