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Centre of mass of uniform circular arc & sector watch

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    How can I find the centre of mass of a uniform circular arc & sector? Tips please!
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    For an arc that subtends an angle  2\alpha (in radians) at the centre of the circle, the centre of mass is a distance  \frac{r\sin \alpha}{\alpha} from the centre of the circle along the axis of symmetry. For a sector, the result becomes  \frac{2r\sin \alpha}{3\alpha} .
    Spoiler:
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    The first result can be proved by considering a general point P on the arc such that an angle of \theta radians between the line PO (O being the centre of the circle) and the axis of symmetry of the arc is made. And then consider the point P', such that an angle of  \delta \theta is made between PO and P'O. Now let an x-axis run (in a positive direction) from O through the axis of symmetry. Then the x-coordinate of the centre of mass of the small arc PP' is  r\cos \theta, and the length of this arc is  r \delta \theta. Now using the fact that:

     \displaystyle\sum mx = \bar{x}\displaystyle\sum m , where  \bar{x} is the centre of mass, we can sum up the masses and centre of masses of all the small arcs to get the centre of mass of the whole arc. (As the arc is considered to be uniform, we can let its length represent its mass.) And hence, to find the centre of mass of the arc, we want:

     \displaystyle\lim_{\delta \theta \to 0} \displaystyle\sum_{\theta = -\alpha}^{\alpha} r\cos \theta \times r \delta \theta = \bar{x} \displaystyle\lim_{\delta \theta \to 0} \displaystyle\sum_{\theta = -\alpha}^{\alpha} r \delta \theta

     \implies r^2 \displaystyle\int_{-\alpha}^{\alpha}\cos \theta \ \mathrm{d}\theta = \bar{x}r \displaystyle\int_{-\alpha}^{\alpha} \ \mathrm{d}\theta

     \implies 2r\sin \alpha = 2\alpha \bar{x} \iff \boxed{\bar{x} = \dfrac{r\sin \alpha}{\alpha}} .

    The proof is exactly the same for the sectors, except, the centre of masses of each arc are taken to be 2/3 of the way from O to P (as the sectors, as  \delta \theta tends to zero, become approoximately triangles).
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    (Original post by Dialetheism)
    How can I find the centre of mass of a uniform circular arc & sector? Tips please!
    Im assuming you need the proof..
    For the arc :
    Draw an arc symmetrical in the x axis such that the x axis bisects the arc at angle alpha.
    Let w be the mass per unit length.
    take a really small section if this arc , which has an angle(d for delta) d theta and length r.dtheta
    The mass of the element is then w.r.dtheta
    find the x coordinate of the c of m) (consider the sum of all w.r.dtheta from apha to -alpha)
    You need to then consider integration for the area ...
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    Thanks! yeah, I needed a proof. I'll try your method. Cheers
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    (Original post by Dialetheism)
    Thanks! yeah, I needed a proof. I'll try your method. Cheers
    I've given a proof in the spoiler of my post above.
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    (Original post by Unbounded)
    For an arc that subtends an angle  2\alpha (in radians) at the centre of the circle, the centre of mass is a distance  \frac{r\sin \alpha}{\alpha} from the centre of the circle along the axis of symmetry. For a sector, the result becomes  \frac{2r\sin \alpha}{3\alpha} .
    Spoiler:
    Show
    The first result can be proved by considering a general point P on the arc such that an angle of \theta radians between the line PO (O being the centre of the circle) and the axis of symmetry of the arc is made. And then consider the point P', such that an angle of  \delta \theta is made between PO and P'O. Now let an x-axis run (in a positive direction) from O through the axis of symmetry. Then the x-coordinate of the centre of mass of the small arc PP' is  r\cos \theta, and the length of this arc is  r \delta \theta. Now using the fact that:

     \displaystyle\sum mx = \bar{x}\displaystyle\sum m , where  \bar{x} is the centre of mass, we can sum up the masses and centre of masses of all the small arcs to get the centre of mass of the whole arc. (As the arc is considered to be uniform, we can let its length represent its mass.) And hence, to find the centre of mass of the arc, we want:

     \displaystyle\lim_{\delta \theta \to 0} \displaystyle\sum_{\theta = -\alpha}^{\alpha} r\cos \theta \times r \delta \theta = \bar{x} \displaystyle\lim_{\delta \theta \to 0} \displaystyle\sum_{\theta = -\alpha}^{\alpha} r \delta \theta

     \implies r^2 \displaystyle\int_{-\alpha}^{\alpha}\cos \theta \ \mathrm{d}\theta = \bar{x}r \displaystyle\int_{-\alpha}^{\alpha} \ \mathrm{d}\theta

     \implies 2r\sin \alpha = 2\alpha \bar{x} \iff \boxed{\bar{x} = \dfrac{r\sin \alpha}{\alpha}} .

    The proof is exactly the same for the sectors, except, the centre of masses of each arc are taken to be 2/3 of the way from O to P (as the sectors, as  \delta \theta tends to zero, become approoximately triangles).

    Thank you this was extremely helpful
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    Thank you soo much ! This helped me lot .
 
 
 

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