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Question Number 24293 by Tinkutara last updated on 15/Nov/17

Find the centre of mass of a uniform  (a) half-disc, (b) quarter-disc.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{a}\:\mathrm{uniform} \\ $$$$\left({a}\right)\:\mathrm{half}-\mathrm{disc},\:\left({b}\right)\:\mathrm{quarter}-\mathrm{disc}. \\ $$

Commented by ajfour last updated on 15/Nov/17

Commented by ajfour last updated on 15/Nov/17

Answered by mrW1 last updated on 15/Nov/17

radius = R  (a) semi disc  θ∈[0,π]  ∫_0 ^( π) (1/2)R×R×((2Rsin θ)/3)dθ=(R^3 /3)∫_0 ^( π) sin θdθ=((2R^3 )/3)  y_C =((2R^3 ×2)/(3×πR^2 ))=((4R)/(3π))    (b) quarter disc  θ∈[(π/4),((3π)/4)]  ∫_(π/4) ^( ((3π)/4)) (1/2)R×R×((2Rsin θ)/3)dθ=(R^3 /3)∫_(π/4) ^( ((3π)/4)) sin θdθ=(((√2)R^3 )/3)  y_C =(((√2)R^3 ×4)/(3×πR^2 ))=((4(√2)R)/(3π))    (c) any sector of disc with angle α  (R^3 /3)∫_((π−α)/2) ^( ((π+α)/2)) sin θ dθ=((2R^3 )/3)sin (α/2)  y_C =((2R^3 sin (α/2))/(3×πR^2 ×(α/(2π))))=((4R)/(3α))×sin (α/2)

$${radius}\:=\:{R} \\ $$$$\left({a}\right)\:{semi}\:{disc} \\ $$$$\theta\in\left[\mathrm{0},\pi\right] \\ $$$$\int_{\mathrm{0}} ^{\:\pi} \frac{\mathrm{1}}{\mathrm{2}}{R}×{R}×\frac{\mathrm{2}{R}\mathrm{sin}\:\theta}{\mathrm{3}}{d}\theta=\frac{{R}^{\mathrm{3}} }{\mathrm{3}}\int_{\mathrm{0}} ^{\:\pi} \mathrm{sin}\:\theta{d}\theta=\frac{\mathrm{2}{R}^{\mathrm{3}} }{\mathrm{3}} \\ $$$${y}_{{C}} =\frac{\mathrm{2}{R}^{\mathrm{3}} ×\mathrm{2}}{\mathrm{3}×\pi{R}^{\mathrm{2}} }=\frac{\mathrm{4}{R}}{\mathrm{3}\pi} \\ $$$$ \\ $$$$\left({b}\right)\:{quarter}\:{disc} \\ $$$$\theta\in\left[\frac{\pi}{\mathrm{4}},\frac{\mathrm{3}\pi}{\mathrm{4}}\right] \\ $$$$\int_{\frac{\pi}{\mathrm{4}}} ^{\:\frac{\mathrm{3}\pi}{\mathrm{4}}} \frac{\mathrm{1}}{\mathrm{2}}{R}×{R}×\frac{\mathrm{2}{R}\mathrm{sin}\:\theta}{\mathrm{3}}{d}\theta=\frac{{R}^{\mathrm{3}} }{\mathrm{3}}\int_{\frac{\pi}{\mathrm{4}}} ^{\:\frac{\mathrm{3}\pi}{\mathrm{4}}} \mathrm{sin}\:\theta{d}\theta=\frac{\sqrt{\mathrm{2}}{R}^{\mathrm{3}} }{\mathrm{3}} \\ $$$${y}_{{C}} =\frac{\sqrt{\mathrm{2}}{R}^{\mathrm{3}} ×\mathrm{4}}{\mathrm{3}×\pi{R}^{\mathrm{2}} }=\frac{\mathrm{4}\sqrt{\mathrm{2}}{R}}{\mathrm{3}\pi} \\ $$$$ \\ $$$$\left({c}\right)\:{any}\:{sector}\:{of}\:{disc}\:{with}\:{angle}\:\alpha \\ $$$$\frac{{R}^{\mathrm{3}} }{\mathrm{3}}\int_{\frac{\pi−\alpha}{\mathrm{2}}} ^{\:\frac{\pi+\alpha}{\mathrm{2}}} \mathrm{sin}\:\theta\:{d}\theta=\frac{\mathrm{2}{R}^{\mathrm{3}} }{\mathrm{3}}\mathrm{sin}\:\frac{\alpha}{\mathrm{2}} \\ $$$${y}_{{C}} =\frac{\mathrm{2}{R}^{\mathrm{3}} \mathrm{sin}\:\frac{\alpha}{\mathrm{2}}}{\mathrm{3}×\pi{R}^{\mathrm{2}} ×\frac{\alpha}{\mathrm{2}\pi}}=\frac{\mathrm{4}{R}}{\mathrm{3}\alpha}×\mathrm{sin}\:\frac{\alpha}{\mathrm{2}} \\ $$

