In the figure, two skaters, each of mass 46.8 kg, approach each other along para

Question

In the figure, two skaters, each of mass 46.8 kg, approach each other along parallel paths separated by 4.26 m. They have opposite velocities of 1.65 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 3.07 m. What then are (d) their angular speed and (e) the kinetic energy of the system?

Explanation / Answer

d=4.26m, m1=m2=m=46.8kg, v=v1=-v2=1.65m/s

a)

ri =d/2=4.26/2=2.13m

b)

w=v/r =1.65/2.13 = 0.775 rad/s

KEnet = KE1+KE2 = ½*m1*v1^2+1/2*m2*v2^2=2*1/2*mv^2 = mv^2 = 46.8*1.65^2 = 127.4 J

d)

rf =3.07/2 = 1.535 m

By law of angular momentum,

Lf=Li

Ii*wi = If*wf

(2*46.8*2.13^2)*0.775 = (2*46.8*1.535^2)*vf

wf=1.49 rad/s

e)

vf=wf*rf = 1.49*1.535 = 2.29 m/s

KEnet = KE1+KE2 = ½*m1*v1f^2+1/2*m2*v2f^2=2*1/2*mvf^2 = mv^2 = 46.8*2.29^2 = 245.42 J

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