Plating a Circle and a Square

Classical Mechanics Level 1

A circular plate of diameter a is kept in contact with a square plate of edge a as shown in the figure. The density of the material and the thickness are same everywhere.

The centre of mass of the composite system will be: .

at the point of contact outside the system inside the circular plate inside the square plate

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1 solution

Since the material density (p) and thickness (d) is the same, the centre of mass depends only on the surface area. The square plate has larger surface area and therefore more mass, and the centre of mass will then be located inside the square.

Mathematically, you can prove the above qualitative statement by considering the centre of the circle as your origin.

So, M c M_{c} = p π ( a / 2 ) 2 d p\pi (a/2)^{2}d = p π a 2 d 4 \frac{p\pi a^{2}d }{4} : Circular Plate

M s M_{s} = p a 2 d pa^{2}d : Square Plate

We realize that M c M_{c} = π 4 \frac{\pi}{4} M s M_{s} = 0.25 π M s 0.25\pi M_{s} ... (1)

Now the x-coordinate of the centre of mass = M c . 0 + M s . a M c + M s \frac{M_{c}.0 + M_{s}.a}{M_{c} + M_{s}}

= M s . a M c + M s \frac{M_{s}.a}{M_{c} + M_{s}}

= M s . a 0.25 π M s + M s \frac{M_{s}.a}{0.25\pi M_{s} + M_{s}} ; Using (1)

= a 0.25 π + 1 \frac{a}{0.25\pi + 1} = 0.56 a 0.56 a

And the y-coordinate of the centre of mass = M c . 0 + M s . 0 M c + M s \frac{M_{c}.0 + M_{s}.0}{M_{c} + M_{s}} = 0

Thus, the centre of mass is located inside the square plate at (0.56 a,0).

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