Find Mass of Object X for Centre of Inertia to Match Vehicle Tyre's Axis

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In summary: This caused the equation to become negative because you were multiplying two vectors together instead of adding them.
  • #1
Valdes
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Homework Statement


The centre of inertia (G) of a heterogeneous vehicle tyre has a distance D=0.1cm from its axis of rotation (the tyre's). The axis is referred to as Delta. The mass of the tyre is M=10kg and its radius is R=20cm. O is the center of the tyre (belongs to the axis)

Homework Equations


Find the mass of an object (dot) X attached to the tyre so that the centre of inertia G is identical to the axis of rotation.

The Attempt at a Solution


This is what I've done so far, but with no avail:

I've considered G' as the centre of inertia of the whole group (the tyre and the object X), and G1 as the centre of inertia of X and "m" as its mass.

Therefore: Vector OG' = (m*vector OG1 + m*vectorOG) / m + M
I substituted G for O since we want it identical to the axis, which results into:
Vector OG' = (m*vector OG1 + m*vecto OO) / m + 10
Vector OG' = (m*vector OX) / m + 10 ( G1 = X itself)
Thus: OG' = (m*20) / m + 10

But I'm not sure where it's going to lead me. Please help me.
 
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  • #2
Valdes said:
Vector OG' = (m*vector OG1 + m*vectorOG) / m + M
I presume you mean /(m+M), and that one of the m's in the numerator should be M, but I there's still something wrong with the vectors. How is your G' different from G?

It seems to me the question is incomplete. It doesn't specify at what radius X is to be attached. Presumably it is at radius R.
 
  • #3
Yes, I meant (m+M) and M, sorry. The G' I considered is the centre of inertia of the entire group, ergo the tyre and the object itself. X is to be attached to the cover of the tyre, the latter is considered as a circle, therefore X is to be attached at the radius R, yes.

I think the centre of mass equation I provided is correct, but I have no idea what to do next.
 
  • #4
Valdes said:
The G' I considered is the centre of inertia of the entire group
Ok, just realized that the question as posed uses G for two different things. I guess it should read:
Find the mass of an object X attached to the tyre so that the centre of inertia G' is identical to the axis of rotation.
Valdes said:
I substituted G for O since we want it identical to the axis
Shouldn't you put G' = O?
 
  • #5
I'm not sure... I think it's a reference to the tyre's G. Could you help me in this case? (G=O)
 
  • #6
Valdes said:
I'm not sure... I think it's a reference to the tyre's G.
No, I think the second "G" clearly means the mass centre of the system, tyre+X. And I don't like "is identical to the axis of rotation". Surely it should say "lies on the axis of rotation". (Note that the question as posed did not mention vectors. You introduced those. Seems to me the whole set-up can be considered planar.)
So you have
Vector OG' = (m*vector OG1 + M*vectorOG) /(m + M)​
and you want G' to lie on the axis of rotation, so G'=O. What does that give you?
 
  • #7
haruspex said:
No, I think the second "G" clearly means the mass centre of the system, tyre+X. And I don't like "is identical to the axis of rotation". Surely it should say "lies on the axis of rotation". (Note that the question as posed did not mention vectors. You introduced those. Seems to me the whole set-up can be considered planar.)
So you have
Vector OG' = (m*vector OG1 + M*vectorOG) /(m + M)​
and you want G' to lie on the axis of rotation, so G'=O. What does that give you?

This is much more clear... the thing is, the question didn't mention which centre of inertia, the tyre's or the system's. Anyway, this is what I got when I replaced G' by O:

OO = (m*OG1 + M*OG) / (m+M)
0 = (m*OX + 10*OG) / (m+10)
0 = (m*20 + 10*0.1) / (m+10)
m*20 + 1 = 0
20m = -1
m = -0.05kg but since a mass can't be negative I considered m = 0.05kg. Is this correct?
 
