Two ice skaters Due in 12 hours and 17 minutes Two ice cream skater hold hands a

Question

Two ice skaters Due in 12 hours and 17 minutes Two ice cream skater hold hands and rotate, making one revolution In 3.0 s. Their masses are 69 kg and 68 kg, and they are separated by 1.4 m. Find the angular momentum of the system about their center of mass. Submit Answer Tries 0/10 If the skaters pull towards each other and decrease their separate to 0.80 m, what is the new period of rotation about the center of mass? Submit Answer Tries 0/10 New Angular Momentum Annonymous 1 Replay (Sat Mar 12 06:00:45 pm 2016 (EST)) Does anyone know how to do this one? I found the centre of mass, then tried to use the net angular momentum and the linear velocity I got using the revolution and time - I got angular speed first, then linear velocity from that - is there something I'm doing wrong that anyone can point Out? NEW Re: Angular Momentum Logan Singrey Reply (Mon Mar 14 12:45:11 pm 2016 (EOT)) Using the center of mass, you would need to Find the radius from the center of mass to each separate mass, Find the linear speed and divide by time for one revolution, calculate each angular momentum separately and then add them together to get the final answer NEW Re: Re: Angular Momentum Anonymous 1 Reply (Tue Mar 15 09:42:07 am 2016 (EOT)) Mercl beaucoup!

Explanation / Answer

1. Angular momentum = Iw
Now com = [69*0 + 68*1.4]/[69+68] = 0.694
so I = 69*0.694^2 + 68*(1.4-0.694)^2 = 67.2118
w = 2*pi/3
Angular momentum = 140.69 kg m^2 s^-1

2. New com = [69*0 + 68*0.8]/[69+68] = 0.397
New I = 69*0.397^2 + 68*(0.8-0.397)^2 = 21.9232
Iw = 140.69 (conservation of angular momentum)
w = 6.417 rad/s
T = 2pi/w = 0.978 s

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