Rotational Motion Find g - Inclined Plane

AI Thread Summary
The discussion focuses on calculating the acceleration due to gravity (g) using a sphere rolling down a 20-degree inclined plane, with the sphere taking 1.6 seconds to travel 3.00 meters. Initial calculations yielded an incorrect value for g, prompting a review of the rotational inertia and torque concepts. The correct approach involves using the parallel-axis theorem to find the moment of inertia and applying both torque and energy conservation methods to derive the acceleration. The importance of friction for rolling motion is also highlighted, along with a side inquiry into the historical development of the concept of work. The thread concludes with a successful resolution of the initial problem and a request for historical context on the formulation of work.
Wellesley
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Rotational Motion Find g - Galileo Inclined Plane

Homework Statement


Galileo measured the acceleration of gravity by rolling a sphere down an inclined plane. Suppose that, starting from rest, a sphere takes 1.6s to roll a distance distance of 3.00 m down a 20 degree inclined plane. What value of g can you deduce from this?

Homework Equations


PE=KE(trans.)+KE (rot.)
I=2/5Mr^2
torque=force*distance
Torque=I*angular acceleration

The Attempt at a Solution


-I've tried to use torque to solve for the acceleration down the plane, and this yielded a=5/7 * g *sin (theta)
I used:
distance=1/2at^2 to solve for a.
a=2.34375m/s^2

When this is plugged back in, I get:
2.34375=5/7 * g *sin (theta)
(2.34375*7)/5=3.28125
3.28125/sin(20)=9.59373
g= 9.59373 m/s2

[STRIKE]This is not close to the answer of 9.6, or the accepted value (9.81). I know I'm doing something significantly wrong, but I can't figure out exactly what the problem is. If anyone could point me in the right direction, I'd really appreciate it. Thanks.[/STRIKE]

The original problem had to do with an incorrect interpretation of the parallel-axis theorem. It should have been I=ICM+md2.
 
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Wellesley said:
-I've tried to use torque to solve for the acceleration down the plane, and this yielded a=5/4 * g *sin (theta)
Show how you arrived at this result. (Note that the acceleration of something sliding down a frictionless plane would be only a = g*sinθ.)
 
Doc Al said:
Show how you arrived at this result. (Note that the acceleration of something sliding down a frictionless plane would be only a = g*sinθ.)
In order for the sphere to roll (as stated in the problem), the plane has to have friction, right?
Anyway, I modeled the derivation from an example in my book:
\tauweight=mgrsin(\theta)

I=mr2+2/5mr2 --> I=7/5mr2

7/5mr2*\alpha=mgrsin(\theta)

\alpha=5/7*1/r*g*sin(\theta)

a=\alpha*r

a=(5/7*1/r*g*sin(\theta))*r

a= 5/7 * g * sin(\theta)
[STRIKE]
Am I on the right track with this approach? I guess I'm confused whether to use torque (like the calculations above), or the conservation of energy.[/STRIKE]
 
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Wellesley said:
I=ICM+2/5mr2 --> I=4/5mr2

That's supposed to be I=ICM+md2, from the parallel-axis theorem.
 
ideasrule said:
That's supposed to be I=ICM+md2, from the parallel-axis theorem.

You're right...I=7/5mr2. When I edited my original post...I got the right answer!

How could I have missed that?! :cry:

Thanks for the help!
And Happy Holidays!
 
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Wellesley said:
In order for the sphere to roll (as stated in the problem), the plane has to have friction, right?
True.
Anyway, I modeled the derivation from an example in my book:
\tauweight=mgrsin(\theta)
OK, you're finding torque with respect to the contact point of the sphere on the plane.
I=ICM+2/5mr2 --> I=4/5mr2
As ideasrule stated, you need the rotational inertia about that contact point, which is found via the parallel axis theorem. Once you have the correct I, your approach will work fine.
Am I on the right track with this approach? I guess I'm confused whether to use torque (like the calculations above), or the conservation of energy.
Either method will work fine. (Use both, then compare!)

Edit: Looks like you figured it out while I was typing this in.
 
Doc Al said:
Edit: Looks like you figured it out while I was typing this in.

Thanks for the help Doc Al!
 
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As an aside I would humbly ask the following of the board:

If we did not know this problem involved rotation, we would have found that the velocity at the bottom of the incline to be much less than expected. If we assumed that the acceleration, whatever it might be, was uniform, could we not find a number for the final velocity at the bottom of the incline? And then using the conservation of energy, we would find a much smaller number for g than 9.6, or whatever, using the posters method.

Which brings me to one more question. I have not found how the idea of work was formulated. I know Joule and others were working with steam engines and such, and were thinking along these lines, but who or what people actually came up with the idea that force applied over a distance was a very meaningful concept. I can't find history on this? Did Newton think about this at all? Any guidance would be appreciated.

Sorry to hijack the post. I just would like to read up on the history of the formulation of certain ideas.
 
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