Mechanics Problem using Polar Coordinates

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SUMMARY

The discussion revolves around solving a mechanics problem using polar coordinates, specifically focusing on the forces acting on a bead on a surface at an angle φ. The user struggles with applying the equations of motion, particularly mv²/r = mg and mr'' = mg, to derive the correct vector equations. Key advice includes identifying the forces acting on the bead, drawing a free body diagram, and expressing these forces in unit vector notation. The user ultimately finds clarity through additional resources, including YouTube videos, which enhance their understanding of inclined plane problems.

PREREQUISITES
  • Understanding of polar coordinates in mechanics
  • Familiarity with Newton's second law of motion
  • Knowledge of free body diagrams
  • Basic vector notation and operations
NEXT STEPS
  • Study the derivation of vector equations in polar coordinates
  • Learn how to construct and analyze free body diagrams
  • Explore advanced mechanics concepts related to inclined planes
  • Watch educational videos on polar coordinate systems and their applications in physics
USEFUL FOR

Students and educators in physics, particularly those tackling mechanics problems involving polar coordinates and inclined planes.

MathDestructor
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Homework Statement
I don't know how to take polar derivatives.
Relevant Equations
r = Rcos(theta) + Rsin(theta)
1580492747754.png


1580493310652.png

This is what I have so far, please need urgent help. I don't understand and know what to do.

For the first part, I got a really long answer, for the second part I am trying in terms of mv^2/r = mg, or mg = m*(answer to first), but I am getting nowhere. PLease help
 
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Please show your work. We don't give answers away but point out mistakes.
 
1580493253059.png
This is all I have, I don't know what to do after, I've been stuck for 2 days, and its due in half an hour.
 
MathDestructor said:
This is all I have, I don't know what to do after, I've been stuck for 2 days, and its due in half an hour.
As long as the bead stays on the surface ##\dot R## = 0. That should simplify things for you.
 
Is my second derivative right?
 
What forces should I use to try and solve the second part.
 
MathDestructor said:
What forces should I use to try and solve the second part.
Please show the revised velocity and acceleration in view of my comments in post #4. You cannot proceed to the second part unless you get the first part right. If you have to post pictures (a practice we discourage) at least make sure they are right side up.
 
1580493893133.png
 
I tried mv^2/r = mg and mr''= mg. None of them are working
 
  • #10
MathDestructor said:
I tried mv^2/r = mg and mr''= mg. None of them are working
You need to write a vector equation. How many different forces act on the bead when it is on the surface at angle φ? Write their vector sum in unit vector notation and set it equal to ##m\vec a## in unit vector notation. Your expressions for the velocity and acceleration are correct.
 
  • #11
kuruman said:
You need to write a vector equation. How many different forces act on the bead when it is on the surface at angle φ? Write their vector sum in unit vector notation and set it equal to ##m\vec a## in unit vector notation. Your expressions for the velocity and acceleration are correct.
I don't know what that means, or how to do that.
 
  • #12
I have 10 mins left, I would really appreciate it if you would help me in a hurry just this once. Please.
 
  • #13
MathDestructor said:
I have 10 mins left, I would really appreciate it if you would help me in a hurry just this once. Please.
Sorry, it is against forum rules. Do your best. Identify the forces that act on the bead, draw a free body diagram and use it to write each force in unit vector notation.
 
  • #14
I know this doesn't give you an answer, but, you can just google for Polar Coordinate Vector System Derivation.
I did that and it all finally made sense.
Some YouTube videos also helped.
It wasn't a part of my curriculum so far, but, learning it helped me understand so many things that, as a result, I have become near-invicible when it comes to inclined plane problems.
 

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