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Two ice-skaters, whose masses are 45 kg and 75 kg, hold hands and rotate about a

ID: 1351279 • Letter: T

Question

Two ice-skaters, whose masses are 45 kg and 75 kg, hold hands and rotate about a vertical axis that passes between them, making one revolution in 3 s. Their centers of mass are separated by 1.7 m, and their center of mass is stationary. Model each skater as a point particle. (a) Find the angular momentum of the system about it's center of mass. ?J·s
(b) Find the total kinetic energy of the system ?J Two ice-skaters, whose masses are 45 kg and 75 kg, hold hands and rotate about a vertical axis that passes between them, making one revolution in 3 s. Their centers of mass are separated by 1.7 m, and their center of mass is stationary. Model each skater as a point particle. (a) Find the angular momentum of the system about it's center of mass. ?J·s
(b) Find the total kinetic energy of the system ?J Two ice-skaters, whose masses are 45 kg and 75 kg, hold hands and rotate about a vertical axis that passes between them, making one revolution in 3 s. Their centers of mass are separated by 1.7 m, and their center of mass is stationary. Model each skater as a point particle. (a) Find the angular momentum of the system about it's center of mass. ?J·s
(b) Find the total kinetic energy of the system ?J

Explanation / Answer

Here ,

let the centre of mass is x m from the 75 Kg mass skater

x = 45 * 3/(75 + 45)

x = 1.125 m

a)

angular velocity , w = 2pi/T

w = 2pi/3

w = 2.094 rad/s

moment of inertia , I = m1*r^2 + m2* r^2

I = 45 * (3 - 1.125)^2 + 75 * 1.125^2

I = 253.125 Kg.m^2

angular momentum = I * w

angular momentum = 253.125 * 2.094

angular momentum = 530.04 Kg.m^2/s

the angular momentum is 530.04 Kg.m^2/s

b)

Total kinetic energy = 0.5 * I * w^2

Total kinetic energy = 0.5 * 253.125 * 2.094^2

Total kinetic energy = 555 J

the total kinetic energy of the system is 555 J

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