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 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.

Answer 1

Answer:

Part a)

$L = 2683.2 kg m^2/s$

Part b)

$v' = 8.60 m/s$

Part c)

$W = 4326.7 J$

Explanation:

Part a)

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

so here we will use

$L = mv r + mvr$

$L = 2(78 \times 4.30 \times 4)$

$L = 2683.2 kg m^2/s$

Part b)

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

so we will have

$2(m v r) = 2( m v' \frac{r}{2})$

$v' = 2v$

$v' = 8.60 m/s$

Part c)

Work done by both of them = change in kinetic energy

so we have

$W = 2(\frac{1}{2}mv'^2 - \frac{1}{2}mv^2)$

$W = m(v'^2 - v^2)$

$W = 78(8.60^2 - 4.3^2)$

$W = 4326.7 J$