Question

A large wooden turntable in the shape of a flat uniform disk has a radius of 2.00 m and a total mass of 120 kg. The turntable is initially rotating at 2.00 rad/s about a vertical axis through its center. Suddenly, a 75.0-kg parachutist makes a soft landing on the turntable at a point near the outer edge.

Part A: Find the angular speed of the turntable after the parachutist lands. (Assume that you can treat the parachutist as a particle.)

Part B: Compute the kinetic energy of the system before the parachutist lands.

Part C: Compute the kinetic energy of the system after the parachutist lands.

Part D: Why are these kinetic energies not equal?

Answer 1

We will use angular momentum conservation because the net torque is zero.

a) Initial angular momentum = Final angular momentum

$I_1\omega _1=I_2\omega _2\\ (120 \times 2^2 \times 0.5)\times 2 = (120 \times 2^2 \times 0.5+75 \times 2^2)\times \omega _2\\ {\color{Red} \omega _2 = 0.89\ \tfrac{rad}{s}}$

b) We will use the formula for rotational kinetic energy:

$K.E = 0.5 \times I_1 \omega ^2_1\\ K.E = 0.5 \times 120 \times 2^2 \times 0.5 \times 2^2\\ {\color{Red} K.E = 480 \ J}$

c) We will use the same formula as part b, but will new angular speed and moment of inertia.

$K.E = 0.5 \times I_1 \omega ^2_1\\ K.E = 0.5 \times (120 \times 2^2 \times 0.5+75 \times 2^2) \times 0.89^2\\ {\color{Red} K.E = 213.867 \ J}$

d) The two kinetic energies are not equal because when the person lands on the lands on the turntable, there is some energy loss because of the friction between the turntable and the person.