- #1
Varaia
- 7
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So I'm sure most everyone has come across the problem where you have a smooth rolling object on an inclined plain that rolls off a roof or something and falls some distance x away from the roof. I can solve that because it usually gives me the distance along the incline and the angle of inclination and we assume conservation of energy and then use free fall equations and blah blah blah. What I can't seem to do is solve the last part of this problem where I'm given:
A solid sphere of of negligable mass and radius so I=2/5M(Rsquared)
Smooth rolling so we assume r(omega)=Vcom
From an initial height of H=6.0 m along a very wavy descent to a final height of h=2.0 m where it falls off the edge and
using the conservation of energy and plugging our stuff in
at the end of the roof the Vcom= 5.29 m/(s) and the average speed=7.23m/s (found first part easy)
I assume that the acceleration is constant (gravity) but that only comes into play on the last part because the normal force cancels it out while it is on the crazy ramp.
so now where do I go from here to find the angle so I can calculate t and then calculate the distance it fell horizontally d to point A that is directly below the edge of the 2.0m roof. The answer is 4.8m but when I can't find my all my values to plug into my quadratic equation it is kinda pointless. I know it is some simple relation that I can't remember. I would like to use the freefall relationships but without my angle it doesn't mean anything. Is there an easier way?
x(not)=0 Vx(not)=Vcos(theta) y=h=2.0m Vy(not)=-Vsin(theta)
y=h-(Vsin(theta))t -1/2g(t(squared))
Distance=Vcos(theta)t
A solid sphere of of negligable mass and radius so I=2/5M(Rsquared)
Smooth rolling so we assume r(omega)=Vcom
From an initial height of H=6.0 m along a very wavy descent to a final height of h=2.0 m where it falls off the edge and
using the conservation of energy and plugging our stuff in
at the end of the roof the Vcom= 5.29 m/(s) and the average speed=7.23m/s (found first part easy)
I assume that the acceleration is constant (gravity) but that only comes into play on the last part because the normal force cancels it out while it is on the crazy ramp.
so now where do I go from here to find the angle so I can calculate t and then calculate the distance it fell horizontally d to point A that is directly below the edge of the 2.0m roof. The answer is 4.8m but when I can't find my all my values to plug into my quadratic equation it is kinda pointless. I know it is some simple relation that I can't remember. I would like to use the freefall relationships but without my angle it doesn't mean anything. Is there an easier way?
x(not)=0 Vx(not)=Vcos(theta) y=h=2.0m Vy(not)=-Vsin(theta)
y=h-(Vsin(theta))t -1/2g(t(squared))
Distance=Vcos(theta)t
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