Solving for the tangential force for a bead on a wire

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Mason Smith
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Homework Statement


Textbook problem 4.32 on a bead threaded on a curved rigid wire, with parts (a), (b) and (c) asking for proofs about speed, tangential force and potential energy.

Figure 4.13: a bead on a stiff wire bent into a double loop-the-loop, with origin O at the left end and arc length s measured along the track from O to the bead.


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.
Equation F_tang = m a_t = m s double-dot, relating tangential force to mass times tangential acceleration, with s as position along the wire.

However, this approach seems a little too simple. Any insight is much appreciated.
 
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  • Textbook problem 4.32 on a bead threaded on a curved rigid wire, with parts (a), (b) and (c) asking for proofs about speed, tangential force and potential energy.
    upload_2019-2-19_13-49-28.png
    26.5 KB · Views: 764
  • Figure 4.13: a bead on a stiff wire bent into a double loop-the-loop, with origin O at the left end and arc length s measured along the track from O to the bead.
    upload_2019-2-19_13-50-17.png
    9.9 KB · Views: 796
  • Equation F_tang = m a_t = m s double-dot, relating tangential force to mass times tangential acceleration, with s as position along the wire.
    upload_2019-2-19_13-53-48.png
    443 bytes · Views: 711
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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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