Newton's second law for rotations

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SUMMARY

This discussion focuses on Newton's second law for rotations, particularly in the context of calculating the minimum force required to initiate the rotation of a homogeneous bar around the x-axis. Participants clarify that in static equilibrium, any net force will result in acceleration, and torque is essential for understanding rotational motion. The conversation emphasizes the importance of recognizing resisting forces, such as friction, and the role of torque in rotational dynamics, specifically referencing the equations F = ma and T = Iα.

PREREQUISITES
  • Understanding of Newton's laws of motion, particularly F = ma and T = Iα.
  • Basic knowledge of torque and moment of inertia.
  • Familiarity with static equilibrium concepts.
  • Awareness of forces acting on rotating bodies, including friction.
NEXT STEPS
  • Study the implications of static equilibrium in rotational systems.
  • Explore the relationship between torque and angular acceleration in detail.
  • Learn about the effects of friction on rotational motion and how to calculate it.
  • Investigate real-world applications of Newton's laws in rotational dynamics.
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Physics students, mechanical engineers, and anyone interested in the principles of rotational dynamics and the application of Newton's laws in real-world scenarios.

pedrovisk
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TL;DR
What is the minimum force to start rotation in a bar?
EDIT: I forgot about Second Newton's law for rotations and this led to a mistake. Anyway, thanks for the people who answered it and remembered me about law of inertia.

I was thinking about how to "make" things to move without rotate the object, then i tried to calculate the minimum force to start a homogeneous bar to rotate in a x axis. I attached a link.
Could someone check if my answer is correct?

 
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:welcome:

What force is resisting rotation?
 
PeroK said:
:welcome:

What force is resisting rotation?
Hi! The bar is at static equilibrium. What I meant with this force F is what "push" I need to do in order to "win" the moment of inertia.
 
pedrovisk said:
Hi! The bar is at static equilibrium. What I meant with this force F is what "push" I need to do in order to "win" the moment of inertia.
I don't know what "win" means in this context. In the absence of a resisting force, any force (no matter how small) will move and rotate the bar.
 
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It is like asking what force is required to make an object accelerate when there is no resisting force. F = ma tells you that any net force will result in an acceleration, just as ##T = I\alpha## tells you any net torque will result in an angular acceleration.
 
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Orodruin said:
It is like asking what force is required to make an object accelerate when there is no resisting force. F = ma tells you that any net force will result in an acceleration, just as ##T = I\alpha## tells you any net torque will result in an angular acceleration.
Can't believe i forgot about this. Basic Newton's first law for rotations.

I think I was confused when I was trying to rotate a bar like that. I probably just forgot about the friction.

Anyway, now I'm curious. Is there any meaning for the pic I attached?
 
PeroK said:
I don't know what "win" means in this context. In the absence of a resisting force, any force (no matter how small) will move and rotate the bar.
Thanks, you are absolutely right. I was having trouble to understand rotations as a extension of Newton's laws.

Asking the same question I did to Orodruin, what is a possible interpretation for the pic? Did it have a meaning or I should just torn it apart the paper and throw it to the trash? The development of the equations looks so smooth.
 
pedrovisk said:
Thanks, you are absolutely right. I was having trouble to understand rotations as a extension of Newton's laws.

Asking the same question I did to Orodruin, what is a possible interpretation for the pic? Did it have a meaning or I should just torn it apart the paper and throw it to the trash? The development of the equations looks so smooth.
You have a torque of ##F\frac l 2##. Somehow you have an opposing torque of ##\frac{mlg}{4}##.
 
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PS gravity acts through the centre of the bar, so it should not produce a torque on the bar about its centre.
 
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  • #10
pedrovisk said:
I was thinking about how to "make" things to move without rotate the object,
Apply all force through the center of mass, otherwise make sure that the torques cancel.
 

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