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# M2 Uniform Lamina watch

1. Struggling with a question here. Any help would be much appreciated
2. If you consider the shape as two separate shapes (a semi circle and a triangle), you should be able to find the centre of mass for each of these. Then you can deduce the centre of mass of the combined shape.

P.S The centre of mass for a sector of a circle is given in the formula booklet
3. Would you consider the y axis to be the line AC? And the x to pass through B and D

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4. (Original post by Frannoooooooo)
If you consider the shape as two separate shapes (a semi circle and a triangle), you should be able to find the centre of mass for each of these. Then you can deduce the centre of mass of the combined shape.

P.S The centre of mass for a sector of a circle is given in the formula booklet
Just found out how to quote :P

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5. (Original post by Connor_leigh)
Would you consider the y axis to be the line AC? And the x to pass through B and D

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The x-axis should pass through BD yes, there's no need to be considering the y coordinates as centre of masses are going to be aligned as they're horizontally symmetrical. As for the y-axis, it doesn't matter too much where you put it - you'll come out with the same answer. But putting it through AC would probably mean slightly fewer calculations are needed, so yes
6. (Original post by Frannoooooooo)
The x-axis should pass through BD yes, there's no need to be considering the y coordinates as centre of masses are going to be aligned as they're horizontally symmetrical. As for the y-axis, it doesn't matter too much where you put it - you'll come out with the same answer. But putting it through AC would probably mean slightly fewer calculations are needed, so yes
Sorry for all the questions I was away when we covered this so I'm just trying to catch up. Your triangles x mid point would be 0+0+3/2 all over 3 which would give you (3/6,2,) and then how would you work out the mid point of the semi circle? I assume just treat it as a sector with theta being pi?

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7. (Original post by Connor_leigh)
Sorry for all the questions I was away when we covered this so I'm just trying to catch up. Your triangles x mid point would be 0+0+3/2 all over 3 which would give you (3/6,2,) and then how would you work out the mid point of the semi circle? I assume just treat it as a sector with theta being pi?

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Yes that's right for the triangle, although there's no need to consider the y coordinates as both centre of masses with lie on y=0. And for the semi circle there's a more simple equation for semi circles which I thought was in the booklet but isn't, it's: 4r/3pi , where r is the radius. This gives the distance of the COM from centre of the circle (on the straight line).
8. (Original post by Frannoooooooo)
Yes that's right for the triangle, although there's no need to consider the y coordinates as both centre of masses with lie on y=0. And for the semi circle there's a more simple equation for semi circles which I thought was in the booklet but isn't, it's: 4r/3pi , where r is the radius. This gives the distance of the COM from centre of the circle (on the straight line).
With these two centre of masses what would you do next? Just find the halfway point?

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9. (Original post by Connor_leigh)
With these two centre of masses what would you do next? Just find the halfway point?

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No because they'll both have different masses. Do you know how to centre the centre of mass of a composite shape, taking moments etc? Take a look at this vid for help https://www.youtube.com/watch?v=Sm7nuBGY094
10. (Original post by Frannoooooooo)
No because they'll both have different masses. Do you know how to centre the centre of mass of a composite shape, taking moments etc? Take a look at this vid for help https://www.youtube.com/watch?v=Sm7nuBGY094
Oh yeah, for some reason I assumed it was just a wire frame. So its (4pi/2 x centre of mass of semicircle) + (3 x centre of mass of triangle)?

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