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Tanaya62
@Tanaya62
September 2021
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Derive how the centre of mass of semicircular disc comes to be 4R/3π
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tanay000000
.Notice, your formula YCM=∫y dmYCM=∫y dm is not correct.
the center of mass is given as
YCM=∫y dm∫dmYCM=∫y dm∫dm
Now, substituting the values y=rsinθy=rsinθ& dm=σrdrdθdm=σrdrdθ, we get
YCM=∫R0∫π0σr2sinθdθ dr∫R0∫π0σrdθ drYCM=∫0R∫0πσr2sinθdθ dr∫0R∫0πσrdθ dr
=∫R0(∫π0sinθdθ)r2 dr∫R0(∫π0dθ)r dr=∫0R(∫0πsinθdθ)r2 dr∫0R(∫0πdθ)r dr
=∫R0(2)r2 dr∫R0(π)r dr=2∫R0r2 drπ∫R0r dr=∫0R(2)r2 dr∫0R(π)r dr=2∫0Rr2 drπ∫0Rr dr
=2[r33]R0π[r22]R0=2[r33]0Rπ[r22]0R
=4R33πR2=4R3π=4R33πR2=4R3π
YCM=4R3
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Answers & Comments
the center of mass is given as
YCM=∫y dm∫dmYCM=∫y dm∫dm
Now, substituting the values y=rsinθy=rsinθ& dm=σrdrdθdm=σrdrdθ, we get
YCM=∫R0∫π0σr2sinθdθ dr∫R0∫π0σrdθ drYCM=∫0R∫0πσr2sinθdθ dr∫0R∫0πσrdθ dr
=∫R0(∫π0sinθdθ)r2 dr∫R0(∫π0dθ)r dr=∫0R(∫0πsinθdθ)r2 dr∫0R(∫0πdθ)r dr
=∫R0(2)r2 dr∫R0(π)r dr=2∫R0r2 drπ∫R0r dr=∫0R(2)r2 dr∫0R(π)r dr=2∫0Rr2 drπ∫0Rr dr
=2[r33]R0π[r22]R0=2[r33]0Rπ[r22]0R
=4R33πR2=4R3π=4R33πR2=4R3π
YCM=4R3