Question

please explain ans is 1 how​

Answer 1

$\large\underline{\underline{\sf Given:}}$

• Radius of Original disc = R

• Centre of Mass of Original disc = C

$\large\underline{\underline{\sf To\:Find:}}$

• Centre of mass after given portion is removed = ?

$\large\underline{\underline{\sf Solution:}}$

Mass per unit Lenght of original disc = $\sigma$

Mass of Original disc (M) = πR²$\sigma$

Radius of small disc = R/2

Mass of small disc (M')

$\implies{\sf π \left(\frac{R}{2}\right)^2\sigma }$

$\large{\sf\frac{M}{4} }$

"O" is the centre of mass of original disc (Given)

Let O' be the centre of mass after the cut .

•°• OO' = R/2

After small disc is cut -

M = Concentrated at O

-M' = (-M/4) Concentrated at O'

Negative sign indicates portion is cut .

We know the relation between centre of masses :-

$\large{\boxed{\sf x=\frac{m_1r_1+m_2r_2}{m_1+m_2} }}$

For the given system we can write it as ,

$\large{\sf x=\frac{M×0+(-M')×R/2}{M+(-M')} }$

$\large\implies{\sf x=\frac{\frac{ M}{4}×\frac{R}{2}}{M-\frac{M}{4}}}$

$\large\implies{\sf x=\frac{\frac{RM}{8}}{\frac{3M}{4}}}$

$\large\implies{\sf x=\frac{RM}{8}×\frac{4}{3M}}$

$\large\implies{\sf x=\frac{R}{6}}$

$\Large\underline{\underline{\sf Answer:}}$

•°•Centre of mass after the disc is cut is R/6