• Advertisement

Archived

This topic is now archived and is closed to further replies.

Change in linear and rotational kinetic energy from force

This topic is 5143 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

If you intended to correct an error in the post then please contact us.

Recommended Posts

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.

Share this post


Link to post
Share on other sites
Advertisement
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)

Share this post


Link to post
Share on other sites
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]

Share this post


Link to post
Share on other sites
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.

Share this post


Link to post
Share on other sites
Guest Anonymous Poster
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);

Share this post


Link to post
Share on other sites
AP got it right....
true, you always get linear motion if you apply a force at a point... sorry for the disinformation.

Share this post


Link to post
Share on other sites
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.

Share this post


Link to post
Share on other sites
Guest Anonymous Poster
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.

Share this post


Link to post
Share on other sites
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]

Share this post


Link to post
Share on other sites
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.

Share this post


Link to post
Share on other sites
imagine the rod balanced on a pin. you hit it 90 degrees, so you get maximum torque. the pin is still gonna feel a force. To cancel the force, you need to apply an opposite torque of identical magnitude and direction on the other side. then the force at the centre of the rod (on the pin head) will cancel out.

Or imagine the rod in space, no gravity. if you hit the extremity with a force, it wont rotate on the spot, but move forward as well. it will rotate on its centre if you apply the opposite torque on the other side.

Share this post


Link to post
Share on other sites
oliii, please re-read my posts. I understand what you''re saying and in fact we both agree that if you apply a torque to a free object it will both move and rotate. However, I''m curious why you and AP think that the linear acceleration is independent of where you hit the object. From my observation, if you hit a rod close to the center, it''ll move a lot and rotate a little. If you hit it at one of its ends, it''ll rotate a lot and move just slightly.

Share this post


Link to post
Share on other sites
think of conservation of momentum... let''s say you''re sitting at one end of the rod and you shoot a bullet in a direction perpendicular to the r0 vector. the bullet has got some momentum, but overall momentum has to stay zero, so you and the rod have to move in the opposite direction.

i think your observation results from the fact that we are not used to do apply a certain force but rather to apply a force over a certain distance, with a certain acceleration. so when you hit the object nearer to the center of mass you apply a larger force to move your hand at some acceleration. when you hit it at the tip, the tip will rotate away and you don''t need that much force to keep the same acceleration of your hand.

Share this post


Link to post
Share on other sites
You guys might be right, and Ga''s explanation sounds valid. If this is really the truth, it''ll definitely simplify my life. Right now I''m busy alternating between physics and weather effects, but I''ll certainly give this a try. Thank you guys for your help.

Share this post


Link to post
Share on other sites
from what I've seen here and in other posts, it seems there are two ways to split in force and torque: the one shown here, that uses the remaining force for torque, and others that use the full applied force for both force and torque.

so, which would be the correct ratio of force and torque when applying a force in a point of a body? That, if we want to conserve energy.

ah, well... seems it's being answered in another thread.




[edited by - vicviper on January 21, 2004 10:37:55 AM]

Share this post


Link to post
Share on other sites

  • Advertisement