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Edexcel M2 Centre of Mass Question

Hi,

I have a homework question and am not sure how to get to the answer. The question is:

Four uniform rods AB, BC, CD, and DA are each 4 metres in length and have masses of 2kg, 3kg, 1kg and 4kg respectively. If they are joined together to form a square framework ABCD, find the position of its centre of gravity.

The answer that the book wants is 1.8 m from AD and AB

I am not sure how to get to that as that is not in the centre and I thought the centre of mass was at the intersection of the lines of symmetry


Many thanks in advance for your help
Original post by bendent1234
Hi,

I have a homework question and am not sure how to get to the answer. The question is:

Four uniform rods AB, BC, CD, and DA are each 4 metres in length and have masses of 2kg, 3kg, 1kg and 4kg respectively. If they are joined together to form a square framework ABCD, find the position of its centre of gravity.

The answer that the book wants is 1.8 m from AD and AB

I am not sure how to get to that as that is not in the centre and I thought the centre of mass was at the intersection of the lines of symmetry


Many thanks in advance for your help


it would be the intersection of the lines of symmetry if all the rods had the same mass, however since they dont so you cant use this method.

firstly I would draw a diagram with A at (0,0) and make a coordinate system. with the other points being at (0,4) (4,4) and (4,0)

then you have to use mixi=Mxˉ \displaystyle \sum m_i x_i = M \bar{x} where mi \displaystyle m_i is the mass of a rod and xi \displaystyle x_i is the x value of the centre of gravity of the rod, M \displaystyle M is the whole mass of the system and xˉ \displaystyle \bar{x} is the C.O.M of the whole system.

Use this to find xˉ \displaystyle \bar{x} . Then use a similar procedure to find yˉ \displaystyle \bar{y}
Original post by DylanJ42
it would be the intersection of the lines of symmetry if all the rods had the same mass, however since they dont so you cant use this method.

firstly I would draw a diagram with A at (0,0) and make a coordinate system. with the other points being at (0,4) (4,4) and (4,0)

then you have to use mixi=Mxˉ \displaystyle \sum m_i x_i = M \bar{x} where mi \displaystyle m_i is the mass of a rod and xi \displaystyle x_i is the x value of the centre of gravity of the rod, M \displaystyle M is the whole mass of the system and xˉ \displaystyle \bar{x} is the C.O.M of the whole system.

Use this to find xˉ \displaystyle \bar{x} . Then use a similar procedure to find yˉ \displaystyle \bar{y}


When I did this I got that Mxˉ M \bar{x} = 20
The whole mass would be 10 (2+3+1+4)
So xˉ \bar{x} would be 2, which is not right
Original post by bendent1234
When I did this I got that Mxˉ M \bar{x} = 20
The whole mass would be 10 (2+3+1+4)
So xˉ \bar{x} would be 2, which is not right


going on the coordinates A(0,0) B(0,4) C(4,4) and (4,0) like so; http://imgur.com/a/2H7rh

x coord of com of AB is 0, mass is 2
x coord of com of BC is 2, mass is 3
x coord of com of CD is 4, mass is 1
x coord of com of AD is 2, mass is 4

do you agree?
Original post by DylanJ42
going on the coordinates A(0,0) B(0,4) C(4,4) and (4,0) like so; http://imgur.com/a/2H7rh

x coord of com of AB is 0, mass is 2
x coord of com of BC is 2, mass is 3
x coord of com of CD is 4, mass is 1
x coord of com of AD is 2, mass is 4

do you agree?


Oh of course, silly me, that makes sense now. So just do the same for yˉ \bar{y} ?
Reply 5
Original post by bendent1234
Hi,

I have a homework question and am not sure how to get to the answer. The question is:

Four uniform rods AB, BC, CD, and DA are each 4 metres in length and have masses of 2kg, 3kg, 1kg and 4kg respectively. If they are joined together to form a square framework ABCD, find the position of its centre of gravity.

The answer that the book wants is 1.8 m from AD and AB

I am not sure how to get to that as that is not in the centre and I thought the centre of mass was at the intersection of the lines of symmetry


Many thanks in advance for your help

If you imagine one side of the framework had mass 50kg and the other side 5kg then it should be clear that you can't assume that the COM will lie in the centre of the square.

For a lamina with uniform mass, the COM will lie on the lines of symmetry.
Original post by bendent1234
Oh of course, silly me, that makes sense now. So just do the same for yˉ \bar{y} ?


yea thatll be you done :biggrin:

Also a note about the form of your answer, always write "a distance 1.8 from AB" etc unless you are given a coordinate system, if you create a coordinate system to make it easier just remember not to quote your answer as coordinates (in case you get a strict marker, it would be harsh to lose marks but you could)
Original post by DylanJ42
yea thatll be you done :biggrin:

Also a note about the form of your answer, always write "a distance 1.8 from AB" etc unless you are given a coordinate system, if you create a coordinate system to make it easier just remember not to quote your answer as coordinates (in case you get a strict marker, it would be harsh to lose marks but you could)


Actually I got the value of yˉ \bar{y} wrong

I did 10 yˉ \bar{y} = (2x4) + (3x2) + (1x0) + (4x2)

So yˉ \bar{y} came out to be 2.2, rather than the needed answer of 1.8. xˉ \bar{x} is fine
@letmethink
Here's a problem for you!
Original post by bendent1234
Actually I got the value of yˉ \bar{y} wrong

I did 10 yˉ \bar{y} = (2x4) + (3x2) + (1x0) + (4x2)

So yˉ \bar{y} came out to be 2.2, rather than the needed answer of 1.8. xˉ \bar{x} is fine


does your diagram look like this?

gg.png

since 4-2.2= 1.8 I would guess that you are correct its just that your y is defined with C and D being at y= 0 and A and B being at y = 4. So y = 2.2 when referring to your diagram means "the COM is a distance 2.2 from DC" which is exactly the same thing as "the COM is 1.8m from AB".

does that make sense?

Note: this is a perfect example of why using coordinates as your answer is a bad idea, so its good this happened and can be cleared up
Original post by DylanJ42
does your diagram look like this?

gg.png

since 4-2.2= 1.8 I would guess that you are correct its just that your y is defined with C and D being at y= 0 and A and B being at y = 4. So y = 2.2 when referring to your diagram means "the COM is a distance 2.2 from DC" which is exactly the same thing as "the COM is 1.8m from AB".

does that make sense?

Note: this is a perfect example of why using coordinates as your answer is a bad idea, so its good this happened and can be cleared up


Thank you very much for all your help with this question
Original post by bendent1234
Thank you very much for all your help with this question


my pleasure :biggrin:

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