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Question-70543

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Question Number 70543 by mr W last updated on 05/Oct/19
Commented by mr W last updated on 05/Oct/19
a semi−cylinder with radius R and mass M rests on a rough table as shown. when a small impulse is given to the semi−cylinder, it begins to oscillate. find the frequency of the oscillation in terms of R.
$${a}\:{semi}−{cylinder}\:{with}\:{radius}\:{R}\:{and} \\ $$$${mass}\:{M}\:{rests}\:{on}\:{a}\:{rough}\:{table}\:{as}\:{shown}. \\ $$$${when}\:{a}\:{small}\:{impulse}\:{is}\:{given}\:{to}\:{the} \\ $$$${semi}−{cylinder},\:{it}\:{begins}\:{to}\:{oscillate}. \\ $$$${find}\:{the}\:{frequency}\:{of}\:{the}\:{oscillation} \\ $$$${in}\:{terms}\:{of}\:{R}. \\ $$
Commented by ajfour last updated on 06/Oct/19
Commented by mr W last updated on 06/Oct/19
dA=((R^2 dθ)/2) y_C A=2∫_0 ^(π/2) ((R^2 dθ)/2)×((2R)/3) sin θ=((2R^3 )/3)∫_0 ^(π/2) sin θ dθ=((2R^3 )/3) y_C =((2R^3 )/(3×((πR^2 )/2)))=((4R)/(3π)) dI_0 =∫_0 ^R ρrdθr^2 dr=((ρR^4 dθ)/4) I_0 =((ρR^4 π)/4)=((ρπR^2 )/2)×(R^2 /2)=((MR^2 )/2) I_0 =I_C +My_C ^2 I_P =I_C +M(R−y_C )^2 ⇒I_P =I_0 +M(R−y_C )^2 −My_C ^2 ⇒I_P =I_0 +MR(R−2y_C ) ⇒I_P =((MR^2 )/2)+MR(R−((8R)/(3π)))=((MR^2 (9π−16))/(6π)) f=(1/(2π))(√((4MgR)/(3πI_P ))) f=(1/(2π))(√((4MgR)/(3π×((MR^2 (9π−16))/(6π))))) ⇒f=(1/π)(√((2g)/((9π−16)R)))
$${dA}=\frac{{R}^{\mathrm{2}} {d}\theta}{\mathrm{2}} \\ $$$${y}_{{C}} {A}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{R}^{\mathrm{2}} {d}\theta}{\mathrm{2}}×\frac{\mathrm{2}{R}}{\mathrm{3}}\:\mathrm{sin}\:\theta=\frac{\mathrm{2}{R}^{\mathrm{3}} }{\mathrm{3}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}\:\theta\:{d}\theta=\frac{\mathrm{2}{R}^{\mathrm{3}} }{\mathrm{3}} \\ $$$${y}_{{C}} =\frac{\mathrm{2}{R}^{\mathrm{3}} }{\mathrm{3}×\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}}}=\frac{\mathrm{4}{R}}{\mathrm{3}\pi} \\ $$$${dI}_{\mathrm{0}} =\int_{\mathrm{0}} ^{{R}} \rho{rd}\theta{r}^{\mathrm{2}} {dr}=\frac{\rho{R}^{\mathrm{4}} {d}\theta}{\mathrm{4}} \\ $$$${I}_{\mathrm{0}} =\frac{\rho{R}^{\mathrm{4}} \pi}{\mathrm{4}}=\frac{\rho\pi{R}^{\mathrm{2}} }{\mathrm{2}}×\frac{{R}^{\mathrm{2}} }{\mathrm{2}}=\frac{{MR}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${I}_{\mathrm{0}} ={I}_{{C}} +{My}_{{C}} ^{\mathrm{2}} \\ $$$${I}_{{P}} ={I}_{{C}} +{M}\left({R}−{y}_{{C}} \right)^{\mathrm{2}} \\ $$$$\Rightarrow{I}_{{P}} ={I}_{\mathrm{0}} +{M}\left({R}−{y}_{{C}} \right)^{\mathrm{2}} −{My}_{{C}} ^{\mathrm{2}} \\ $$$$\Rightarrow{I}_{{P}} ={I}_{\mathrm{0}} +{MR}\left({R}−\mathrm{2}{y}_{{C}} \right) \\ $$$$\Rightarrow{I}_{{P}} =\frac{{MR}^{\mathrm{2}} }{\mathrm{2}}+{MR}\left({R}−\frac{\mathrm{8}{R}}{\mathrm{3}\pi}\right)=\frac{{MR}^{\mathrm{2}} \left(\mathrm{9}\pi−\mathrm{16}\right)}{\mathrm{6}\pi} \\ $$$${f}=\frac{\mathrm{1}}{\mathrm{2}\pi}\sqrt{\frac{\mathrm{4}{MgR}}{\mathrm{3}\pi{I}_{{P}} }} \\ $$$${f}=\frac{\mathrm{1}}{\mathrm{2}\pi}\sqrt{\frac{\mathrm{4}{MgR}}{\mathrm{3}\pi×\frac{{MR}^{\mathrm{2}} \left(\mathrm{9}\pi−\mathrm{16}\right)}{\mathrm{6}\pi}}} \\ $$$$\Rightarrow{f}=\frac{\mathrm{1}}{\pi}\sqrt{\frac{\mathrm{2}{g}}{\left(\mathrm{9}\pi−\mathrm{16}\right){R}}} \\ $$
Commented by mr W last updated on 06/Oct/19
Commented by ajfour last updated on 06/Oct/19
Sir, you missed 2 from 2^(nd) to 3^(rd) line. y_C =((4R)/(3π)) And thanks, i realise I_P isn′t that difficult to obtain.
$${Sir},\:{you}\:{missed}\:\mathrm{2}\:{from}\:\mathrm{2}^{{nd}} \:{to} \\ $$$$\mathrm{3}^{{rd}} \:{line}.\:\:{y}_{{C}} =\frac{\mathrm{4}{R}}{\mathrm{3}\pi}\:\: \\ $$$${And}\:{thanks},\:{i}\:{realise}\:{I}_{{P}} \:\:{isn}'{t} \\ $$$${that}\:{difficult}\:{to}\:{obtain}. \\ $$
Commented by mr W last updated on 06/Oct/19
thanks too sir! now it should be correct.
$${thanks}\:{too}\:{sir}! \\ $$$${now}\:{it}\:{should}\:{be}\:{correct}. \\ $$

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