A large wooden turntable in the shape of a flat disk has a radius of 2.50 m and a total mass of 130 kg. The turntable is initially rotating at 3.30 rad/s about a vertical axis through its center. Suddenly, a 74.0 kg parachutist makes a soft landing on the turntable at a point on its outer edge.
A) Find the angular speed of the turntable after the parachutist lands. (Assume that you can treat the parachutist as a particle.)B) Compute the kinetic energy of the system before the parachutist lands. C) Compute the kinetic energy of the system after the parachutist lands. D) Why are these kinetic energies not equal?

a) 1.54 rad/s

b) 2212 J

c) 1030J

Explanation:

Given:

Radius  'r' = 2.5 m

Total mass of turnable 'M'= 130 kg

Speed of rotation of turntable ' w' = 3.3 rad/s

Mass of parachutist 'm' = 74 kg

A) Assuming the system of turntable and the parachutist. As we know that the system of turn table and parachutist is isolated and there is no external force on this system.

By applying law of conservation of momentum i.e the angular momentum is the same before and after landing. Therefore,

=

Where,

is moment of inertia of the disk and is moment of inertia of the parachute and the turntable

= 1/2MR² => 1/2 x 130 x 2.5²

= 65 x 6.25

= 406.25 kgm²

= 1/2MR² + mR² =>  406.25 + (74 x 2.5²)

= 406.25 + 74 x 6.25 =>406.25 + 462.5

= 868.75 kgm²

=

= /

 = (406.25 x 3.3) / 868.75 =>1340.625 / 868.75

 = 1.54 rad/s

B) For initial kinetic energy :

= 1/2² => 1/2 x 406.25 x 3.3²

= 1/2 x 406.25 x 10.89

= 2212 J

C)For final kinetic energy:

= 1/2² => 1/2 x 868.75 x 1.54²

= 1/2 x 868.75 x 2.3716

 = 1030 J

D) Because of the negative work done by the parachutist on the turntable, during the softlanding we have different KE. Energy is decreased due to friction between parachutist and the disc.

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.