Two skaters, each of mass 55 kg, approach each other along parallel paths separa

Question

Two skaters, each of mass 55 kg, approach each other along parallel paths separated by 10.9 m. They have equal and opposite velocities of 2.1 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.

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

Part a

As there is no net external force acting on system(2 skaters) so from Newton’s second law –

            F=ma

so        a=0

acceleration of center of mass of 2 skaters is zero .

so they will move in a circle centered at C.M. after they have become connected by pole.

Part a

Mass of each skaters, m=55kg

Speed of each skaters, v=2.1 m/s

Radius of circle, r=10.9/2=5.45 m.

Angular speed, =v/r

            =2.1/5.45 =0.385 rad/s.

so, Angular speed is 0.385 rad/s

Part b

Radius of new circle, r’=1.0/2=0.5 m.

As there is no external force on the system,

So, from law conservation of angular momentum-

            mr2=m’r’2

                        ’=r2/r’2

            ’=0.385*(5.45/0.5)2 = 45.74 rad/s.

            ’=45.74 rad/s.

Part c

As kinetic energy is(K) =( ½) I2

So

            (Kf/Ki)= ( ½) I’’2 /( ½) I2 =r’2 ’2/r2 2

                        =0.52*45.742/5.452*0.3852 =118.80

Ratio of final kinetic energy to original kinetic energy is =118.80

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