Question

A person of mass 80 kg stands at the center of a rotating merry-go-round platform of radius 3.5 m and moment of inertia 950 kg*m^2 . The platform rotates without friction with angular velocity 0.85 rad/s . The person walks radially to the edge of the platform.
1. Calculate the angular velocity when the person reaches the edge.
w=______________ rad/s
2. Calculate the rotational kinetic energy of the system of platform plus person before the person's walk.
Ki=____________ J
3. Calculate the rotational kinetic energy of the system of platform plus person after the person's walk.
Kf=__________ J

Answer 1

Answer with Explanation:

We are given that

Mass of person,M=80 kg

Radius,r=3.5 m

Moment of inertia of merry,I=$950 kgm^2$

Angular velocity of platform=$\omega=0.85rad/s$

1.Law of conservation of angular momentum

$I\omega=I'\omega'$

Where$I'=950+80(3.5)^2$

Substitute the values

$950\times 0.85=(950+80(3.5)^2)\omega'$

$\omega'=\frac{950\times 0.85}{950+80(3.5)^2}=0.418rad/s$

Angular velocity when the person reaches the edge =0.418 rad/s

2.Rotational kinetic energy of the system  before the person's walk

$K_i=\frac{1}{2}I\omega^2=\frac{1}{2}(950)(0.85)^2=343.2 J$

3.Rotational kinetic energy of the system  after the person's walk

$K_f=\frac{1}{2}(950+80(3.5)^2)\times (0.418)^2$

$K_f=168.6 J$