A merry-go-round is stationary. A dog is running on the ground justoutside its c
ID: 1722201 • Letter: A
Question
A merry-go-round is stationary. A dog is running on the ground justoutside its circumference, moving with a constant angular speed of0.705 rad/s. The dog does not changehis pace when he sees what he has been looking for: a bone restingon the 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 0.0160 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
write position equation for each. Make dog start at positionzero, bone starts at 2/3 (one third ofa full rev) . So... each is finalposition = initial pos + initial vel *time + (1/2) acc * t2 . Dog: final position = 0 + 0.705 t . Bone: final position = 2/3 + 0 + (1/2) * 0.0155 * t2 . Set the final positions equal... and solve for time. . 0.705t = 2/3 + 0.00775 t2 . Quadratic equation... divide everything by 0.00775 . 0 = t2 - 90.96774t + 270.2452 . Solve using quadraticformula... t = 3.07seconds and 87.9seconds (answers for part a and partb) . Quadratic equation... divide everything by 0.00775 . 0 = t2 - 90.96774t + 270.2452 . Solve using quadraticformula... t = 3.07seconds and 87.9seconds (answers for part a and partb)Related Questions
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