## Circular Platform

A horizontal circular platform (M = 112.1 kg, r = 3.11 m) rotates about a frictionless vertical axle. A 95.3 kg student walks slowly from the rim of the platform toward the center. The angular velocity of the system is 2.1 rad/s when the student is at the rim.

Find the angular velocity of the system when the student is 1.39 m from the center.

4.23 rad/s

Find the angular velocity of the system when the student is 1.39 m from the center.

4.23 rad/s

## Discussion

**NEW**

*Anonymous 1*

How do you do this problem?

**NEW**

m= mass of student

M= mass of platform

r1= radius of platform

r2= new radius (m from center)

w1= angular velocity

w2= ??? (new angular velocity)

Solve for 1st I of the student at rim using

1/2(m*r^2) + M*r^2

[Mine was 1333.29 kg*m^2]

Find the 2nd I of student at new location

1/2(m*r^2) + M*r2^2 <- 2nd half uses r2 not r

[Mine was 570.13 kg*m^2]

Then solve for w2

w2=I1*w1/I2

[Mine was 7.72 rad/s]

Hope that helps :3

**NEW**

**Help plz**

Could anybody tell me what I'm doing wrong, I'm following

all the right steps and nothing is working.

M=73.1

m=65.3

r=3.11

r2=1.27

wi=4.1

Ii=.5(73.1(3.11)^2 + (65.3)(3.11)^2

=981

If= 5(73.1(.3.11)^2 + (65.3)(1.27)^2

=448.8

wf=4.1(981/448.8)=8.96rad/s

but it is wrong...

**NEW**

the equations are correct but the masses are switch: for I1 the equation is

1/2(M*r^2)+m*r^2

For I2 the equation is

1/2(M*r^2)+m*r2^2

then use

w2=I1w1/I2

to get the answer

**NEW**

Shoot, sorry about that! =/

**NEW**

Thanks Deliah and Greg! That was a big help!