Mechanics prob: energy conservation with a swing pls help... ?
My answer for the lowest point works.
But I'm having trouble doing the highest point one..?
Explanation with answer pls :/
- husoskiLv 71 year agoFavorite Answer
Is this a trick question? The ball *starts* at the highest point in its swing, and the speed is zero there!
The potential energy at a distance h below the rest position, relative to that starting position, is -mgh, right? So conservation of energy and the work-energy theorem say that all of that P.E. lost goes to kinetic energy:
mgh = (1/2)mv²
v = √(2gh)
At the bottom of both circles, h = L = 1.080 m, so the speed is:
v = √(2 * 9.81 * 1.080) ~~ 4.60 m/s
At the top of the small circle, h = d - r = d - (L - d) = 2d - L = (2*0.756 + 1.080) m
h = 0.432 m
At that position, the speed is:
v = √(2 * 9.81 * 0.432) ~~ 2.91 m/s
- Born YesterdayLv 71 year ago
108 - 75.6 = 32.4 = r (from image)
108 - 2×32.4 = 43.2 cm (distance below
original starting point)
From energy conservation:
m/2×v^2 = mgh
v = (2gh)^.5 = (19.6×.432)^.5 = 2.91m/s
Check: (19.6×1.08)^.5 = 4.6 m/s
Update if you need more, and please choose
a best answer from among those you get.