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## Homework Statement

A large wooden turntable in the shape of a flat uniform disk has a radius of 2.00m and a total mass of 140kg . The turntable is initially rotating at 2.00rad/s about a vertical axis through its center. Suddenly, a 65.0-kg parachutist makes a soft landing on the turntable at a point near the outer edge.

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

## Homework Equations

[tex]\tau[/tex]=Fd

[tex]\tau[/tex]=I[tex]\alpha[/tex]

[tex]\alpha[/tex]=v

^{2}/r

v=r[tex]\omega[/tex]

## The Attempt at a Solution

Okay this is how I tried to solve this (i could be way off)

[tex]\tau[/tex]=Fd

[tex]\tau[/tex]=(637)(2) < this is the weight of the parachutist multiplied by the radius

[tex]\tau[/tex] = 1274

[tex]\tau[/tex]=I[tex]\alpha[/tex]

1274 = mr

^{2}[tex]\alpha[/tex]

1274 = 205(2)

^{2}[tex]\alpha[/tex] < this is the torque i found in the previous equation. I added the mass of the the parachutist to the mass of the turntable times the radius squared and solved for [tex]\alpha[/tex]

[tex]\alpha[/tex]=1.5

[tex]\alpha[/tex]=v

^{2}/r

v = 1.76

v = r[tex]\omega[/tex]

[tex]\omega[/tex] = .88

Can someone steer me in the right direction? Tell me what I did wrong here? Or a easier approach?