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Find the centre of mass of a uniform : (a) half-disc, (b) quarter-disc. |
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Answer» `((4R)/(3pi),(3R)/(3pi))` ,`((4R)/(3pi),(3R)/(3pi))` `m=(M)/((1)/(2)piR^(2))=(2M)/(piR^(2))`  (a)thehalf -disc can besupposed to beconsistsof a large numberof semicircularrings ofmass dm andthicknessdrand rangingfromr=0 to r=R Surfaceareaof semicircular ringof radiusr andof thickness `dr=(1)/(2)2 pi r XX dr =pi r dr` `therefore`Massof thiselemetary ring`dm=pirdrxx(2M)/(piR^(2))` `dm =(2M)/(R^(2))rdr` If (x,y) arecoordinates of centreof massof thiselement, then, `(x,y)=(0,(2r)/(pi))` therefore , `x=0and y=(2r)/(pi)` Let`x_(cM)`and`y_(CM)` be thecoordinates of thecentreofthecentre ofthesemicircular disc. then `x_(CM)=(1)/(M)int_(0)^(R)xdm =(1)/(M)int_(0)^(R)dm=0` ` y_(CM)=(1)/(M)int_(0)^(R)ydm=(1)/(M)int_(0)^(R)(2l)/(pi)xx((2M)/(R^(2))rdr)` `=(4)/(piR^(2))int_(0)^(R)r^(2)dr=(4)/(piR^(2))[(r^(3))/(3)]_(0)^(R)` ` =(4)/(piR^(2))xx((R^(3))/(3)-0)=(4R)/(3pi)` `therefore ` Centre of mass of thesemiciculardisc `=(9,(4R)/(3pi))` ( b)Centreof mass a unifromquarter disc, Massper unitareaof thequarter disc `=(M)/((piR^(2))/(4))=(4M)/(3PiR^(3))` Usingsymmetry for a halfalongy - axiscentreof masswill beat `x=(4R)/(3pi)` Fora half- discalongx-axiscentre of masswill beat `x=($R)/(3pi)` Hence, forthequarter disc centreof mass `=((4R)/(3pi),(3R)/(3pi))` 
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