A merry-go-round is stationary. A dog is running on the ground just outside its

Question

A merry-go-round is stationary. A dog is running on the ground just outside its circumference, moving with a constant angular speed of 0.735 rad/s. The dog does not change his pace when he sees what he has been looking for: a bone resting on the 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 0.0135 rad/s2.

(a) At what time will the dog reach the bone?

_____________s

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

_____________s

Explanation / Answer

The merry go round angular displacement is m =(2pi/3)+(1/2)(0.0135 rad/s2.)t2

Similalry the dog's angular displacement is d =(0.735rad/s)t

Now the time taken for the dog to reach the bone is

m =d    then (2pi/3)+(1/2)(0.0135 rad/s2.)t2 = (0.735rad/s)t

                        6.75*10-3t2 -0.735t+2.093 =0

Solving the quadratic equation we get

then the time taken is t =2.926s

b)

Now the merry go round angular displacement is m = m- 2pi

                                                     (2pi/3)+(1/2)(0.0135 rad/s2.)t2 = (0.735rad/s)t-2pi

6.75*10-3t2 -0.735t+8.373 =0

Then t =12.926s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.