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 78.0 kg. The rope of negligible mass is 4.0 m long and they move at a speed of 4.30 m/s.
(a) What is the magnitude, in kg · m2/s, 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, in m/s, 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, in joules, did they do?
____J.

Question

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Respuesta :

aristocles aristocles · Answer 1 · 2019-12-01T14:04:38+01:00

Answer:

Part a)

[tex]L = 2683.2 kg m^2/s[/tex]

Part b)

[tex]v' = 8.60 m/s[/tex]

Part c)

[tex]W = 4326.7 J[/tex]

Explanation:

Part a)

As we know that there is no external torque on the system of two twins

so here we will use

[tex]L = mv r + mvr[/tex]

[tex]L = 2(78 \times 4.30 \times 4)[/tex]

[tex]L = 2683.2 kg m^2/s[/tex]

Part b)

Since angular momentum is conserved here as there is no external torque

so we will have

[tex]2(m v r) = 2( m v' \frac{r}{2})[/tex]

[tex]v' = 2v[/tex]

[tex]v' = 8.60 m/s[/tex]

Part c)

Work done by both of them = change in kinetic energy

so we have

[tex]W = 2(\frac{1}{2}mv'^2 - \frac{1}{2}mv^2)[/tex]

[tex]W = m(v'^2 - v^2)[/tex]

[tex]W = 78(8.60^2 - 4.3^2)[/tex]

[tex]W = 4326.7 J[/tex]