In the figure one fourth part of a uniform di | vedclass

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Question

Now, $\sigma=\frac{M}{\frac{\pi R^2}{4}}=\frac{4 M}{\pi R^2}=\frac{d m}{d A}$

Now, $d A=\frac{\pi r}{2} d \gamma$

Now, $X_{c m}=\int \frac{x d m

Vedclass · Accepted Answer

$\frac{{4R}}{{3\pi }}$

Vedclass · Answer

Now, $\sigma=\frac{M}{\frac{\pi R^2}{4}}=\frac{4 M}{\pi R^2}=\frac{d m}{d A}$

Now, $d A=\frac{\pi r}{2} d \gamma$

Now, $X_{c m}=\int \frac{x d m}{M}=0$

$Y_{c m}=\frac{\int y d m}{M} =\frac{\int \frac{2 \gamma}{\pi} \cdot \frac{4 M}{\pi R^2} \cdot d A}{M}$

$=\frac{\int \frac{2 \gamma}{\pi} \cdot \frac{4 M}{\pi R^2} \cdot \frac{\pi \gamma}{2} d \gamma}{M}$

$=\frac{4}{\pi R^r} \int_0^R \gamma^2 d \gamma$

$=\frac{4}{\pi R^2} \cdot \frac{R^3}{3}$

$Y_{c M} =\frac{4 R}{3 \pi}$

In the figure one fourth part of $a$ uniform disc of radius $R$ is shown. The distance of the centre of mass of this object from centre $‘O’$ is:

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