Solving for the tangential force for a bead on a wire

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SUMMARY

The discussion focuses on calculating the tangential force for a bead on a wire, specifically addressing the relationship between mass, tangential acceleration, and the second time derivative of position along the wire. The participant correctly identifies that the tangential force is the product of the bead's mass and its tangential acceleration, represented as the second time derivative of position (denoted as ##\ddot s##). However, the participant expresses concern that their solution appears overly simplistic and seeks further validation or a more formal derivation of the tangential component of acceleration.

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Mason Smith
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Homework Statement


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Homework Equations

The Attempt at a Solution


I am pretty sure that I solved part a correctly. However, I feel as though my solutions for parts b and c are not quite correct because they seem simple. For instance, my solution for part b argues that the tangential force is just the product of the mass of the bead and the tangential acceleration. The tangential acceleration is equal to the second time derivative of the position along the length of the wire. Therefore, the tangential force is equal to the product of the mass of the bead and the second time derivative of the position along the length of the wire.
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However, this approach seems a little too simple. Any insight is much appreciated.
 

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Mason Smith said:

Homework Statement


View attachment 239004
View attachment 239005

Homework Equations

The Attempt at a Solution


I am pretty sure that I solved part a correctly. However, I feel as though my solutions for parts b and c are not quite correct because they seem simple. For instance, my solution for part b argues that the tangential force is just the product of the mass of the bead and the tangential acceleration. The tangential acceleration is equal to the second time derivative of the position along the length of the wire. Therefore, the tangential force is equal to the product of the mass of the bead and the second time derivative of the position along the length of the wire.
View attachment 239007
However, this approach seems a little too simple. Any insight is much appreciated.
It seems simple because you are thinking like a physicist. :oldsmile: Good! (I'm just having some fun.)

Maybe they want a more formal derivation. In particular, can you justify your claim that the tangential component of acceleration is given by ##\ddot s##? Try the suggestion of taking the time derivative of the identity ##v^2 = \vec v \cdot \vec v##.
 
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