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Two ice skaters hold hands and rotate, making one revolution in 7.6 s. Their masses are 15 kg and 50 kg, and they are separated...Asked by Amber
Two ice skaters hold hands and rotate, making one revolution in 7.6 s. Their masses are 15 kg and 50 kg, and they are separated by 4.6 m.
Find the angular momentum of the system about their center of mass.
Answer in units of J · s
Find the total kinetic energy of the system.
Answer in units of J
Find the angular momentum of the system about their center of mass.
Answer in units of J · s
Find the total kinetic energy of the system.
Answer in units of J
Answers
Answered by
Damon
These are pretty crazy measurements for human ice skaters but anyway
w = 2 pi/7.6 = .827 radians/second
find center of mass at x from big one
x(50)=(4.6-x)(15)
50 x = 69 - 15 x
65 x = 69
x = 1.06 m from big one and 3.54 from little one
I = 50(1.06)^2 + 15(3.54)^2
= 244 kg m^2
L = ang momentum = I w
= 244 (.827)
= 202 kg m^2/s
w = 2 pi/7.6 = .827 radians/second
find center of mass at x from big one
x(50)=(4.6-x)(15)
50 x = 69 - 15 x
65 x = 69
x = 1.06 m from big one and 3.54 from little one
I = 50(1.06)^2 + 15(3.54)^2
= 244 kg m^2
L = ang momentum = I w
= 244 (.827)
= 202 kg m^2/s
Answered by
Cesar
for KE
since you already have all the numbers...
KE1 + KE2 = (.5*I1*W1^2) + (.5*I2*W2^2)
since you already have all the numbers...
KE1 + KE2 = (.5*I1*W1^2) + (.5*I2*W2^2)
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