Why Do We Use Ramp Length Instead of Height in Rolling Sphere Calculations?

jcruise322
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Homework Statement


A sphere of radius .06 m and mass .5 kg rolls down a ramp that is angled 30 degrees down the incline. It starts rolling from a height of 7 feet and does not slip

What is its final linear velocity?
Now, I used mgh=translational +rotational KE and found that the final velocity was 9.9, but I didn't want to do that.

I wanted to use the formula acm=(g*sinθ)/(1+β) where β is the coefficient in front of mr^2 (for a sphere, .4 or 2/5).
So, Vf^2=vo^2+2*a*Δx.

My question is, which Δx do I use? I initially used the height and my answer was wrong...I was supposed to use the ramp, which by trig is 14 meters.

Homework Equations


acm=(g*sin(θ))/(1+β)
KE and PE equations.
V^2=Vo^2+2*a*Δx[/B]

The Attempt at a Solution



I already know the answer. I just want to know why the length of the ramp is used for v^2=vo^2+2adelta x equation.

Thanks, this should only take a moment. Appreciate any input guys! :) :)[/B]
 
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You should use the length of the ramp because the force you used to obtain the acceleration is in the downramp direction (g*sinθ), not vertical.
 
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Aha, that is what I suspected. By the way Haruspex, you have commented on all of the posts that I have made on physics forums so far. Appreciate the help!​
 
jcruise322 said:
Aha, that is what I suspected. By the way Haruspex, you have commented on all of the posts that I have made on physics forums so far. Appreciate the help!​
Maybe our timezones match.
 

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