Rotational Motion: Force on Rod, Translation vs Rotation

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SUMMARY

The discussion centers on the dynamics of a rod subjected to an external force, specifically analyzing the effects of translation and rotation. When a force is applied at one end of a rod, it will rotate about its center of mass while also translating. The acceleration of both rotation and translation can be determined by decomposing the external force vector into components: one that passes through the center of mass (CM) contributing to translational acceleration, and another perpendicular to it that generates torque. In the example of a seesaw with specific weights, the forces acting on the fulcrum are calculated based on the net torque and the weights of the objects involved.

PREREQUISITES
  • Understanding of rotational dynamics and torque
  • Familiarity with center of mass concepts
  • Basic knowledge of force decomposition in physics
  • Ability to calculate net forces and torques in static and dynamic systems
NEXT STEPS
  • Study the principles of torque and rotational motion in detail
  • Learn how to calculate center of mass for various shapes and configurations
  • Explore force decomposition techniques in physics problems
  • Investigate the dynamics of levers and seesaws in practical applications
USEFUL FOR

This discussion is beneficial for physics students, educators, and engineers who are interested in understanding the principles of rotational motion and force interactions in rigid bodies.

Moose352
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Imagine a rod in space. If I exert a force at one end, will the rod translate, rotate, or both? How do I determine what it will do?
 
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It will rotate about it's centre of mass, and it's center of mass will move in a translation. Essentially, you are applying a moment and an unbalanced force to the body.
 
But what will be the acceleration of its rotation and translation?
 
Decompose the external force vector into components at the point of application. One componet which passes through the CM of the body the other perpendicular to it. The component through the CM will become a translational acceleration the other component times the distance to the CM will be the torque which cause rotation.
 
So does that imply that when a force is acted upon a lever, like a seesaw, not all of the force acts in the rotation?

Okay, if I had a seesaw of 2 kg, which was 10 meters long, with a child (4kg) at each end, then what would be the force exerted on the fulcrum. Would it not be 2g + 4g + 4g?

What would be the force be if the seesaw became to rotate (that is, the net torque is not 0).
 
Originally posted by Moose352
Okay, if I had a seesaw of 2 kg, which was 10 meters long, with a child (4kg) at each end, then what would be the force exerted on the fulcrum. Would it not be 2g + 4g + 4g?
Yes, it would be.
What would be the force be if the seesaw became to rotate (that is, the net torque is not 0).
Say for example one kid weighs 2kg (small kid). Then the balanced force acting on the fulcrum is 2+2+2 and the rotational force (moment) is 2.
 
So the translation force is equal to: net force - rotational moment = balanced force? In that case, in the earlier experiment, would the translation force be 0 since the force is not balanced?
 

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