Rigid Body - Linear (Center of Mass) motion/Rotational Motion

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1. The problem statement, all variables and given/known data

This is a very simple problem to state, yet I've not found any answers on PF or anywhere else online.
So let's say we have a simple rigid body like a rod. We apply an external force F_ext for a fixed time dt to the rod in two different ways: (1) directly on the center of mass and (2) on the side of the rod. In both cases, we have the same motion of the center of mass, and hence the same kinetic energy associated with the center of mass. However, in (1) the rod does not rotate, whereas in (2) it does as we've applied a torque relative to the CM. So in (2) we have an extra kinetic energy of rotation in addition to the kinetic energy for the CM's motion. How is this possible, given that we applied the same force for the same time in the two instances? Doesn't this violate the fact that we did the same amount of work "input" in the two cases?

2. Relevant equations

F_ext= MA_CM
T =0.5mv²

3. The attempt at a solution

N/A


Thanks for all the help.

ehild

The motion of a rigid body consist of a translation and a rotation about the CM. If the force exerted has nonzero torque about the CM the body will rotate. The work "input" equals to the change of KE, and it is translational KE + rotational KE so no "fact" is violated.

ehild

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Okay, thanks, but suppose we don't invoke the work-kinetic energy theorem, and instead want to calculated the work "input" directly from the integral of F*dr. Then doesn't it seem like we've imparted the same amount of work to the rod since the "dr" in (1) and (2) are the same?

Also, we've imparted the same amount of momentum F*dt to the body. Doesn't this mean that their kinetic energies should be the same, since Kinetic Energy = p²/2m?

ehild

The same work done on that rigid body will change the kinetic energy by equal amounts. But the kinetic energy includes both the translational KE 1/2 mV² (V is the velocity of the centre of mass) and the rotational kinetic energy 1/2 Iω² ( I is the moment of inerta and ω is the angular speed). The dr term in the work is the displacement of the point of attack of the force. It is not the same as the displacement of the CM in the second case.

If a body rolls down a slope the time it reaches the ground depends on the shape. A solid sphere will move faster than a hollow one in spite the equal works done by gravity in both cases.

The force acting for dt time changes the momentum by Fdt. If it has torque τ the angular momentum will change, too, by τdt.

ehild

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Thanks, ehild. This has been revelatory.

ehild

You are welcome :)

ehild

aaaakiller

I had the same question in my mind. So based on what i could understand can i say that the force acted in case 2 will be more 'efficient' compared to force acted in case 1? Because it will create more total kinetic energy.