Part of bike wheel is cut out, find period of oscillations

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The discussion revolves around calculating the period of oscillations for a bike wheel with a segment cut out. The center of mass of the wheel is affected by the cut, with an increase in height expressed as $$R\phi \sin(\alpha/2)$$. Participants express confusion about the interpretation of the problem, particularly regarding the placement of the cut-out segment and its impact on the center of mass. There is a focus on deriving potential energy from the center of mass adjustment, while kinetic energy is suggested to be expressed in terms of the rotation rate $$\omega$$. The conversation emphasizes the need for clarity in understanding the mechanics involved in the oscillation calculations.
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Homework Statement
Part of bike wheel with angle ##\alpha## is cut out. All the mass of wheel is concetrated in wheel rim.
Relevant Equations
Find period of wheel's oscillations, if attach wheel's center to rod.
Assume the part of wheel with angle ##\phi## not is cut out, but is located next to cut out part. Turn this from right to left. Then wheel's center of mass will be higher by $$R\phi \sin(\alpha/2)$$. It allows to express potential energy. But I don't know how I can express kinetic energy in this case. Help me. please
 
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alalalash_kachok said:
Homework Statement: Part of bike wheel with angle ##\alpha## is cut out. All the mass of wheel is concetrated in wheel rim.
Relevant Equations: Find period of wheel's oscillations, if attach wheel's center to rod.

Assume the part of wheel with angle ##\phi## not is cut out, but is located next to cut out part. Turn this from right to left. Then wheel's center of mass will be higher by $$R\phi \sin(\alpha/2)$$.
Let us see your calculation for this center of mass increase. It looks wrong to me.

Also, if we are cutting out a hole at a positive (first or second quadrant) angle ##\phi_0##, does that mean that the new center of mass is going to be higher or lower than the original center?

Possibly I am misunderstanding your interpretation of the problem statement. We have cut out a piece of the wheel but then pasted it back in next to where we cut it out? And then flipped the wheel in a mirror image fashion from right to left? That seems unnecessarily baroque.

alalalash_kachok said:
It allows to express potential energy. But I don't know how I can express kinetic energy in this case. Help me. please
Once we have potential energy as a function of ##\phi##, we can look at kinetic energy in terms of rotation rate ##\omega##.

You should be able to write that formula now.
 
Last edited:
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My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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