A merry-go-round is stationary. A dog is running on the ground justoutside its c

Question

A merry-go-round is stationary. A dog is running on the ground justoutside its circumference, moving with a constant angular speed of0.735 rad/s. The dog does not change hispace when he sees what he has been looking for: a bone resting onthe merry-go-round one third of a revolution in front of him. Atthe instant the dog sees the bone (t = 0), themerry-go-round begins to move in the direction the dog is running,with a constant angular acceleration of 0.0150 rad/s2. (a) At what time will the dog reach thebone?
1 s

(b) The confused dog keeps running and passes the bone. How longafter the merry-go-round starts to turn do the dog and the bonedraw even with each other for the second time?
2 s
(a) At what time will the dog reach thebone?
1 s

(b) The confused dog keeps running and passes the bone. How longafter the merry-go-round starts to turn do the dog and the bonedraw even with each other for the second time?
2 s

Explanation / Answer

   from the theory we know that     = d / dt    so the location of the dog is described by    d = (0.750 rad / s) t    the equation for the bone will be    b = (1 / 3) 2 rad +(1 / 2) (0.0135 rad / s2) t2    so we get a solution as    0.75 t = (2 / 3) + 0.006745t2    0 = 0.006745 t2 - 0.75 t + 2.09    we can see that this is a quadratic expressionlike ax2 + b x + c    t = [0.75 ± (0.752- 4 (0.006745) 2.09)] / 0.0135      = ...... s or ....... s    now the dog will also pass if    0.75 t = (2 / 3) - 2 +0.006745 t2    if 0.75 t = (2 / 3) + 2 + 0.006745 t2    that is if the dog or the turn table gain alap on the other    if 0.75 t = (2 / 3) + 2 + 0.006745 t2    that is if the dog or the turn table gain alap on the other

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