Angular Momentum: Rotating Object

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Homework Help Overview

The discussion revolves around understanding angular momentum, specifically focusing on the calculation of torque and the role of the cosine theta term in determining the moment arm in a seesaw scenario.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between the angle theta and the calculation of torque, questioning the use of sine and cosine in the context of moment arms and force components.

Discussion Status

Some participants have offered insights into the use of cosine for determining the moment arm, suggesting multiple interpretations of the relationship between force and distance vectors. There appears to be a productive exchange of ideas, with acknowledgment of different approaches leading to the same conclusion.

Contextual Notes

Participants are navigating the complexities of angular momentum calculations and the specific geometry involved in the problem setup, including the orientation of the seesaw and the angles between vectors.

Speedking96
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Homework Statement



I'm trying to understand an example from my textbook about angular momentum. This is the example given:
upload_2014-12-11_16-18-1.png


For the part in red: I don't understand where the cosine theta term came from. When you're calculating the magnitudes of torques, don't you just use FRsin(theta)? If someone could clear that up for me, it would be great! Thank you.
upload_2014-12-11_16-18-1.png

upload_2014-12-11_16-18-1.png
 
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Speedking96 said:
I don't understand where the cosine theta term came from. When you're calculating the magnitudes of torques, don't you just use FRsin(theta)?
If the theta is the angle between the force vector and the distance vector, yes. But in the diagram, the angle between the vectors is the angle between the vertical and the seesaw. Theta is the angle between the horizontal and the seesaw.
 
Ok. I think I get it; the cosine is used to get the moment arm in this case, correct?
 
Speedking96 said:
Ok. I think I get it; the cosine is used to get the moment arm in this case, correct?
That's one way to look at it. There are at least 3 ways, leading to the same answer:
- distance cos (theta) = moment arm.
- force cos (theta) = component of force perpendicular to distance
- vector product of force and distance = force * distance * sin(90-theta)
 
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Ok, I understand. Thank you very much.
 

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