Two friends hold on to a rope, one at each end, on a smooth, frictionless ice su

Question

Two friends hold on to a rope, one at each end, on a smooth, frictionless ice surface. They skate in a circle about an axis through the center of the rope and perpendicular to the ice. The mass of one friend is 83.0 kg and the other has a mass of 125.0 kg. The rope of negligible mass is 4.5 m long and they move at a speed of 4.40 m/s.

Two friends hold on to a rope, one at each end, on a smooth, frictionless ice surface. They skate in a circle about an axis through the center of the rope and perpendicular to the ice. The mass of one friend is 83.0 kg and the other has a mass of 125.0 kg. The rope of negligible mass is 4.5 m long and they move at a speed of 4.40 m/s. They now pull on the rope and move closer to each other so that the rope between them is now half as long. Determine the speed with which they move now. The two friends have to do work in order to move closer to each other. How much work did they do?

Explanation / Answer

b)

Initial moment of inertia=83*r^2+125*r^2 =208 r^2

Final moment of inertia =83*(r/2)^2+125*(r/2)^2=52 r^2

The angular momentum will remain conserved,

Initial momentum=(208 r^2)(4.4/r)

Let final velocity be v,

Final momentum =(52 r^2)(v/r)

Initial momentum= Final momentum

(208 r^2)(4.4/r)=(52 r^2)(v/r)

v=(208*4.4)/52=17.6 m/s

c)

Work done= Change in kinetic energy

Initial kinetic energy =I w^2=(208 r^2)(4.4/r)^2=(208)*4.4^2 =4026.88 J

Final Kinetic energy= I w^2=(52 r^2)(1.76*2/r)^2=(208)*1.76^2= 644.3008 J

Work Done=4026.88-644.3008=3382.5792 J

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