Force & Torque in Classic Mechanics: Calculating Accelerations

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In classic mechanics, when a force is applied to a disk at its top, it generates both linear and rotational accelerations. The same force is used in the equations for both types of acceleration, with linear acceleration calculated as force divided by mass, and rotational acceleration determined by the torque, which is the force cross-multiplied by the radius vector and divided by the moment of inertia. The torque's calculation depends on the specific components of the applied force and the position of application. The disk's behavior may resemble that of a pendulum due to its momentum during acceleration. Clarification confirms that the same force indeed causes both linear and rotational accelerations, essential for accurate physics simulations.
tom_backton
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This is a question about classic mechanics.

Let's say there is a disk in mid-air. There are only 2 dimentions, which is why it's a disk and not a ball...anyway, force is applied to the disk, exactly at the disk's top. Vector r goes from the disks center to the pint to which the force is applied, so r=(0,rx) . F=(Fx,Fy) . The torque causing rotaion is the xy plane (which is the torque vector's z component...) is equal to rx*Fy-Fx*ry . The force causes both linear and rotational acceleration. The question is how exactly I calculate them. Do I use the same force for both? If not, how do I calculate the accelerations?
 
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hi tom! :smile:

yes, you use the same force in ∑ F = ma, to get the linear (vector) acceleration a.
 
tom_backton said:
anyway, force is applied at the disk's top. Vector r goes from the disks center to the point where the force is applied, so r=(0,rx) .
If the force is applied at the top then shouldn't r=(0,ry)?

F=(Fx,Fy) . The torque is equal to rx*Fy-Fx*ry.
For the initial state, since rx = 0, then only the -Fx*ry component matters for the torque component.

The force causes both linear and rotational acceleration.
As mentioned above, the linear acceleration corresponds to linear force divided by mass. The rotational reaction doesn't affect the linear acceleration, but it will require that point of application of that force to move faster, to account for the power component used to increase rotational energy. (power = force x speed).

Assuming that the force is applied through the equivalent of a pin at the edge of the disk so the force can "pull" the disk as well as "push" it, and assuming the force remains constant in both Fx and Fy components, then it's path can be determined by the reaction of the disk to linear and angular acceleration. The math seems complicated, and momentum of the disk will cause it to behave somewhat like a pendulum while being accelerated.
 
So I use the same force F for both the linear acceleration (force F divided by mass m) and the rotational acceleration (force F cross-multiplied by r and then divided by the moment of inertia I) ? Does the same force cause the two accelerations? I just need to be sure...it's for a physics computer simulation program...
 
tom_backton said:
So I use the same force F for both the linear acceleration (force F divided by mass m) and the rotational acceleration (force F cross-multiplied by r and then divided by the moment of inertia I) ? Does the same force cause the two accelerations? I just need to be sure...it's for a physics computer simulation program...

Yes … the Fs in those formulas are both called force because they are the same thing …

we'd call them something else if that didn't work! :wink:
 
Thanks for the help :)
 

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