Two identical twins hold on to a rope, one at each end, on a smooth, frictionles

Question

Two identical twins hold on to a rope, one at each end, on a smooth, frictionless ice surface. They skate in a circle about the center of the rope (the center of mass of the two-body system) and perpendicular to the ice. The mass of each twin is 76.0 kg. The rope of negligible mass is 2.5 m long and they move at a speed of 5.00 m/s center of mass (a) What is the magnitude of the angular momentum of the system comprised of the two twins? Treat the two twins as point particles. kg·m2/s (b) They now pull on the rope and move closer to each other so that the rope between them is now half as long of Determine the speed with which they move now m/s (c) The two twins have to do work in order to move closer to each other. How much work did they do? Additional Materials

Explanation / Answer

(a) angular momentum of system= angular momnetum of two twins

they have same angular momentum, so L= 2 m vr

r= 2.5/2 =1.25

L = 2 (76)(5)(1.25) = 950 kgm2 /s

(b) when the length of rope reduces to half, then r= 1.25/2 = 0.625 m

let Vf be the final speed

using law of conservation of angular momentum

Lf= Li

2(76)(0.625)(Vf) = 950

Vf= 10 m/s

(c) work done= change in kinetic energy

w= Kf- Ki

w= 2(1/2 mVf2 -1/2m Vi2 )

w= m(Vf2 -Vi2 )

w= 76{ (10)2 - (5)2 }

w= 5700 joules

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