A solid sphere of radius 15 cm and mass 5 kg starts from rest and
rolls without slipping a distance of 8 m down a roof that is
inclined at 25˚. It leaves the roof a distance of 5 m above the
ground. How far from the edge of the house does the cylinder hit
the level ground?
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answers:
Steve say: Solid sphere: I = k*m*r² where k = 0.4
Vi = √[2*g*h/(1+k)] = 6.88 m/s
Vyi = Vi*sinΘ = 2.908 m/s down
H = 5 = Vi*t + ½g*t² → t = 0.7561 sec to fall 5 m
Vxi = Vi*cosΘ = 6.235 m/s
X = Vxi*t = 4.71 m from edge of house
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Stan say: you Guys are Amazing.
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Robert say: Steve and oldprof are pretty much correct. The initial formula is based on conservation of energy as indicated by Electron 1, but his calculation of time is incorrect. Everyone has failed to notice that the bottom of the sphere (which is what hits the ground) is off the edge of the roof when the sphere loses contact with the roof. Considering that, then the sphere hits the ground 4.778 meters from a point below the edge the roof.
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electron1 say: Since the solid sphere rolls without slipping, it has linear and rotational kinetic energy. As the solid sphere rolls 8 meters down the roof, all of its potential energy will be converted into these two types of kinetic energy.
Linear KE = ½ * m * v^2 = 2.5 * v^2
Rotational KE = ½ * I * ω^2
For a solid sphere, I = 2/5 * m * r^2 = 0.045
ω^2 = v^2 ÷ r^2
Rotational KE = ½ * 0.045 * v^2 ÷ 0.15^2 = v^2
Total KE = 3.5 * v^2
Rotational KE = ½ * 2/5 * m * r^2 * v^2 ÷ r^2 = 0.2 * m * v^2
Linear KE = ½ * m * v^2
Total KE = 0.7 * m * v^2 = 3.5 * v^2
This is exactly the same answer.
PE = m * g * h h = L * sin θ
PE = 5 * 9.8 * 8 * sin 25 = 392 * sin 25
This is approximately 165.7 J.
3.5 * v^2 = 392 * sin 25
v^2 =112 * sin 25