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Question Number 134249 by abdurehime last updated on 01/Mar/21

Answered by mr W last updated on 01/Mar/21

Commented by mr W last updated on 01/Mar/21

1: stretched position 2: unstretched position of spring 3: rest position ky=mgsin θ ⇒y=((mgsin θ)/k) at position 1: E=mg(x+y)sin θ+(1/2)kx^2 at position 3: E=(1/2)Iω^2 +(1/2)m(Rω)^2 +(1/2)ky^2 (1/2)Iω^2 +(1/2)m(Rω)^2 +(1/2)ky^2 =mg(x+y)sin θ+(1/2)kx^2 (I+mR^2 )ω^2 =2mgxsin θ+kx^2 +(((mgsin θ)^2 )/k) ω=(√((2mgxsin θ+kx^2 +(((mgsin θ)^2 )/k))/(I+mR^2 ))) =(√((2×0.5×9.71×0.2×sin 37°+50×0.2^2 +(((0.5×9.81×sin 37°)^2 )/(50)))/(1+0.5×0.3^2 ))) =1.792 rad/s

$$\mathrm{1}:\:{stretched}\:{position} \\ $$$$\mathrm{2}:\:{unstretched}\:{position}\:{of}\:{spring} \\ $$$$\mathrm{3}:\:{rest}\:{position} \\ $$$${ky}={mg}\mathrm{sin}\:\theta \\ $$$$\Rightarrow{y}=\frac{{mg}\mathrm{sin}\:\theta}{{k}} \\ $$$${at}\:{position}\:\mathrm{1}: \\ $$$${E}={mg}\left({x}+{y}\right)\mathrm{sin}\:\theta+\frac{\mathrm{1}}{\mathrm{2}}{kx}^{\mathrm{2}} \\ $$$${at}\:{position}\:\mathrm{3}: \\ $$$${E}=\frac{\mathrm{1}}{\mathrm{2}}{I}\omega^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{m}\left({R}\omega\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{ky}^{\mathrm{2}} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{I}\omega^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{m}\left({R}\omega\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{ky}^{\mathrm{2}} ={mg}\left({x}+{y}\right)\mathrm{sin}\:\theta+\frac{\mathrm{1}}{\mathrm{2}}{kx}^{\mathrm{2}} \\ $$$$\left({I}+{mR}^{\mathrm{2}} \right)\omega^{\mathrm{2}} =\mathrm{2}{mgx}\mathrm{sin}\:\theta+{kx}^{\mathrm{2}} +\frac{\left({mg}\mathrm{sin}\:\theta\right)^{\mathrm{2}} }{{k}} \\ $$$$\omega=\sqrt{\frac{\mathrm{2}{mgx}\mathrm{sin}\:\theta+{kx}^{\mathrm{2}} +\frac{\left({mg}\mathrm{sin}\:\theta\right)^{\mathrm{2}} }{{k}}}{{I}+{mR}^{\mathrm{2}} }} \\ $$$$=\sqrt{\frac{\mathrm{2}×\mathrm{0}.\mathrm{5}×\mathrm{9}.\mathrm{71}×\mathrm{0}.\mathrm{2}×\mathrm{sin}\:\mathrm{37}°+\mathrm{50}×\mathrm{0}.\mathrm{2}^{\mathrm{2}} +\frac{\left(\mathrm{0}.\mathrm{5}×\mathrm{9}.\mathrm{81}×\mathrm{sin}\:\mathrm{37}°\right)^{\mathrm{2}} }{\mathrm{50}}}{\mathrm{1}+\mathrm{0}.\mathrm{5}×\mathrm{0}.\mathrm{3}^{\mathrm{2}} }} \\ $$$$=\mathrm{1}.\mathrm{792}\:{rad}/{s} \\ $$

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