Two skater, each of mass 75 kg, approach each other along parallel paths speedst

Question

Two skater, each of mass 75 kg, approach each other along parallel paths speedster by 6.1m. They have equal velocities of 2.2 m/s. The first skater carries one end of a long pole with negligable mass, and the second skater grabs the other end of it as she passes. Assume frictionless ice. Describe quantitatively the motion of the skaters after they have become connected by the pole. A) What is their angular speed? B) By pulling on the pole, the skaters reduce their separation to 1.0 m. What is their angular speed then? C) Calculate the ratio of the final kinetic energy to the original kinetic energy. Two skater, each of mass 75 kg, approach each other along parallel paths speedster by 6.1m. They have equal velocities of 2.2 m/s. The first skater carries one end of a long pole with negligable mass, and the second skater grabs the other end of it as she passes. Assume frictionless ice. Describe quantitatively the motion of the skaters after they have become connected by the pole. A) What is their angular speed? B) By pulling on the pole, the skaters reduce their separation to 1.0 m. What is their angular speed then? C) Calculate the ratio of the final kinetic energy to the original kinetic energy. Two skater, each of mass 75 kg, approach each other along parallel paths speedster by 6.1m. They have equal velocities of 2.2 m/s. The first skater carries one end of a long pole with negligable mass, and the second skater grabs the other end of it as she passes. Assume frictionless ice. Describe quantitatively the motion of the skaters after they have become connected by the pole. A) What is their angular speed? B) By pulling on the pole, the skaters reduce their separation to 1.0 m. What is their angular speed then? C) Calculate the ratio of the final kinetic energy to the original kinetic energy.

Explanation / Answer

relation between angular speed and lnear speed

v= rw

w= v/r

= 2.2/3.05

= 0.721 rad/s

after pulling rope

L=I

I=I1

mR2=mR121

1=R2/R12=(6.1/2)20.721/(1/2)2

1=26.82rad/sec

initial kinetic energy

KE i = 1/2 I wi^2

= 1/2 ( 2 mr^2) wi^2

= ( 75)( 3.05)^2 ( 0.721)^2

=362.68 J

KEf = 1/2 ( 2 mr^2) wf^2

= ( 75)( 2.55)^2 ( 26.82)^2

=350909.66 J

KEf/KEi = 967.54

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