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

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alalalash_kachok
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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.
 
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FAQ: Part of bike wheel is cut out, find period of oscillations

What is the physical setup of the problem?

The problem involves a bicycle wheel with a section removed, making it incomplete. The wheel is then suspended in such a way that it can oscillate back and forth like a pendulum. The goal is to determine the period of these oscillations.

How does the missing part of the wheel affect its oscillation?

The missing part of the wheel changes the distribution of mass, affecting the wheel's moment of inertia. This altered moment of inertia influences the wheel's oscillatory motion and, consequently, the period of oscillation.

What is the formula for the period of oscillation of a physical pendulum?

The period of oscillation \(T\) of a physical pendulum is given by \(T = 2\pi \sqrt{\frac{I}{mgd}}\), where \(I\) is the moment of inertia of the pendulum about the pivot point, \(m\) is the mass of the pendulum, \(g\) is the acceleration due to gravity, and \(d\) is the distance from the pivot to the center of mass.

How do you calculate the moment of inertia for the modified wheel?

The moment of inertia for the modified wheel can be calculated by integrating the mass distribution of the remaining part of the wheel. This involves subtracting the moment of inertia of the missing section from the moment of inertia of the complete wheel.

What additional information is needed to solve for the period of oscillation?

To solve for the period of oscillation, you need the mass of the wheel, the radius of the wheel, the size and position of the missing section, the location of the pivot point, and the acceleration due to gravity. With this information, you can determine the center of mass and the moment of inertia, which are necessary for calculating the period.

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