Kinematics Rolling motion off a cliff

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SUMMARY

The discussion centers on solving a kinematics problem involving a solid sphere rolling off a roof. The initial height is 6.0 m, and the final height is 2.0 m, with the sphere achieving a center of mass velocity (Vcom) of 5.29 m/s. The user struggles to find the angle of inclination necessary to calculate the horizontal distance fallen (d) to point A directly below the roof. Ultimately, the solution involves applying basic horizontal motion equations from Chapter 3 of Halliday/Resnick's Fundamentals of Physics, leading to the conclusion that the horizontal distance is 4.8 m.

PREREQUISITES
  • Understanding of kinematics, specifically projectile motion
  • Familiarity with conservation of energy principles
  • Knowledge of rotational dynamics, particularly the moment of inertia for a solid sphere
  • Proficiency in solving quadratic equations
NEXT STEPS
  • Review the equations of motion for projectile motion
  • Study the conservation of energy in rolling motion scenarios
  • Learn about the moment of inertia and its applications in rotational dynamics
  • Explore problem-solving techniques for kinematics in physics textbooks, particularly Halliday/Resnick
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of kinematics and energy conservation in rolling motion scenarios.

Varaia
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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
 
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The angle at which it leaves the roof? If that isn't given somewhere in the geometry of the situation, there isn't any way to calculate it. Exactly what is given?

And what in the world do you mean by "the distance it fell horizontally d to point A that is directly below the edge of the 2.0m roof"? If it feel to a point directly below the roof isn't the horizontal distance 0? Or do you not really mean "directly below"?
 
If some how one can find the distance it traveled along the roof , the angle of inclination can be found since we know the vertical displacement of the center of mass.
 
Here is the problem and I attached a picture (you'll love my mad photoshop skills)
In the picture a solid ball rolls smoothly from rest (starting at height H=6.0m) until it leaves the horizontal section at the end of the track at height h=2.0m. How far horizontally from point A does the ball hit the floor?

If anyone has spent the money to buy the student solutions manual to the 7th ed of Halliday/Resnicks Fundamentals of Physics this problem is in there as #9 in Chp 11

I should know how to do it by tomorrow and then I'll post the process because dang this was a tuffie.
 

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Got it! and boy do I feel dumb

:-p So it was a much easier problem than I was making it out to be, had to go back all the way to Chp 3 for simple horizontal motion equation and then figured it out from there. Thanks for helping.
 
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