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# Motion of object, angular speed, and amount of work done?

A woman whose mass is 54.7 kg stands at the rim of a horizontal turntable which has a moment of inertia of

338 kg m^2 about the axis of rotation and a radius of 2.26 m. The system is initially at rest and the turntable is free to rotate about a frictionless, vertical axle through its center. The woman then starts walking around the rim in a clockwise direction (viewed from above) at a constant speed of 0.873 m/s relative to the ground.

What will the motion of the turntable be, relative to the ground?

(pick best one)

1. at rest, non-rotating

2. rotating counterclockwise

3. rotating clockwise

What is its angular speed? Answer in units of rad/s.

How much work does the woman do to set the system into motion? Answer in units of J.

Which one is the angular speed? it's saying that 0.37 isn't correct, but that's how i worked it out as well

### 1 Answer

- odu83Lv 71 decade agoFavorite Answer
Even though the woman does work on the system the system will obey conservation of momentum.

So Lw+Lt=0

where Lw is the angular momentum of the woman and Lt is the angular momentum of the turntable.

since Lw=-Lt and

L=I*w

then the turntable rotates counter clockwise.

we know here angular speed to be

.873/2.26

=0.386 rad/sec

I will treat her moment of inertia as a point mass

54.7*2.26=338*w

where w is the angular speed of the turntable

w=54.7*2.26/338

=0.366 rad/sec counter clockwise

The energy can be calculated looking at the total rotational kinetic energy of the system.

Rotational energy is

.5*I*w^2

since this is a magnitude, simply add the turntable and woman energy together

.5*(54.7*2.26*0.386^2+

338*0.366^2)

=31.85 J

j