Express h in Terms of r and theta: Rotational Physics

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Discussion Overview

The discussion revolves around expressing the height \( h \) in terms of the radius \( r \) and the angle \( \theta \) for a disc that is rotating due to a block falling. The context includes concepts from rotational physics and energy conservation, as well as geometric relationships.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant presents an energy conservation approach, equating potential energy lost to kinetic energy gained, leading to an expression for \( h \) that includes \( g \) and \( t \).
  • Another participant argues that the problem is purely geometric, suggesting that energy considerations are unnecessary.
  • A subsequent reply proposes a simplified geometric relationship \( h = r \cdot \theta \) as a potential solution.
  • A later response confirms the geometric relationship as valid, referencing it as a standard constraint in rotational motion.

Areas of Agreement / Disagreement

Participants express differing views on whether energy conservation principles are relevant to the problem, with some advocating for a geometric interpretation while others maintain an energy-based approach. The discussion remains unresolved regarding the best method to express \( h \) in terms of \( r \) and \( \theta \).

Contextual Notes

The discussion highlights the dependence on interpretations of the problem, with some participants focusing on energy dynamics while others emphasize geometric relationships. There are unresolved assumptions regarding the applicability of energy conservation in this context.

scifan
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So the question is this:

A disc is released from rest. A block is causing it to rotate. After a time t the block has fallen a height h and the disc has rotated through an angle theta. (in rad)
Express h in terms of r (the radius of the part of the hub around which the string is wound) and theta.

I came up with this answer:

PE lost = KE gained
mgh = (1/2)mv^2 + (1/2)I(omega)^2
mgh = (1/2)m(r*omega)^2 + (1/2)*(mr^2)omega^2
gh = (1/2)r^2*omega^2 + (1/2)omega^2*r^2
gh = omega^2*r^2
omega = d(theta)/dt
subs for omega and solve for h: h = (r^2*theta^2)/(t^2*g)

The question asks me to express h in terms of r and theta but my answer has g and t in it. Any help?

Thanks in advance!
 
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welcome to pf!

hi scifan! welcome to pf! :smile:

(have a theta: θ and try using the X2 icon just above the Reply box :wink:)
scifan said:
A disc is released from rest. A block is causing it to rotate. After a time t the block has fallen a height h and the disc has rotated through an angle theta. (in rad)
Express h in terms of r (the radius of the part of the hub around which the string is wound) and theta.

read the question! :rolleyes:

energy has nothing to do with it …

this is just geometry!​
 


tiny-tim said:
hi scifan! welcome to pf! :smile:

(have a theta: θ and try using the X2 icon just above the Reply box :wink:)


read the question! :rolleyes:

energy has nothing to do with it …

this is just geometry!​


Would that be correct if I say h = r*theta then?
 

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