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# Change in linear and rotational kinetic energy from force

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I''m trying to figure out how much a force acting on an object contributes to the object''s linear and rotational motion. The force and its distance from the center of mass is given, as is the mass and moment of inertia of the object. I''m using the Newtonian equations F = ma and t = Ia. If the force is acting on the center of mass, it''s easy. I can get linear acceleration from the first equation and torque is zero, so there''s no change in rotation. But if the force is producing a torque I get confused. I think it''s correct to use the 2nd equation to calculate the angular acceleration (I just multiply the force by its distance from the center of mass and the sinus of the angle between the force and the rotation axis). However, I clearly can''t just use the first equation, F = ma, because the force isn''t acting on the center. The question is, how much does the force contribute to linear acceleration? I''ve read about conservation of energy and stuff like KE(total) = KE(linear) + KE(rotational). Since I can calculate angular acceleration from above, I can calculate the change in angular (or rotational) KE. But how do I calculate the total KE to get linear KE? Perhaps there''s another approach to get linear acceleration? Please share your ideas.

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you have to split the force into a linear accelerating and an angular acclerating part. If r is the vector from the object''s center of mass to the point on which the force acts and r0 = r/|r|, the linear accelerating part is

F_l = (F*r0)*r0

where (F*r0) is a dot product (it''s simply the projection of the force vector onto the radial vector r)

now the rest is the angular accelerating force

F_r = F - (F*r0)*r0

and the torque is

t = r x F_r

where x means a cross product

energies:
rotational energy = I*w²/2 (with the angular velocity w)
linear kinetic energy = m*v²/2
rotational energy increment dE_R = t*dphi (torque times change of angle)
linear kinetic energy increment dE_L = F_l*ds (force times change in position)

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Thanks for the reply. Splitting the force makes sense, but the way you split it doesn't entirely make sense to me. Ok, so r0 is the normalized vector from the center of mass to the point on which the force acts. You're saying that if the force and r0 are at 90 degrees, the force doesn't contribute to linear acceleration at all? That doesn't seem right. Suppose you hit a vertically-hanging pencil horizontally and you hit it at some point other than its center. The pencil will rotate but it will also fly forward and the closer you hit it to center, the more it will fly forward, yet in this case F and r0 are always at 90 degrees no matter how far from the center F acts unless of course it acts exactly on the center.

[edited by - Michael_S on January 12, 2004 5:29:06 PM]

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i think this is because it''s very hard to apply the force only perpendicular to r0. as you touch the pencil you apply a force which will cause it to rotate slightly. then you are still applying the force in the same direction but now it''s not exactly perpendicular to r0 so you get a forward movement. if you''re nearer to the center of mass the the angle of the pencil is bigger for a given distance which the point where you apply the force has moved so it''s harder to apply only forces perpendicular to r0.

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You can calculate linear motion and angular motion independently.

For example, given a RigidBody RB with mass m, inertia I and centermass in cm. Given a force F actuating in r1 of RB surface.

Linear Motion
-------------

a = F / m

Angular Motion
---------------

a = Torque * I ^(-1)
a = ((r1 - cm) x F) * I ^(-1);

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AP got it right....
true, you always get linear motion if you apply a force at a point... sorry for the disinformation.

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I''m sorry but both of you are incorrect. AP, you cannot use the F = ma equation like that because F is assumed to be acting on the center of mass, and in my problem it''s not. Ga, I actually used your idea a long time ago, but quickly realized it''s false. It becomes clear when you have an object like a very thin rod, maybe just 1 mm in thickness. You can see that F and r0 will always be at almost 90 degrees, even when F is close to the center of mass, yet there will definitely be linear acceleration.

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Michael_S trust me. What i''m saying is true.
I can give you an explanation if you want. But that is something you will find in the literature.

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AP is right. it's the way to do it.

a small rigid body demo here

[edited by - oliii on January 14, 2004 3:31:59 PM]

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AP, yes please explain. Also explain why it makes no difference in linear acceleration whether you hit a rod at the center of mass or at the edge. My observation from experiments tells me otherwise.

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