In the figure below, two skaters, each of mass 55 kg, approach each other along

Question

In the figure below, two skaters, each of mass 55 kg, approach each other along parallel paths separated by 2.5 m. They have opposite velocities of 1.8 m/s each. One skater carries one end of a long pole with negligible mass, and the other skater grabs the other end as she passes. The skaters then rotate around the center of the pole. Assume that the friction between skates and ice is negligible. (a) What is the radius of the circle? (b) What is the angular speed of the skaters? (c) What is the kinetic energy of the two-skater system? Next, the skaters pull along the pole until they are separated by 1.1 m. (d) What then is their angular speed? (e) What then is the kinetic energy of the system? (f) What provided the energy for the increased kinetic energy? (Select all that apply.) friction gravity

Explanation / Answer

Here ,

a)

radius of the circle = distance between them/2

radius of the circle = 2.5/2

radius of the circle = 1.25 m

b)

let the angular speed is w

v = r * w

1.8 = 1.25 * w

w = 1.44 rad/s

the angular speed of the skator is 1.44 rad/s

c)

kinetic energy of the skators = 2 * 0.50 * m * v^2

kinetic energy of the skators = 2 * 0.50 * 55 * 1.8^2

kinetic energy of the skators = 178.2 J

d)

new radius , r2 = 1.1/2 = 0.55 m

let the final angular speed is w2

Using conseravtion of angular momentum

2 * 55 * 1.25^2 * 1.44 = 2 * 55 * 0.55^2 * wf

wf = 7.44 rad/s

e)

new kinetic energy of the system = 0.50 * I * wf^2

new kinetic energy of the system = 0.50 * 2 * 55 * 0.55^2 * 7.44^2

new kinetic energy of the system = 921 J

f)
work is done by frictional force

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