Answer:
86.43
25.72
Explanation:
Location of center of mass from 59 Kg
[tex]r_1=\frac {xm_2}{m1+m2}=\frac {1.1*73}{59+73}= 0.608333m[/tex]
Location of center of mass from 73 Kg
[tex]r_2=\frac {xm_1}{m1+m2}=\ frac {1.1*59}{59+73}= 0.491667m[/tex]
Momentum of inertia
[tex]I=m1r_1^{2}+m2r_2^{2}=59*0.608333^{2}+73*0.491667^{2}=21.8341 +17.64674=39.48083 Kgm^{2}[/tex]
Angular moment
[tex]L=I\omega=I(\frac {2\pi}{T})=39.48083(\frac {2*\pi}{2.87 s})=86.43393Js[/tex]
(b)
When separation is 0.6 m
Location of centre of mass from 59 Kg
[tex]r_1=\frac {xm_2}{m1+m2}=\frac {0.6*73}{59+73}= 0.331818 m[/tex]
Location of centre of mass from 73 Kg
[tex]r_2=\frac {xm_1}{m1+m2}=\ frac {0.6*59}{59+73}= 0.268182 m[/tex]
Momentum of intertia
[tex]I=m1r_1^{2}+m2r_2^{2}=59*0.331818^{2}+73*0.268182^{2}=6.496095 + 5.250269=11.74636 Kgm^{2}[/tex]
Angular moment
[tex]L=I\omega=I(\frac {2\pi}{T})=11.74636(\frac {2*\pi}{2.87 s})=25.71588 Js[/tex]
