Moment of inertia of ring with 3 masses on

In summary, the moment of inertia of a system consisting of three particles with mass m = 0.5 kg fixed on a uniform circular ring of mass M = 2 kg, radius a = 0.6 m and centre O, to form an equilateral triangle is 1.26 kgm2. This is calculated using the Parallel Axis Theorem and adding the contribution from the ring.
  • #1
j3dwards
32
0

Homework Statement


Three particles, each with mass m = 0.5 kg, are fixed on a uniform circular ring of mass M = 2 kg, radius a = 0.6 m and centre O, to form the corners of an equilateral triangle. What is the moment of inertia of this system?

Homework Equations


I = Σma2

The Attempt at a Solution


I = 3(2 x 0.62) = 2.16 kgm2

I'm confused because it is either that or zero and I can never tell when it is zero.

Thank you.
 
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  • #2
j3dwards said:

Homework Statement


Three particles, each with mass m = 0.5 kg, are fixed on a uniform circular ring of mass M = 2 kg, radius a = 0.6 m and centre O, to form the corners of an equilateral triangle. What is the moment of inertia of this system?

Homework Equations


I = Σma2

The Attempt at a Solution


I = 3(2 x 0.6) = 1.728 kgm

I'm confused because it is either that or zero and I can never tell when it is zero.

Thank you.
The moment of inertia is never zero, unless the mass is zero.

Draw a sketch of the ring and the masses. Use the Parallel Axis Theorem to calculate the MOI of this assembly.
 
  • #3
SteamKing said:
The moment of inertia is never zero, unless the mass is zero.

Draw a sketch of the ring and the masses. Use the Parallel Axis Theorem to calculate the MOI of this assembly.

So the centre of mass is at O so is the answer just I = 3(2 x 0.62) = 2.16 kgm2
 
  • #4
j3dwards said:
So the centre of mass is at O so is the answer just I = 3(2 x 0.62) = 2.16 kgm2
Don't forget to include the contribution to the MOI from the ring.
 
  • #5
SteamKing said:
Don't forget to include the contribution to the MOI from the ring.

Oh I've been doing this very wrong. So is it: I = Σma2 + MR2 = 3(0.5 x 0.62) + (2 x 0.62) = 1.26 kgm2
 
  • #6
j3dwards said:
Oh I've been doing this very wrong. So is it: I = Σma2 + MR2 = 3(0.5 x 0.62) + (2 x 0.62) = 1.26 kgm2
That looks much better.
 
  • #7
SteamKing said:
That looks much better.
Thank you so much!
 

Related to Moment of inertia of ring with 3 masses on

1. What is moment of inertia?

Moment of inertia is a measure of an object's resistance to changes in its rotational motion. It is similar to mass in linear motion and depends on the mass distribution and shape of the object.

2. How is moment of inertia calculated for a ring with 3 masses?

The moment of inertia for a ring with 3 masses can be calculated by adding the individual moments of inertia of each mass. The formula is I = mr², where I is the moment of inertia, m is the mass, and r is the distance of the mass from the axis of rotation.

3. What is the axis of rotation for a ring with 3 masses?

The axis of rotation for a ring with 3 masses is a line that passes through the center of the ring and is perpendicular to its plane. This is also known as the rotational axis or the axis of symmetry.

4. Does the placement of the masses affect the moment of inertia of the ring?

Yes, the placement of the masses does affect the moment of inertia of the ring. The farther the masses are from the axis of rotation, the higher the moment of inertia will be. Placing all three masses at equal distances from the axis of rotation will result in a higher moment of inertia compared to placing them at unequal distances.

5. How does the moment of inertia of a ring with 3 masses affect its rotational motion?

The moment of inertia of a ring with 3 masses affects its rotational motion in the same way that mass affects linear motion. Objects with higher moment of inertia will require more force to accelerate and will rotate at a slower speed compared to objects with lower moment of inertia. It also determines the stability and ease of rotation of the ring.

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