Two skaters, each of mass 50 kg , approach each other along parallel paths separ

Question

Two skaters, each of mass 50 kg, approach each other along parallel paths separated by 5.9 m. They have equal and opposite velocities of 1.6 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?

Explanation / Answer

here,

mass of each skater , m = 50 kg

velocity of skier , v = 1.6 m/s

distance between them , d = 5.9 m

a)

the angular speed , w = v/( d/2)

w = 0.54 rad/s

the angular speed is 0.54 m/s

(b)

new distance , d' = 1 m

let the angular speed be w'

using conservation of angular momentum

m * (d'/2) * w' = m *( d/2) * w

( 1/2) * w' = ( 5.9/2 ) *0.54

w' = 3.19 rad/s

the new angular speed is 3.19 rad/s

(c)

the ration of the final kinetic energy to the original kinetic energy ,

r = (0.5 * ( 2 * m* (d'/2)^2) * w'^2 ) / (0.5 * ( 2 * m* (d/2)^2) * w^2 )

r = ( (1/2)^2) * 3.19^2 ) / ( (5.9/2)^2) * 0.54^2 )

r = 1.003 :1

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