Question

Two ice skaters hold hands and rotate, making one revolution in 2.87 s. Their masses are 59.0 kg and 73.0 kg, and they are separated by 1.10 m. Find the angular momentum of the system about their center of mass. If the skaters pull towards each other and decrease their separation to 0.60 m, what is the new period of rotation about the center of mass?

Answer 1

Answer:

86.43

25.72

Explanation:

Location of center of mass from 59 Kg

$r_1=\frac {xm_2}{m1+m2}=\frac {1.1*73}{59+73}= 0.608333m$

Location of center of mass from 73 Kg

$r_2=\frac {xm_1}{m1+m2}=\ frac {1.1*59}{59+73}= 0.491667m$

Momentum of inertia

$I=m1r_1^{2}+m2r_2^{2}=59*0.608333^{2}+73*0.491667^{2}=21.8341 +17.64674=39.48083 Kgm^{2}$

Angular moment

$L=I\omega=I(\frac {2\pi}{T})=39.48083(\frac {2*\pi}{2.87 s})=86.43393Js$

(b)

When separation is 0.6 m

Location of centre of mass from 59 Kg

$r_1=\frac {xm_2}{m1+m2}=\frac {0.6*73}{59+73}= 0.331818 m$

Location of centre of mass from 73 Kg

$r_2=\frac {xm_1}{m1+m2}=\ frac {0.6*59}{59+73}= 0.268182 m$

Momentum of intertia

$I=m1r_1^{2}+m2r_2^{2}=59*0.331818^{2}+73*0.268182^{2}=6.496095 + 5.250269=11.74636 Kgm^{2}$

Angular moment

$L=I\omega=I(\frac {2\pi}{T})=11.74636(\frac {2*\pi}{2.87 s})=25.71588 Js$