7.16 From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body

Radius of the original disc = R

Mass of the smaller disc = $\tau$ = (1/4) $\omega$ = M/4

Let O and O’ be the respective centers of the original disc and the cut out disc respectively. As per the definition of the centre of mass, the centre of mass of the original disc is supposed to be concentrated at O, while that of the smaller disc is supposed to be concentrated at O’.

It is given, OO’ = R/2

After the smaller disc has been cut from the original, the remaining portion is considered to be a system of two masses. The two masses are

M – concentrated at O and M/4 concentrated at O’

Let x be the distance thro

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Radius of the original disc = R

Mass of the smaller disc = $\tau$ = (1/4) $\omega$ = M/4

Let O and O’ be the respective centers of the original disc and the cut out disc respectively. As per the definition of the centre of mass, the centre of mass of the original disc is supposed to be concentrated at O, while that of the smaller disc is supposed to be concentrated at O’.

It is given, OO’ = R/2

After the smaller disc has been cut from the original, the remaining portion is considered to be a system of two masses. The two masses are

M – concentrated at O and M/4 concentrated at O’

Let x be the distance through which the centers of masses of two masses shifts from O.

The relation between the centers of masses is given as

x = (m₁r₁ + m₂r₂)/ (m₁ +m₂)

= $\pi R^2\sigma$ = $\pi {\left(\frac{R}{2}\right)}^2\sigma$ = $\pi R^2\sigma$ x $\frac{Mx0-M^{\text{'}}x\left(\frac{R}{2}\right)}{M+(-M^{\text{'}})}$ = $\frac{-\left(\frac{M}{4}\right)x\left(\frac{R}{2}\right)}{M-M/4}$

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<p>Radius of the original disc = R</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1734436905phpwgrCxx_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1734436905phpwgrCxx.jpeg" alt="" width="258" height="38"></picture></div></div><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1729070589phpTwvPy8_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1729070589phpTwvPy8.jpeg" alt="" width="190" height="166"></picture></div></div><p>Mass of the smaller disc =&nbsp;<span title="Click to copy mathml"><math><mi>τ</mi></math></span>&nbsp;= (1/4)&nbsp;<span title="Click to copy mathml"><math><mi>ω</mi></math></span>&nbsp;= M/4</p><p>Let O and O’ be the respective centers of the original disc and the cut out disc respectively. As per the definition of the centre of mass, the centre of mass of the original disc is supposed to be concentrated at O, while that of the smaller disc is supposed to be concentrated at O’.</p><p>It is given, OO’ = R/2</p><p>After the smaller disc has been cut from the original, the remaining portion is considered to be a system of two masses. The two masses are</p><p>M – concentrated at O and M/4 concentrated at O’</p><p>Let x be the distance through which the centers of masses of two masses shifts from O.</p><p>The relation between the centers of masses is given as</p><p>x = (m₁r₁ + m₂r₂)/ (m₁ +m₂)</p><p>=&nbsp;<span title="Click to copy mathml"><math><mi>π</mi><mi></mi><msup><mrow><mrow><mi>R</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mi></mi><mi>σ</mi></math></span>&nbsp;=&nbsp;<span title="Click to copy mathml"><math><mi>π</mi><mi></mi><msup><mrow><mrow><mo> (</mo><mfrac><mrow><mrow><mi>R</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac><mo>)</mo></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mi></mi><mi>σ</mi></math></span>&nbsp;=&nbsp;<span title="Click to copy mathml"><math><mi>π</mi><mi></mi><msup><mrow><mrow><mi>R</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mi></mi><mi>σ</mi></math></span>&nbsp;x&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mi>M</mi><mi></mi><mi>x</mi><mi></mi><mn>0</mn><mo>-</mo><msup><mrow><mrow><mi>M</mi></mrow></mrow><mrow><mrow><mi>'</mi></mrow></mrow></msup><mi>x</mi><mi></mi><mo> (</mo><mfrac><mrow><mrow><mi>R</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac><mo>)</mo></mrow></mrow><mrow><mrow><mi>M</mi><mo>+</mo><mo> (</mo><mo>-</mo><msup><mrow><mrow><mi>M</mi></mrow></mrow><mrow><mrow><mi>'</mi></mrow></mrow></msup><mo>)</mo></mrow></mrow></mfrac></math></span>&nbsp;=&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mo>-</mo><mfenced separators="|"><mrow><mrow><mfrac><mrow><mrow><mi>M</mi></mrow></mrow><mrow><mrow><mn>4</mn></mrow></mrow></mfrac></mrow></mrow></mfenced><mi>x</mi><mi></mi><mo> (</mo><mfrac><mrow><mrow><mi>R</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac><mo>)</mo></mrow></mrow><mrow><mrow><mi>M</mi><mo>-</mo><mi>M</mi><mo>/</mo><mn>4</mn></mrow></mrow></mfrac></math></span></p>

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