Two skaters, each of mass 65 kg , approach each other along parallel paths separ

Question

Two skaters, each of mass 65 kg, approach each other along parallel paths separated by 12.1 m. They have equal and opposite velocities of 1.3 m/s. The first skater carries one end of a long pole with negligible mass, and the second skater grabs the other end of it as she passes; see the figure. Assume frictionless ice. Describe quantitatively the motion of the skaters after they have become connected by the pole. What is their angular speed? By pulling on the pole, the skaters reduce their separation to 1.0 m. What is their angular speed then?

Explanation / Answer

Here ,

mass , m = 65 Kg

v = 1.3 m/s

d = 12.1 m

let the angular velocity is wi

Using conservation of angular momentum

I * w = 2 * m * v * (d/2)

(2 * m * (d/2)^2 * w ) = 2 * m * v * (d/2)

2 * 65 * (12.1/2)^2 * w = 2 * 65 * 1.3 * (12.1)

solving for w

w = 0.43 rad/s

the angular speed of the skaters is 0.43 rad/s

-------------------------------

let the anglar speed is wf

Using conservation of angular momentum

(2 * m * (d1/2)^2 * w ) = 2 * m * (d2/2)^2 * wf

(12.1/2)^2 * 0.43 = (1/2)^2 * wf

wf = 63 rad/s

the angular speed then is 63 rad/s

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