  • #8
Valdes said:
This is much more clear... the thing is, the question didn't mention which centre of inertia, the tyre's or the system's. Anyway, this is what I got when I replaced G' by O:

OO = (m*OG1 + M*OG) / (m+M)
0 = (m*OX + 10*OG) / (m+10)
0 = (m*20 + 10*0.1) / (m+10)
m*20 + 1 = 0
20m = -1
m = -0.05kg but since a mass can't be negative I considered m = 0.05kg. Is this correct?
Numerically that's right, but you need to figure out why it went negative.
The false step is here:
0 = (m*OX + 10*OG) / (m+10)
0 = (m*20 + 10*0.1) / (m+10)​
You replaced two vectors by ... what?
 
  • #9
They're not vectors in that step, because if a vector A equals a vector B, then their distances are equal. That's what I did there, if the null vector equals an operation including vectors, their distances are equal. Am I right?
 
  • #10
Valdes said:
They're not vectors in that step, because if a vector A equals a vector B, then their distances are equal. That's what I did there, if the null vector equals an operation including vectors, their distances are equal. Am I right?
No, I'm referring to the way you replaced vectors OX and OG by 20 and 0.1.
 
  • #11
OX and OG are distances.

If: vector 0 = (m*vector OX + M*vector OG) / (m+M)

Then: 0 = (m*OX + M*OG) / (m+M)

This is mathematically correct.
 
  • #12
Valdes said:
OX and OG are distances.

If: vector 0 = (m*vector OX + M*vector OG) / (m+M)

Then: 0 = (m*OX + M*OG) / (m+M)

This is mathematically correct.
OK, I was interpreting OX and OG as vectors. If you are using them to stand for distances then the problem is in going from 'vector OX' to OX etc. It is not mathematically correct. What operation do you perform on a vector to get a corresponding distance?
It might become clearer if you multiply out the first equation above to get the relationship between the two vectors.
 
  • #13
All right, this is what I get when I leave them as vectors:

m*vector OX + M*vector OG = vector 0
m*vector OX = -M*vector OG
m*vector OX = M*vector GO
m*OX = M*GO
m*20 = 1
m = 0.05kg

Is this what you mean?
 
  • #14
Not really. This step is not valid:
Valdes said:
m*vector OX = -M*vector OG
m*vector OX = M*vector GO
On what basis do you simply throw away the minus sign?
You want to get rid of the vectors. The correct procedure is to take the modulus of each side:
m*vector OX = -M*vector OG
|m*vector OX| = |-M*vector GO|
|m| * |vector OX| = |M|*|vector GO|
|m| |OX| = |M| |GO|
Since masses are positive:
m |OX| = M |GO|​
 

1. How do I calculate the mass of an object for the centre of inertia to match the vehicle tyre's axis?

To calculate the mass of an object for the centre of inertia to match the vehicle tyre's axis, you will need to use the formula: M = I/ R^2. M is the mass of the object, I is the moment of inertia, and R is the distance between the object's centre of mass and the tyre's axis.

2. What is the centre of inertia and why is it important for vehicle tyres?

The centre of inertia is the point at which the mass of an object is evenly distributed. In the case of vehicle tyres, it refers to the point at which the weight is evenly distributed, which is crucial for maintaining balance and stability while driving.

3. How do I find the moment of inertia for an object?

The moment of inertia for an object can be found by using the formula: I = MR^2. M is the mass of the object and R is the distance from the object's centre of mass to the axis of rotation.

4. What factors can affect the centre of inertia for an object?

The centre of inertia for an object can be affected by its shape, mass distribution, and orientation. Objects with irregular shapes or uneven mass distribution will have a different centre of inertia compared to objects with symmetrical shapes and even mass distribution.

5. How can I adjust the mass of an object to match the vehicle tyre's axis?

You can adjust the mass of an object by adding or removing weight from different areas of the object. By manipulating the distribution of mass, you can change the centre of inertia and align it with the vehicle tyre's axis.

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