A solid ball rolls smoothly from rest (starting at height H = 8.0 m) until it leaves the horizontal section at the end of the track, at height h = 2.0 m. How far horizontally from point A does the ball hit the floor?
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To find where the ball lands, we need to know its speed as it leaves the track (using conservation of energy).
Ki + Ui = Kf + Uf
where
Ki = initial kinetic energy = 0
Kf = final kinetic energy = K(translation) + K(rotation)
Ui = initial potential energy = mgH
Uf = final potential energy = mgh
Ki + Ui = Kf + Uf
0 + mgH = [K(translation) + K(rotation)] + mgH
0 + mgH = [(1/2 mv^2) + (1/2 I ω^2)] + mgH
0 + mgH = [(1/2 mv^2) + (1/2)(2/5 mR^2)(v^2 / R^2)] + mgH
0 + mgH = [(1/2 mv^2) + (1/5 mv^2)] + mgH
0 + mgH = 7/10 mv^2 + mgH
gH = 7/10 v^2 + gH
v^2 = gH - gh / 0 7
v = √g(H - h) / 0.7
v = √(9.8)(8 - 2) / 0.7
v = 9.17 m / s
To find the horizontal distance, we first need to find the value of "t" and then solve for x, using the equations for one dimensional motion:
h = 1/2 g t^2
2.0 = 1/2 (9.8) t^2
t^2 = 0.408
t = 0.64 s
Finally,
x = v t
x = (9.17)(0.64)
x = 5.86 m
So, the the ball hits the floor 5.86 m horizontally from point A.
Hope this helps...good luck!
Ki + Ui = Kf + Uf
where
Ki = initial kinetic energy = 0
Kf = final kinetic energy = K(translation) + K(rotation)
Ui = initial potential energy = mgH
Uf = final potential energy = mgh
Ki + Ui = Kf + Uf
0 + mgH = [K(translation) + K(rotation)] + mgH
0 + mgH = [(1/2 mv^2) + (1/2 I ω^2)] + mgH
0 + mgH = [(1/2 mv^2) + (1/2)(2/5 mR^2)(v^2 / R^2)] + mgH
0 + mgH = [(1/2 mv^2) + (1/5 mv^2)] + mgH
0 + mgH = 7/10 mv^2 + mgH
gH = 7/10 v^2 + gH
v^2 = gH - gh / 0 7
v = √g(H - h) / 0.7
v = √(9.8)(8 - 2) / 0.7
v = 9.17 m / s
To find the horizontal distance, we first need to find the value of "t" and then solve for x, using the equations for one dimensional motion:
h = 1/2 g t^2
2.0 = 1/2 (9.8) t^2
t^2 = 0.408
t = 0.64 s
Finally,
x = v t
x = (9.17)(0.64)
x = 5.86 m
So, the the ball hits the floor 5.86 m horizontally from point A.
Hope this helps...good luck!