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thegenuwineone asked:


constant angular speed of .750 rad/s. The dog does not change his pace when he sees what he has been lolking for: a bone resting on the edge of teh merry go round one third of a revolution in front of him. At the instant the dog sees the bone (t=0) the merry go round begins to move in the direction the dog is running, with a constant angular acceleration of .015 rad/s^2.
a. At what time will the dog reach the bone?
b. the confused dog keeps running and passes the bone. How long after the merry go round starts to turn will the dog and the bone draw even with each other for the second time?

i know that the dogs initial (theta) is 0 and his final (theta) is 2pi/3. omega is .750 and alpha is 0 due to constant velocity.

also for part b. i know that final theta for bone is 5pi/3 and initial theta is 0. and alpha is .015

i just dont know how to do this problem.

using the equation (theta)initial = (theta)final + omega(t) + 1/2(alpha)t^2

Berniece

2 Responses to “A merry go round is stationary. A dog is running on the ground just outside its circumference, moving with a?”

1. lindsey_osborne Says:
   February 12th, 2010 at 4:25 pm

   Tyra

   You have all the information you need. Plug the numbers into the equation.

2. usdaddy Says:
   February 15th, 2010 at 7:43 am

   Earlene

   You should break this problem up into parts and list it. Give 10 points for each part and you might get some takers.