Moment of Inertia for a Triangle with Masses at the Vertices

In summary: Systems with the masses more spread out from an axis will have higher rotational inertia. If you sit on a swivel chair and hold two heavy books with your arms stretched out, it's going to be harder (i.e. require more torque) to produce some value of angular acceleration than if you sit with the two masses right next to you. The total mass of the system might be the same, but the moment of inertia differs in the two scenarios because of the distribution of mass.
  • #1
Rongeet Banerjee
45
6
Homework Statement
Three particles each of mass m gram,are situated at the vertices of an equilateral triangle ABC of side l cm(as shown in the figure).The moment of inertia of the system about a line AX perpendicular to AB and in the plane of ABC in gcm² units will be:
1. (3/4)ml²
2. 2ml²
3. (5/4)ml²
Relevant Equations
Moment of Inertia I=summation
of(MR²)
1587020374828118998531.jpg

If I take the three masses individually and try to calculate the moment of inertia of the system separately then
I=(m*0²)+(m*(l/2)²)+(m*l²)
=ml²/4 +ml²=(5/4)ml²
But If I try to calculate Moment of Inertia of the system using its Centre of mass then
As centre of mass is located at the the centroid,
I=3m*(l/2)²=(3/4)ml²
Acc to my book option 3 is correct.Why is my second method wrong?
 
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  • #2
I do not understand your second method. How do you arrive at 3m*(l/2)²? Are you familiar with the parallel axis theorem?
 
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  • #3
I have simply considered the entire mass of the system to be located at the C.O.M. and then multiplied with the square of distance.
I am familiar with the parallel axis theorem but why do I need to apply that here?
 
  • #4
Rongeet Banerjee said:
I have simply considered the entire mass of the system to be located at the C.O.M. and then multiplied with the square of distance.
I am familiar with the parallel axis theorem but why do I need to apply that here?
what is the MoI of the trio of points about a line through its mass centre and parallel with AX?
 
  • #5
haruspex said:
what is the MoI of the trio of points about a line through its mass centre and parallel with AX?
Is it:
m*(l/2)² +m*(l/2)²=ml²/2
 
  • #6
Rongeet Banerjee said:
Is it:
m*(l/2)² +m*(l/2)²=ml²/2
But I still had to use method 1
i.e. to calculate m.o.i. of all the paticles separately.
 
  • #7
Rongeet Banerjee said:
Is it:
m*(l/2)² +m*(l/2)²=ml²/2

`That's right, that is the MoI of the system about an axis parallel to AX passing through the centre of mass. To get the MoI about AX, as @haruspex said you need to use the parallel axis theorem.

This is just saying that if you know the MoI about an axis passing through the CoM to be ##I_0##, then the MoI about an axis parallel to that one is ##I = I_0 + m_t d^2## where ##d## is the distance between the axes and ##m_t## is the total mass of the system.

Hopefully you will get the same result as before for method 1, otherwise Physics has failed us!
 
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  • #8
Rongeet Banerjee said:
But I still had to use method 1
i.e. to calculate m.o.i. of all the paticles separately.
Sure, because treating the whole mass as concentrated at the mass centre is simply not a valid way to find the MoI.
 
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  • #9
Exactly .That was my question!
 
  • #10
etotheipi said:
`That's right, that is the MoI of the system about an axis parallel to AX passing through the centre of mass. To get the MoI about AX, as @haruspex said you need to use the parallel axis theorem.

This is just saying that if you know the MoI about an axis passing through the CoM to be ##I_0##, then the MoI about an axis parallel to that one is ##I = I_0 + m_t d^2## where ##d## is the distance between the axes and ##m_t## is the total mass of the system.

Hopefully you will get the same result as before for method 1, otherwise Physics has failed us!
Yes. I=ml²/2 +3m*(l/2)²=(5/4)ml²
So that is the correct Method 2
 
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  • #11
Rongeet Banerjee said:
Exactly .That was my question!

Systems with the masses more spread out from an axis will have higher rotational inertia. If you sit on a swivel chair and hold two heavy books with your arms stretched out, it's going to be harder (i.e. require more torque) to produce some value of angular acceleration than if you sit with the two masses right next to you. The total mass of the system might be the same, but the moment of inertia differs in the two scenarios because of the distribution of mass.

Treating a system with the mass concentrated at the centre of mass would sort of defeat the point!
 
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Related to Moment of Inertia for a Triangle with Masses at the Vertices

1. What is moment of inertia for a triangle with masses at the vertices?

The moment of inertia for a triangle with masses at the vertices is a measure of its resistance to rotational motion. It takes into account the mass, shape, and distribution of the masses of the triangle.

2. How is moment of inertia calculated for a triangle with masses at the vertices?

The moment of inertia for a triangle with masses at the vertices can be calculated using the formula I = Σmr², where I is the moment of inertia, Σm is the sum of the masses at the vertices, and r is the perpendicular distance from the mass to the axis of rotation.

3. What is the unit of measurement for moment of inertia?

The unit of measurement for moment of inertia depends on the system of units being used. In the SI system, the unit is kg·m². In the imperial system, the unit is slug·ft².

4. How does the distribution of masses affect the moment of inertia for a triangle?

The distribution of masses affects the moment of inertia for a triangle by changing the distance of the masses from the axis of rotation. The farther the masses are from the axis, the higher the moment of inertia will be.

5. What is the significance of moment of inertia for a triangle with masses at the vertices?

The moment of inertia for a triangle with masses at the vertices is important in understanding the rotational motion of the triangle. It is used in various applications, such as in designing rotating machinery and predicting the behavior of objects in rotational motion.

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