Commented by mrW1 last updated on 15/Nov/17

Commented by mrW1 last updated on 16/Nov/17

I gave a general method to determine  the center of mass. It is always on  the symmetric line. You only need to  know the distance from the COM to  the center of circle. You can see I have  placed the disc such that its symmetric  line is the y−axis. Then we only need  y_C , since x_C =0. In case of quarter−disc,  y_C =((4(√2)R)/(3π)). If you at all costs want to  use your coordinate system then you  will get x_C ′=((4(√2)R)/(3π))×((√2)/2)=((4R)/(3π)),   y_c ′=((4(√2)R)/(3π))×((√2)/2)=((4R)/(3π)).    Using my method you can easily  determine the COM of any angled  disc sector, and you only need to  know one value instead of two.

$${I}\:{gave}\:{a}\:{general}\:{method}\:{to}\:{determine} \\ $$$${the}\:{center}\:{of}\:{mass}.\:{It}\:{is}\:{always}\:{on} \\ $$$${the}\:{symmetric}\:{line}.\:{You}\:{only}\:{need}\:{to} \\ $$$${know}\:{the}\:{distance}\:{from}\:{the}\:{COM}\:{to} \\ $$$${the}\:{center}\:{of}\:{circle}.\:{You}\:{can}\:{see}\:{I}\:{have} \\ $$$${placed}\:{the}\:{disc}\:{such}\:{that}\:{its}\:{symmetric} \\ $$$${line}\:{is}\:{the}\:{y}−{axis}.\:{Then}\:{we}\:{only}\:{need} \\ $$$${y}_{{C}} ,\:{since}\:{x}_{{C}} =\mathrm{0}.\:{In}\:{case}\:{of}\:{quarter}−{disc}, \\ $$$${y}_{{C}} =\frac{\mathrm{4}\sqrt{\mathrm{2}}{R}}{\mathrm{3}\pi}.\:{If}\:{you}\:{at}\:{all}\:{costs}\:{want}\:{to} \\ $$$${use}\:{your}\:{coordinate}\:{system}\:{then}\:{you} \\ $$$${will}\:{get}\:{x}_{{C}} '=\frac{\mathrm{4}\sqrt{\mathrm{2}}{R}}{\mathrm{3}\pi}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}=\frac{\mathrm{4}{R}}{\mathrm{3}\pi},\: \\ $$$${y}_{{c}} '=\frac{\mathrm{4}\sqrt{\mathrm{2}}{R}}{\mathrm{3}\pi}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}=\frac{\mathrm{4}{R}}{\mathrm{3}\pi}. \\ $$$$ \\ $$$${Using}\:{my}\:{method}\:{you}\:{can}\:{easily} \\ $$$${determine}\:{the}\:{COM}\:{of}\:{any}\:{angled} \\ $$$${disc}\:{sector},\:{and}\:{you}\:{only}\:{need}\:{to} \\ $$$${know}\:{one}\:{value}\:{instead}\:{of}\:{two}. \\ $$

Commented by mrW1 last updated on 16/Nov/17

Commented by Tinkutara last updated on 16/Nov/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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