Acceleration of a 2D circle due to Gravity.


by clm222
Tags: circle geometry, gravity acceleration, inclined plane
clm222
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Jul7-12, 08:21 PM
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Hello
Im wondering how to calculate the acceleration of a circle down an inclined plane (due to gravity). I am familiar with caclulating the acceleration of a body sliding down a inclined plane, but not a circle. How do you determine the acceleration of a circle (preffer rotation per second, if possible).
I also would like to know how to calculate the fall of a top-heavy object that has enough mass to fall, yes has a base on the ground (ie-a wooden plank with one end in the air, the other on the ground)
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Jul7-12, 09:13 PM
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Quote Quote by clm222 View Post
I also would like to know how to calculate the fall of a top-heavy object that has enough mass to fall, yes has a base on the ground (ie-a wooden plank with one end in the air, the other on the ground)
The gravitational force (weight) on the plank can be considered to act at its centre of mass. If the plank is not completely vertical, then the centre of mass is not lined up with the contact point between the floor and the plank (at the base). This results in the gravitational force producing a torque around this contact point. If you know the torque and the moment of inertia of the plank, you can then compute the angular acceleration and hence the rotation angle of the plank vs. time.
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Jul8-12, 02:16 AM
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Quote Quote by clm222 View Post
Hello
Im wondering how to calculate the acceleration of a circle down an inclined plane (due to gravity). I am familiar with caculating the acceleration of a body sliding down a inclined plane, but not a circle. How do you determine the acceleration of a circle (prefer rotation per second, if possible).
I assume you mean either a uniform thin ring or a uniform disc, radius r, rolling down a plane.
Such an object acquires both linear velocity, v, and angular velocity, ω. Because it is rolling, these are related by v = rω. Likewise the accelerations, [itex]\dot{v} = r\dot{ω}[/itex].
There will be frictional force F acting up the plane. If the angle of the plane to horizontal is θ:
[itex]m\dot{v} = mgsin(θ) - F[/itex] (resolving parallel to plane)
[itex]M\dot{ω} = Fr[/itex] (moments about centre of object)
where M is the moment of inertia of the object. For a thin ring that's mr2; for a disc it's half that.
You can solve between the three equations.
I also would like to know how to calculate the fall of a top-heavy object that has enough mass to fall, yes has a base on the ground (ie-a wooden plank with one end in the air, the other on the ground)
That's more complex than it sounds. As the plank falls sideways it acquires a horizontal velocity. At some point before impact the point of contact is going to move. Precisely when depends on the coefficient of friction.
It may even become airborne at some point.

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Jul8-12, 01:11 PM
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Acceleration of a 2D circle due to Gravity.


Thanks, I'll have to learn 3D vectors/inertia/torque, I was planning to do so anyways, thank you
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Jul8-12, 09:42 PM
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Will the weight be acting on the top of the plank, or the center? Which do you usually use in physics?
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Jul8-12, 09:57 PM
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Quote Quote by clm222 View Post
Will the weight be acting on the top of the plank, or the center? Which do you usually use in physics?
Quote Quote by cepheid View Post
The gravitational force (weight) on the plank can be considered to act at its centre of mass.
It's not actually a choice or a convention. The weight really does act here. If you try to support a horizontal bar at its centre of mass, it will be balanced. If you try to support it at one of its ends, it will pivot around that point, because the weight (which acts in the centre) produces a torque around the pivot point.
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Jul8-12, 10:21 PM
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Quote Quote by clm222 View Post
Will the weight be acting on the top of the plank, or the center? Which do you usually use in physics?
Assuming the plank is uniform, you can treat it as acting at the centre. Of course, it really acts uniformly right through the plank, but unless you want to take into account bending you don't need to worry about that.


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