Forces Acting on a Rod in Empty Space: Motion & Kinetic Energy

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A metal rod in empty space experiences both translational and rotational motion when a perpendicular force is applied at a distance from its center of mass. The rod will rotate about the point where the force acts while also translating in the direction of the force. The work done on the rod will be divided between translational and rotational kinetic energy, with the exact distribution depending on the distance from the center of mass and the duration of force application. The angular acceleration is determined by the force's effect at that distance, leading to a specific relationship between linear and angular motion. Understanding these dynamics is crucial for analyzing the rod's behavior over the time interval Δt.
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Suppose I have a metal rod in empty space with a uniform mass distribution. Suppose a force acts perpendicular to the rod at some distance R from the rod's centre of mass over some time Δt. There are no other forces acting on the rod.

(1) What type of motion will the rod have over Δt? Will the rod rotate about the point where the force acts, or will it have some other type of motion?

(2) How much of the work done on the rod will be converted to translational kinetic energy and how much will be converted to rotational kinetic energy?

Note that the force does not change direction even as the rod rotates - ie it only acts perpendicular to the rod at the start of its motion.

Thank You.

BTW - This question is theoretical in nature - ie it doesn't require a calculation so I don't think it counts as a "homework" type question. I've been thinking about this for some time now. Nevertheless, I apologise if I've placed it in the wrong section.
 
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HelloWorld2 said:
Suppose I have a metal rod in empty space with a uniform mass distribution. Suppose a force acts perpendicular to the rod at some distance R from the rod's centre of mass over some time Δt. There are no other forces acting on the rod.

(1) What type of motion will the rod have over Δt? Will the rod rotate about the point where the force acts, or will it have some other type of motion?

(2) How much of the work done on the rod will be converted to translational kinetic energy and how much will be converted to rotational kinetic energy?

Note that the force does not change direction even as the rod rotates - ie it only acts perpendicular to the rod at the start of its motion.

Thank You.

BTW - This question is theoretical in nature - ie it doesn't require a calculation so I don't think it counts as a "homework" type question. I've been thinking about this for some time now. Nevertheless, I apologise if I've placed it in the wrong section.

Welcome to the PF.

Are you familiar with Free Body Diagrams (FBDs)? Are you familiar with moments and the Moment of Inertia (MOI) of a rod?
 
We've just covered basic mechanics at uni (I'm doing first year physics). We've finished all topics in both the translational and rotational aspects of mechanics, including force diagrams and rotational inertia.

I know that the angular acceleration of the rod will be the quotient of the acceleration and the distance the force was exerted from the centre of mass. From here we can find the angular velocity after Δt, hence the rod's rotational kinetic energy. We can then subtract the rotational kinetic energy from the total work done to the system to find the translational kinetic energy.

After I've removed the force the rod will move in a straight line and rotate about its centre of mass, but what will be its motion over Δt? And why does the rod gain rotational acceleration equal to α / R? I can calculate how much rotational energy the system gains only once I know that this statement is true (I want to know why its true).

Thank you
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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