# Rotational Motion Find g - Inclined Plane

• Wellesley
In summary: Thanks!In summary, Galileo measured the acceleration of gravity by rolling a sphere down an inclined plane. He found that the velocity at the bottom of the incline was much less than expected, and he was able to deduce the value of g using the conservation of energy.
Wellesley
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:
$$\tau$$weight=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]

Last edited:
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?!

Thanks for the help!
And Happy Holidays!

Last edited:
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:
$$\tau$$weight=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!

Last edited:
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.

## 1. What is rotational motion and how is it different from linear motion?

Rotational motion refers to the movement of an object around an axis or pivot point. It is different from linear motion, which involves movement in a straight line. In rotational motion, the position and orientation of the object changes as it rotates, whereas in linear motion, the object moves in a single direction without changing its orientation.

## 2. What is the role of gravity in rotational motion on an inclined plane?

Gravity plays a crucial role in rotational motion on an inclined plane. It is the force that pulls the object down the incline, causing it to accelerate. This acceleration, along with the angle of the incline, determines the speed and direction of the object's rotational motion.

## 3. How is the acceleration due to gravity (g) related to rotational motion on an inclined plane?

The acceleration due to gravity (g) is related to rotational motion on an inclined plane through the formula g = a/r, where a is the linear acceleration of the object down the incline and r is the radius of the circular motion. This means that the steeper the incline, the greater the acceleration due to gravity and the faster the object will rotate.

## 4. What is the importance of the angle of inclination in rotational motion on an inclined plane?

The angle of inclination plays a significant role in rotational motion on an inclined plane. It affects the speed and direction of the object's rotation, as well as the amount of force required to keep the object in motion. A steeper incline will result in a faster rotation and a larger force needed to overcome friction and maintain the object's motion.

## 5. How can the mass of an object affect its rotational motion on an inclined plane?

The mass of an object can affect its rotational motion on an inclined plane in two ways. First, a heavier object will have a greater gravitational force pulling it down the incline, resulting in a faster rotation. Second, a heavier object will require more force to overcome friction and maintain its motion, making it more difficult to rotate at a constant speed.

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