Moment of Inertia for a Triangle with Masses at the Vertices

[Rongeet Banerjee]

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)

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?

 

[haruspex]

I do not understand your second method. How do you arrive at 3m*(l/2)²? Are you familiar with the parallel axis theorem?

 

[Rongeet Banerjee]

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?

 

[haruspex]

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?

 

[Rongeet Banerjee]

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

 

[Rongeet Banerjee]

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.

 

[etotheipi]

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!

 

[haruspex]

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.

 

[Rongeet Banerjee]

Exactly .That was my question!

 

[Rongeet Banerjee]

`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

 

[etotheipi]

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!