Angular momentum along a sloping line

AI Thread Summary
The discussion revolves around calculating the angular momentum of a 1.0 kg particle moving at 3.5 m/s along the line defined by the equation y=0.62x + 1.4. Participants clarify the correct approach to substituting values for x and y, emphasizing the importance of evaluating angular momentum at a specific point along the line rather than at x=0. The conversation highlights the conservation of angular momentum along the particle's trajectory and suggests using the point where the line crosses the y-axis for simplification. Additionally, a formula for the perpendicular distance from a point to a line is mentioned as a useful tool for future calculations. The discussion concludes with an appreciation for the collaborative problem-solving process.
rpthomps
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Homework Statement



A 1.0 kg particle is moving at a constant 3.5 m/s along the line y=0.62x +1.4, where x and y are in meters and where the motion is toward the positive x and y directions. Find its angular momentum about the origin2. Attempt at a solution##L=Iw\\\\L=myv\\\\L=(1)(0.62x+1.4)(3.5)##Not sure what to do with x though. If I set x=0, this just evaluates the momentum at a point not over the line. The line is infinite, so I would have thought the momentum evaluates to infinity as well but the answer is 4.2
 
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Firstly, how did you get this formula? By the definition?
Rethink your substitution for y. What is ##y## by definition?
 
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Suraj M said:
Firstly, how did you get this formula? By the definition?
Rethink your substitution for y. What is ##y## by definition?

You're right. There is a problem with my relationship.

Physics_Question.jpg


The trig part doesn't seem to simplify to nicely though...
 
rpthomps said:
You're right. There is a problem with my relationship.

Physics_Question.jpg


The trig part doesn't seem to simplify to nicely though...
What is this point P you have chosen? Just consider the point where the trajectory crosses the y axis.
 
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Can I use that position because angular momentum will be conserved for the whole trip and thus will be the same along the path of the mass and the position you suggested is the simplest to calculate?
 
rpthomps said:
Can I use that position because angular momentum will be conserved for the whole trip and thus will be the same along the path of the mass and the position you suggested is the simplest to calculate?
Yes.
 
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Then thank you sir for your help! Really appreciated.
 
OP, since you've got the answer, it might help you in the future to know the formula for the perpendicular distance of a point from a line, which would simplify the calculation as there would be no angle involved in the calculation.
Do you happen to have a formula like that? if you did you'll get your d and hence answer would just be mvd.
 
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Thank you.
 

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