Question

from a uniform circular plate of diameter 1m,circular portion of diameter 0.5m is cut out as shown in the figure.Then the portion of centre of mass from the centre of original plate (OG2)

Answer 1

Answer:

1/12

Explanation:

YCM=m1x1+m2x2/M1+M2

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your answer should be 1/12

Answer 2

Given :

Diameter of the circular plate = 1 m

Diameter of the cut out portion = 0.5 m

To Find :

The distance of centre of mass from the centre of original plate

Solution :

Let M be the mass of the circular plate and M₁ be the mass of the removed portion of the plate.

If mass per unit area is m then,

Mass of original plate (M) = m x π x (0.5)²

                                          = m x π x 0.25

Mass of removed portion (M₁) = m x π x (0.25)²

                                                 = m x π x 0.0625

Mass of the remaining part = M – M₁

Now,

Centre of mass position of the original plate (r)= 0 (at origin)

Centre of mass position of the removed disc (r₁) = 1 – 0.5 = 0.5 m

Centre of mass of the remaining part= r₂

Therefore,

⇒ M×0 = M₁r₁ + (M – M₁)r₂

⇒ M₁r₁  =  – (M – M₁)r₂

⇒  r₂    =  $\frac{M_1r_1}{-(M - M_1)}$

⇒  r₂    =   $\frac{m * \pi * 0.0625 * 0.5}{-(m * \pi * 0.25 - m * \pi * 0.0625)}$

⇒  r₂    =  $\frac{m * \pi * 0.03125}{ - (0.25m * \pi - 0.0625 m\pi ) }$

⇒  r₂    =  $\frac{0.03125m\pi }{0.1875m\pi }$

∴   r₂    = - 0.167 m

Therefore, the centre of mass of the remaining portion is 0.167 m to the left of the centre of the bigger plate.