Period of Small Amplitude Oscillation for Hoop of Radius 50cm

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In summary, the period of oscillation for a circular hoop with a radius of 50 cm, hanging on a narrow horizontal rod and allowed to swing, can be calculated using the equation T=2*pi(I/(MgD))^.5. The value of D is the distance from the center of mass to the pivot point, and the moment of inertia should be found with a factor of M in it. The length of the rod and its rotation should also be taken into consideration.
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Homework Statement


"A circular hoop of radiusm 50 cm is hung on a narrow horizontal rod and allowed to swing in the plane of the hoop. What is the period of its oscillation, assuming that the amplitude is small?

Homework Equations



T= 2*pi(I/(MgD))^.5

The Attempt at a Solution



Ok... so if M is not given for the hoop and the rod, how do I go about figured it out? What is the value of D? And do I factor the moment of inertia for both the rod and hoop, and if so... how do I figure out the length of the rod, if it is not given.

Thanks.
 
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  • #2
Find the moment of inertia- it will have a factor of M in it.
I believe D is the distance from the center of mass to the pivot point.
Is the rod rotating?
 
  • #3
robb_ said:
Find the moment of inertia- it will have a factor of M in it.
I believe D is the distance from the center of mass to the pivot point.
Is the rod rotating?

Yes.. it is swinging from the hoop
 

1. What is the period of small amplitude oscillation for a hoop of radius 50cm?

The period of small amplitude oscillation for a hoop of radius 50cm is approximately 1.45 seconds. This is the time it takes for the hoop to complete one full oscillation, or back-and-forth motion. This calculation is based on the physical properties of the hoop, including its mass, radius, and moment of inertia.

2. How is the period of small amplitude oscillation calculated for a hoop?

The period of small amplitude oscillation for a hoop can be calculated using the formula T=2π√(I/mgR), where T is the period, I is the moment of inertia, m is the mass, g is the acceleration due to gravity, and R is the radius of the hoop. This formula is derived from the principles of simple harmonic motion.

3. Does the period of small amplitude oscillation change with different hoop sizes?

Yes, the period of small amplitude oscillation is affected by the size of the hoop. The larger the hoop's radius, the longer the period will be. This is because a larger hoop has a greater moment of inertia, which slows down the oscillation.

4. How does the period of small amplitude oscillation for a hoop change with different masses?

The period of small amplitude oscillation is directly proportional to the square root of the moment of inertia. This means that as the mass of the hoop increases, the period will also increase. This is because a heavier hoop has a larger moment of inertia, which slows down the oscillation.

5. Can the period of small amplitude oscillation be affected by external factors?

Yes, the period of small amplitude oscillation can be affected by external factors such as air resistance, friction, and the elasticity of the material the hoop is made of. These factors can alter the moment of inertia and thus impact the period of oscillation. However, in ideal conditions with no external influences, the period of small amplitude oscillation will remain constant for a given hoop.

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