(hrwc11p47) Two skaters, each of mass 65 kg , approach each other along parallel

Question

(hrwc11p47) Two skaters, each of mass 65 kg, approach each other along parallel paths separated by 4.3 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?

Calculate the ration of the final kinetic energy to the original kinetic energy.

Explanation / Answer

here

ma = mb = 65 kg
D = 4.3 m
v = 1.3m/s

We assume that from the moment of grabbing the stick onward, they maintain rigid postures so that the system can be analyzed as a symmetrical rigid body with center of mass midway between the skaters

Angular Speed of System will :

w1 = v/r
w1 = 1.3/2.15
w1 = 0.604 rad/s

Angular speed when d = 4.3 - 1 = 3.3
w2 = v/r
w2 = 4.3/(3.3/2)
w2 = 2.606 rad/s

For kinetic energy ratios :

Moment of inetia of rod :
I1 = Mr^2
I1 = 2 * 65 * (4.3/2)^2
I1 = 600.925 kg.m^2

Ki = 0.5 * I1 * w1^2
Ki = 0.5* 600.925 * 0.604^2
Ki = 109.613 J

when d is reduced = 4.3 -1 = 3.3 m
I2 = 2mr^2
I2 = 2*65*(3.3/2)^2
I2 = 353.925 kg.m^2

kf = 0.5 * I2 *w2^2
kf = 0.5 * 353.925 * 2.606^2
kf = 1201.794 J

Therefore
K = 1201.794 /103.613
k = 111.6 J

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