Physics, asked by aditi17039, 9 months ago

pls try this ques anybody..!!! ​

Attachments:

Answers

Answered by Rajshuklakld
1

Solution;-

If that small circle doesn't taken out,then,

Xcm about O=

 \frac{mass \: of \: cicle \times distance \: of \: centre \: of \: circle \: from \: o + mass \: of \: square \times 0}{mass \: of \: circle \:  + mass \: of \: square}  \\ clearly \: distance \: of \: centre \: of \: circle \\ from \: o = 4 + 2 = 6cm \\ let \: th \: mass \: per \: unit \: area \: of \: square \: and \\ circlee \:  \: be \: p \\ then \: mass \: of \: circle = p \times \pi \times 16 \\ mass \: of \: square = 400 \times p \\ puttiing \: this \: value \: we \: get \\ centre \: of \: mass \: if \:circle \: is \: not \: taken \: out \\  =  \frac{0 + p \times 96\pi}{16\pi \times p  + 400 \times p } \\  =  \frac{16\pi}{16\pi + 400} \\ but \: as \: the \: mass \: of \: circle \: is \: taken \: out \: so  \\ to \: get \: the \: centre \: of \: mass \: of \: remaining \: part \\ take \: the \: mass \: of \: circle \: as \: negative \\ required \: centre \: of \: mass \\  =  \frac{ - 96\pi}{400 - 16\pi} = -  . 73 \\ so \: centre \: of \: mass \: will \:be \: on \: .73 \: left \: of \: o

Note:-(-) sign denotes the left of O

{hope it helps you}

Answered by ᎷíssGℓαмσƦσυs
4

that small circle doesn't taken out,then,

Xcm about O=

\begin{lgathered}\frac{mass \: of \: cicle \times distance \: of \: centre \: of \: circle \: from \: o + mass \: of \: square \times 0}{mass \: of \: circle \: + mass \: of \: square} \\ clearly \: distance \: of \: centre \: of \: circle \\ from \: o = 4 + 2 = 6cm \\ let \: th \: mass \: per \: unit \: area \: of \: square \: and \\ circlee \: \: be \: p \\ then \: mass \: of \: circle = p \times \pi \times 16 \\ mass \: of \: square = 400 \times p \\ puttiing \: this \: value \: we \: get \\ centre \: of \: mass \: if \:circle \: is \: not \: taken \: out \\ = \frac{0 + p \times 96\pi}{16\pi \times p + 400 \times p } \\ = \frac{16\pi}{16\pi + 400} \\ but \: as \: the \: mass \: of \: circle \: is \: taken \: out \: so \\ to \: get \: the \: centre \: of \: mass \: of \: remaining \: part \\ take \: the \: mass \: of \: circle \: as \: negative \\ required \: centre \: of \: mass \\ = \frac{ - 96\pi}{400 - 16\pi} = - . 73 \\ so \: centre \: of \: mass \: will \:be \: on \: .73 \: left \: of \: o\end{lgathered}

massofcircle+massofsquare

massofcicle×distanceofcentreofcirclefromo+massofsquare×0

clearlydistanceofcentreofcircle

fromo=4+2=6cm

letthmassperunitareaofsquareand

circleebep

thenmassofcircle=p×π×16

massofsquare=400×p

puttiingthisvalueweget

centreofmassifcircleisnottakenout

=

16π×p+400×p

0+p×96π

=

16π+400

16π

butasthemassofcircleistakenoutso

togetthecentreofmassofremainingpart

takethemassofcircleasnegative

requiredcentreofmass

=

400−16π

−96π

=−.73

Similar questions