In the figure, two skaters, each of mass 48.3 kg, approach each other along para

Question

In the figure, two skaters, each of mass 48.3 kg, approach each other along parallel paths separated by 3.99 m. They have opposite velocities of 1.34 m/s each. One skater carries one end of a long pole of negligible mass, and the other skater grabs the other end as she passes. The skaters then rotate about the center of the pole. Assume that the friction between skates and ice is negligible. What are (a) the radius of the circle, (b) the angular speed of the skaters, and (c) the kinetic energy of the two-skater system? Next, the skaters each pull along the pole until they are separated by 1.99 m. What then are (d) their angular speed and (e) the kinetic energy of the system?

Explanation / Answer

given that

m = 48.3 kg

distance b/w skaters d = 3.99 m

v =1.34 m/s

part(a)

Rotation is about their combined centre of mass so center of mass equals to radius of circle .and they have equal masses .

so the mid-point of the bar is the center of mass of this system .

COM = r = 3.99 / 2 = 1.995 m

part(b)

for each skater ,linear momentum (mv) transfers to angular momentum (Iw) .

m*v = I*w

I = moment of inertia as a point mass at radius r = m*r^2

w = angular velocity ,so

m*v = m*r^2*w

put the values in above equation

48.3 *1.34 = 48.3 *(1.995)^2 *w

w = 0.33 rad /sec

part(c)

we know that rotational kinetic energy

KE(rotational) = 1/2*I*w^2 = 1/2*m*r^2*w^2

KE(rotational) = 1/2*48.3*(1.995)^2 * (0.33)^2

KE(rotational) = 10.46 J

kinetic energy of two skater system = 2* KE(rotational)

kinetic energy of two skater system = 2*10.46 = 20.92 J

part(d)

now the distance b/w skaters = 1.99m

so the center of mass = 1.99/2 = 0.995 m

which is equal to radius of circle so R = 0.995 m

from the conservation of momentum

initial momentum = final momentum

Ii*wi = If*wf

m*r^2*w^2 = m*R^2*wf

initial radius of circle is r and anglular velocity is w.

final radius of circle is R and anlular velocity is wf.

put the values

(1.995)^2 *0.33 = (0.995)^2 *wf

wf = 1.326

part(e)

kinetic energy of system = 2*1/2*I*wf^2

kinetic energy of system = m*R^2*wf^2

kinetic energy of system = 48.3*(0.995)^2*(1.326)^2

kinetic energy of system = 84.07 J

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