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
Elena
Distance between the skaters is d, frequency is n =1/7.6 s^-1
Location of the center of mass
of the second skater of mass m2 is
x1 = d •m2/(m1+m2),
x2 = d - x1,
Moment of inertia of the system is
I= I1 +I2 =
= m1•x1^2 + m2•x2^2.
Angular momentum is
L = I•ω= I•2•π•n.
KE = I• ω^2/2 = I•(2•π•n)^2/2.
Check your given data. It seems to me that the distance 4.6 m is too much if the “skaters hold hands”. Moreover, the mass m1 = 15 kg is improbable for skater (this is the mass of the child of three).
Location of the center of mass
of the second skater of mass m2 is
x1 = d •m2/(m1+m2),
x2 = d - x1,
Moment of inertia of the system is
I= I1 +I2 =
= m1•x1^2 + m2•x2^2.
Angular momentum is
L = I•ω= I•2•π•n.
KE = I• ω^2/2 = I•(2•π•n)^2/2.
Check your given data. It seems to me that the distance 4.6 m is too much if the “skaters hold hands”. Moreover, the mass m1 = 15 kg is improbable for skater (this is the mass of the child of three).
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