# To locate the centre of gravity of a rod

• gnits
In summary: I rewrite this as:0 < a < (2/5)LI can see that this equation says that the distance of COG from A is between 0 and (2/5)L.
gnits
Homework Statement
To locate the centre of gravity of a rod
Relevant Equations
moments
Could I please ask for a help on how to attack this question?

A heavy rod AB of length L can be made to balance across a small smooth peg C when a weight of 2W is suspended from A. Alternatively, it can be made to balance across the peg with a weight of 3W suspended from B. If the distance AC in the first case is the same as the distance BC in the second, show that the distance of the centre of gravity of the rod from A lies between (2/5)L and (1/2)L.

Here is my diagram for the two cases.

(I have called the weight of the rod ω)

In each diagram I have labelled ω as acting at the same point, but not at the centre point of AB, as I take it from this question that the rod is not of uniform density.

Well, if I take moments about C in the first case I get:

2Wd=ω*x1 giving x1 = 2Wd/ω

So the distance of the centre of gravity of the rod from A is:

d + x1 = d + 2Wd/ω.

Well, I can do the same for the second case giving distance of centre of gravity of the rod from A as:

L - d - x2 = L - d - 3Wd/ω

Thanks for any pointer towards a solution,
Mitch.

Delta2
The problem is telling you that the actual center of mass of the rod is located between 40% and 50% of L, when measured from end A.

gnits said:
Homework Statement:: To locate the centre of gravity of a rod
Relevant Equations:: moments

Could I please ask for a help on how to attack this question?

A heavy rod AB of length L can be made to balance across a small smooth peg C when a weight of 2W is suspended from A. Alternatively, it can be made to balance across the peg with a weight of 3W suspended from B. If the distance AC in the first case is the same as the distance BC in the second, show that the distance of the centre of gravity of the rod from A lies between (2/5)L and (1/2)L.

Here is my diagram for the two cases.

View attachment 293869
(I have called the weight of the rod ω)

In each diagram I have labelled ω as acting at the same point, but not at the centre point of AB, as I take it from this question that the rod is not of uniform density.

Well, if I take moments about C in the first case I get:

2Wd=ω*x1 giving x1 = 2Wd/ω

So the distance of the centre of gravity of the rod from A is:

d + x1 = d + 2Wd/ω.

Well, I can do the same for the second case giving distance of centre of gravity of the rod from A as:

L - d - x2 = L - d - 3Wd/ω

Thanks for any pointer towards a solution,
Mitch.
It's a tricky problem. A useful idea is the introduce the dimensionless constant ##\mu = \frac{W}{w}##, which is the ratio of the weight of the rod and the the other unknown weight ##W##.

The position of the CoM of the rod must depend on this ratio.

The variable you must eliminate is ##d##, leaving the distance from A to the CoM in terms of ##L## and ##\mu##.

I also suggest you need a symbol for this distance from A to the CoM. E.g. ##a##.

Delta2
PeroK said:
It's a tricky problem. A useful idea is the introduce the dimensionless constant ##\mu = \frac{W}{w}##, which is the ratio of the weight of the rod and the the other unknown weight ##W##.

The position of the CoM of the rod must depend on this ratio.

The variable you must eliminate is ##d##, leaving the distance from A to the CoM in terms of ##L## and ##\mu##.

I also suggest you need a symbol for this distance from A to the CoM. E.g. ##a##.
Thank you very much for your reply. I would never have thought of that approach. So here is my previous diagram, with 'a' added:

I can combine my previously derived equations with your suggested terminology to give:

(am using u for the symbol 'mu')

a = d(1 + 2u)

and

a = L - d(1 + 3u)

and I eliminate d from these to give:

a = L - a * (1+3u)/(1+2u)

(Call this "Equation 1")

and can arrange this for a like so:

a = L(1 + 2u)/(2 + 5u)

(Call this "Equation 2")

So, tantalizingly, I can see that from "Equation 2" 'a' is not defined when u = -2/5 and from "Equation 1" 'a' is not defined when u = -1/2

I can't yet though see how to relate this to the required answer - required answer being that the distance of COG from A >(2/5)L and <(1/2)L

EDIT:
OK so looking at "Equation 2". I need 0 < a < L so I need:

0 < (1 + 2u)/(2 + 5u) < 1

Solving 0 < (1 + 2u)/(2 + 5u) leads to:

-1/2 > u > -2/5

and solving (1 + 2u)/(2 + 5u) < 1 leads to:

-2/5 > u > -1/3

Remembering that u = W/ω I can't yet see how to get from these limits on W/ω to limits for distance of COG from A.

I'll keep looking,
Thanks for any further help,
Mitch.

Last edited:
PeroK
gnits said:
Thank you very much for your reply. I would never have thought of that approach. So here is my previous diagram, with 'a' added:View attachment 293930
I can combine my previously derived equations with your suggested terminology to give:

(am using u for the symbol 'mu')

a = d(1 + 2u)

and

a = L - d(1 + 3u)

and I eliminate d from these to give:

a = L - a * (1+3u)/(1+2u)

(Call this "Equation 1")

and can arrange this for a like so:

a = L(1 + 2u)/(2 + 5u)

(Call this "Equation 2")
Once you have equation 2 you are nearly finished. ##u = \frac{W}{w}## is the ratio of the weights and can take any positive value. You only need to calculate the maximum and minimum for ##a##.

gnits said:
So, tantalizingly, I can see that from "Equation 2" 'a' is not defined when u = -2/5 and from "Equation 1" 'a' is not defined when u = -1/2
##u## cannot be negative as it's a ratio of weights/masses.

Lnewqban
PeroK said:
Once you have equation 2 you are nearly finished. ##u = \frac{W}{w}## is the ratio of the weights and can take any positive value. You only need to calculate the maximum and minimum for ##a##.##u## cannot be negative as it's a ratio of weights/masses.

I see it!

I would never have gotten it without your help. In particular, setting the equations in terms of W/ω, I don't see how I would ever of thought of that.

Thank you very much indeed.

Mitch.

Lnewqban and PeroK

## 1. How do you determine the centre of gravity of a rod?

The centre of gravity of a rod can be determined by suspending it from different points and finding the point at which it balances perfectly. This point will be the centre of gravity.

## 2. Why is it important to locate the centre of gravity of a rod?

Locating the centre of gravity of a rod is important because it helps in understanding the stability and balance of the rod. It also helps in predicting how the rod will behave when subjected to external forces.

## 3. What factors affect the location of the centre of gravity of a rod?

The location of the centre of gravity of a rod is affected by its shape, size, and distribution of mass. A rod with a larger mass at one end will have its centre of gravity closer to that end.

## 4. How can you find the centre of gravity of an irregularly shaped rod?

To find the centre of gravity of an irregularly shaped rod, you can use a plumb line or a similar tool to determine the vertical line passing through the centre of mass. The intersection of this line with the horizontal line passing through the rod's axis will give you the centre of gravity.

## 5. Can the centre of gravity of a rod be outside of the physical object?

Yes, the centre of gravity of a rod can be outside of the physical object if the rod is not uniform in shape or mass distribution. In such cases, the centre of gravity may lie in the empty space around the rod.

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