Applying force to a triangle, does distance from mid point make a difference?

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6 comments, last by Dmytry 18 years, 10 months ago
Hi, say you have a triangle and you are pushing at one of its edges, does the distance from the mid point have any effect on the amount of force. eg M being midPoint vector, d1 being distance value F1 being force vector P1 point where force1 is being applied F2 etc d2 etc P2 etc So from that I would say with scenario1, the triangle's linear velocity would be faster, and angular velocity slower, then with scenario 2's, after the forces had been applied. So my question is, what do I do with the distances to calculate the new linear velocity. eg what I have so far is: linearVelocity+= force/mass; //so where does distance come into this? Thanks
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What your are asking about is called "rigid body dynamics". If you search for that, you will get thousands of hits.

Here is a series of articles for game devleopers by Chris Hecker that explain the fundamentals pretty well: Rigid Body Dynamics
John BoltonLocomotive Games (THQ)Current Project: Destroy All Humans (Wii). IN STORES NOW!
If this is the only force being applied then there will only be a linear component of acceleration. It's completely counter-intuitive and you'll have a hard time comprehending it, but it's true.

If there is another force (such as friction) then you will have to combine the forces to produce a torque (search for force couples in Google) in order to work out any changes in angular velocity.
In both cases, linear acceleration of center of mass will be the same.
Your linearVelocity+= force/mass; is correct.

Angular acceleration of triangle will be bigger in second case.
(there will be angular acceleration, BTW.)
Assuming 2D case, angular acceleration is equal to
force x distance_vector / [rotational inertia of triangle]

where x denotes cross product.
In 2 dimensions, it's just
z = a x b = ax*by - ay*bx
(sometimes it's named 2D pseudo-cross-product)
Of course in case force is applied at distance, point of application travels bigger distance, and force does more work.

Quote:
If this is the only force being applied then there will only be a linear component of acceleration. It's completely counter-intuitive and you'll have a hard time comprehending it, but it's true.

it's not "counterintuitive", it's precisely what "intuition" told you, and it is wrong. There will be rotational acceleration even with one force, that could be computed as above. What might be counterintuitive is that there will be both rotational and linear acceleration [linear acceleration of center of mass], and that linear acceleration of center of mass does not depend to position of application of force, as long as force is the same.
Quote:Original post by Dmytry
Quote:
If this is the only force being applied then there will only be a linear component of acceleration. It's completely counter-intuitive and you'll have a hard time comprehending it, but it's true.

it's not "counterintuitive", it's precisely what "intuition" told you, and it is wrong. There will be rotational acceleration even with one force, that could be computed as above.


So you would only require one force to make it rotate?


Quote:Original post by Dmytry
Angular acceleration of triangle will be bigger in second case.
(there will be angular acceleration, BTW.)
Assuming 2D case, angular acceleration is equal to
force x distance_vector / [rotational inertia of triangle]

where x denotes cross product.
In 2 dimensions, it's just
z = a x b = ax*by - ay*bx
(sometimes it's named 2D pseudo-cross-product)
Of course in case force is applied at distance, point of application travels bigger distance, and force does more work.


So doesnt the object's mass do anything to the angular velocity? eg torque/mass?

Quote:Original post by johnnyBravo
Quote:Original post by Dmytry
Quote:
If this is the only force being applied then there will only be a linear component of acceleration. It's completely counter-intuitive and you'll have a hard time comprehending it, but it's true.

it's not "counterintuitive", it's precisely what "intuition" told you, and it is wrong. There will be rotational acceleration even with one force, that could be computed as above.


So you would only require one force to make it rotate?

yes. One force would cause it to accelerate linearly AND have rotational acceleration. With two forces you can make it rotate without accelerating linearly.
On spaceship, if you'll for short time fire thruster so line of thrust is not on center of mass, spaceship will start rotating, and will also accelerate in direction of thrust. Linear acceleration of center of mass is often unwanted, so two thrusters are used, on opposite sides of spacecraft, and firing in opposite directions.
Quote:


Quote:Original post by Dmytry
Angular acceleration of triangle will be bigger in second case.
(there will be angular acceleration, BTW.)
Assuming 2D case, angular acceleration is equal to
force x distance_vector / [rotational inertia of triangle]

where x denotes cross product.
In 2 dimensions, it's just
z = a x b = ax*by - ay*bx
(sometimes it's named 2D pseudo-cross-product)
Of course in case force is applied at distance, point of application travels bigger distance, and force does more work.


So doesnt the object's mass do anything to the angular velocity? eg torque/mass?


It does, just in not so simple way. Moment of inertia* is proportional to mass, but also depends to shape of object.
I would also recommend you to learn more about rigid body dynamics.

* I forgot correct English term for that. by "rotational inertia" I meant moment of inertia. You can google for it.
Quote:Original post by Dmytryyes. One force would cause it to accelerate linearly AND have rotational acceleration. With two forces you can make it rotate without accelerating linearly.


With the diagram I drew, I thought on scenario 1, there would be more linear acceleration, then the angular acceleration, as there is less force going into rotating the object, where as in scenario 2 the force is going to make the triangle spin alot more, so i thought the linear acceleration would be less?

Quote:Original post by Dmytry

I would also recommend you to learn more about rigid body dynamics.


Yes I am reading up on rigid body dynamics, but right now im relearning some other maths so i can understand all the symbols etc.
Quote:Original post by johnnyBravo
Quote:Original post by Dmytryyes. One force would cause it to accelerate linearly AND have rotational acceleration. With two forces you can make it rotate without accelerating linearly.


With the diagram I drew, I thought on scenario 1, there would be more linear acceleration, then the angular acceleration, as there is less force going into rotating the object, where as in scenario 2 the force is going to make the triangle spin alot more, so i thought the linear acceleration would be less?

Not. Common misunderstanding, comes from confusing forces and energies.
In second scenario, force will also do more work (work=force*distance) as distance will be bigger because it'll both rotate and accelerate. (so point of application of force will move faster.)

Also, linear momentum conserves. If it would be like you describe, then we could use a spring between 2 objects with equal mass, at one object connected to center, at other to the side. Spring applies equal forces to both things. Now, if linear acceleration of second thing would be smaller, net linear momentum would change and linear momentum conservation would be violated. In fact, it would be possible to make reactionless space engine, for example [grin]

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