What is the Period of a Physical Pendulum with a Horizontal Knife-Edge Pivot?

In summary, the problem involves a thin ring of mass 0.50 kg and radius 0.60 m hanging from a horizontal knife-edge pivot, acting as a physical pendulum. The question asks for the period when the amplitude of the motion is small. The relevant equation is T = 2pi / ω, where ω can be found using the parallel-axis theorem.
  • #1
Joshua A
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Homework Statement



A thin 0.50-kg ring of radius R = 0.60 m hangs vertically from a horizontal knife-edge pivot about which the ring can oscillate freely.

If the amplitude of the motion is kept small, what is the period?

Homework Equations



T = 2pi / ω

Not sure what others...

The Attempt at a Solution



I have no idea what a "horizontal knife-edge pivot" is (I tried to Google it with no success) so I actually don't know where to begin with this problem because I cannot understand what is happening. I'd appreciate an explanation as to what the question means.
 
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  • #2
Joshua A said:

Homework Statement



A thin 0.50-kg ring of radius R = 0.60 m hangs vertically from a horizontal knife-edge pivot about which the ring can oscillate freely.

If the amplitude of the motion is kept small, what is the period?

Homework Equations



T = 2pi / ω

Not sure what others...

The Attempt at a Solution



I have no idea what a "horizontal knife-edge pivot" is (I tried to Google it with no success) so I actually don't know where to begin with this problem because I cannot understand what is happening. I'd appreciate an explanation as to what the question means.

This is the arrangement. The ring hangs from a knife, the edge of the knife is the pivot. It is a "physical pendulum". http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html

upload_2017-11-25_9-15-36.png
 

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  • #3
ehild said:
This is the arrangement. The ring hangs from a knife, the edge of the knife is the pivot. It is a "physical pendulum". http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html

View attachment 215582

Thank you! I figured out the question now.

Hint for anyone who may stumble upon this thread in the future: when using the equation on the HyperPhysics page, note that it says Isupport. Use the parallel-axis theorem to help you find this.
 

1. What is the Period for Oscillating Ring?

The period for oscillating ring is the amount of time it takes for the ring to complete one full back-and-forth motion, also known as one oscillation. This is affected by the mass, length, and tension of the ring.

2. How is the Period for Oscillating Ring calculated?

The period for oscillating ring can be calculated using the equation T=2π√(m/g) where T is the period, m is the mass of the ring, and g is the acceleration due to gravity. This assumes that the ring is in a simple harmonic motion.

3. What factors affect the Period for Oscillating Ring?

The period for oscillating ring is affected by the mass, length, and tension of the ring. It is also affected by any external forces acting on the ring, such as air resistance or friction.

4. How does the Period for Oscillating Ring change with different masses?

According to the equation T=2π√(m/g), the period for oscillating ring is directly proportional to the square root of the mass. This means that as the mass increases, the period will also increase.

5. Can the Period for Oscillating Ring be affected by temperature?

In most cases, temperature does not significantly affect the period for oscillating ring. However, in some materials, changes in temperature can alter the tension and elasticity of the ring, which can indirectly impact the period.

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