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# M2 Centres of Mass Help Please Watch

1. Hi all,

Having a bit of trouble with a problem from the M2 textbook, it's Miscellaneous Exercise 5, Question 1.

"The structure having the shape shown in the diagram, consisting of a right-angled triangle inscribed in a circle, is made of uniform wire. Calculate the distance of the centre of mass of the structure from AB."

I'll happily draw it if you can on TSR, but it's a circle with a (5 12 13) right angled triangle inside it. AB is the 5cm length, and I'm assuming that the 13cm length AC is a diameter. Any help would be appreciated
2. (Original post by confusedmathser)
Hi all,

Having a bit of trouble with a problem from the M2 textbook, it's Miscellaneous Exercise 5, Question 1.

"The structure having the shape shown in the diagram, consisting of a right-angled triangle inscribed in a circle, is made of uniform wire. Calculate the distance of the centre of mass of the structure from AB."

I'll happily draw it if you can on TSR, but it's a circle with a (5 12 13) right angled triangle inside it. AB is the 5cm length, and I'm assuming that the 13cm length AC is a diameter. Any help would be appreciated
Refering to the diagram.
You need the mass and the distance of the centre of mass from AB for each part.
You can do the triangle in three parts, and the circle as the fourth part.

Have a go, and post with details if you're stuck.

3. (Original post by ghostwalker)
Refering to the diagram.
You need the mass and the distance of the centre of mass from AB for each part.
You can do the triangle in three parts, and the circle as the fourth part.

Have a go, and post with details if you're stuck.

Thanks for responding Part of my problem was not understanding that inscribed had a different meaning in Maths, I had to look it up because that isn't a term I've ever heard before.

Once I got that, I used the triangle and circle as two parts, why would the triangle need to be treated as three seperate parts? Also, for a question like this is the mass just the length of wire used? I think I got 5.1... and the answer was 5.5...., I'll double check that though.
4. (Original post by confusedmathser)
Thanks for responding Part of my problem was not understanding that inscribed had a different meaning in Maths, I had to look it up because that isn't a term I've ever heard before.

Once I got that, I used the triangle and circle as two parts, why would the triangle need to be treated as three seperate parts?
If you can do the triangle as one part that's fine; but it's a wire frame, not a lamina.

Also, for a question like this is the mass just the length of wire used? I think I got 5.1... and the answer was 5.5...., I'll double check that though.
Mass is proportional to length, so you can work with length. You could multply by the mass per unit length, and it will cancel out in the end.

I got 5.58 (3 sig.fig.)
5. (Original post by ghostwalker)
If you can do the triangle as one part that's fine; but it's a wire frame, not a lamina.

Mass is proportional to length, so you can work with length. You could multply by the mass per unit length, and it will cancel out in the end.

I got 5.58 (3 sig.fig.)
Alright cheers, I'll have another go
6. (Original post by ghostwalker)
If you can do the triangle as one part that's fine; but it's a wire frame, not a lamina.

Mass is proportional to length, so you can work with length. You could multply by the mass per unit length, and it will cancel out in the end.

I got 5.58 (3 sig.fig.)
Thanks for the help, I was treating the triange with the same law as a lamina. When I treated each side as an individual wire I got the answer

What is the law for a hollow triangle? Tbh it's probably easier to do each side individually, but I'd be interested to know how it being hollow changes where the CoM lies.
7. (Original post by confusedmathser)
Thanks for the help, I was treating the triange with the same law as a lamina. When I treated each side as an individual wire I got the answer

What is the law for a hollow triangle? Tbh it's probably easier to do each side individually, but I'd be interested to know how it being hollow changes where the CoM lies.
There is no simple rule for a wire frame triangle, that I'm aware of. You could certainly work out something though, given the coordinates of the three vertices, but I doubt if it's worth the effort.

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