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 85.0 kg. The rope of negligible mass is 3.0 m long and they move at a speed of 5.00 m/s. (a) What is the magnitude of the angular momentum of the system comprised of the two twins? 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. 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? J

Answer 1

Answer with Explanation:

We are given that

Mass of each twin=85 kg

Length of rope=l=3 m

$r=\frac{l}{2}=\frac{3}{2}=1.5 m$

Initial speed,u=5 m/s

a.Angular momentum of the system=$l=2mvr$

Using the formula

$l=2\times 85\times 5\times 1.5=1275kgm^2/s$

b.According to law of conservation of angular momentum

$L_f=L_i$

$L_i=1275 kgm^2/s$

$r'=\frac{1.5}{2}=0.75 m$

$L_f=2mvr'$

$2(85)(1.25)v=1275$

$v=\frac{1275}{2(85)\times 0.75}=10 m/s$

c.According to work energy theorem

Work done by twins=Change in kinetic energy

$W=2\times \frac{1}{2}m(v^2-u^2}=85((10)^2-(5)^2)=6375 J$