How to Calculate Angular Speed of a Rotating Turntable with Added Mass

[elsternj]

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

\tau=Fd
\tau=I\alpha
\alpha=v²/r
v=r\omega

The Attempt at a Solution

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

\tau=Fd
\tau=(637)(2) < this is the weight of the parachutist multiplied by the radius
\tau = 1274

\tau=I\alpha
1274 = mr²\alpha
1274 = 205(2)²\alpha < 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 \alpha
\alpha=1.5

\alpha=v²/r
v = 1.76

v = r\omega
\omega = .88

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

 

[RobL14]

The weight of the parachutist doesn't really exert a torque on the system about the rotation axis. Why? The force vector points perpendicularly into the disk, so there is no component of his weight that contributes to rotational torque about that central axis.

The key to this problem is angular momentum, which is conserved because there are only internal torques acting on the system. Remember: angular momentum is conserved about an inertial rotational axis provided that no external torque acts on the system about that axis.

So, now that you know Li = Lf, take it from here.

 

[ideasrule]

\tau=Fd
\tau=(637)(2) < this is the weight of the parachutist multiplied by the radius
\tau = 1274

The F in \tau=Fd is the component of the force that goes into rotating the object. The weight of the parachutist pushes directly downwards, and does not directly apply any torque. Instead, the turntable tries to accelerate the parachutist after he touches down, and static friction between him and the turntable applies the torque.

However, the forces involved are much too complicated to model. Instead, try using the conservation of angular momentum, since there are no sources of external torque acting on the turntable+parachutist system.

EDIT: oops, sorry Rob

 

[elsternj]

okay that seems simple enough but still not coming up with the right answer.

L = I\omega
I = mr²/2

L = (140)(2)²/2(2)
L = 560

560 = mr²/2(\omega)
560 = (205)(2)²/2(\omega)
\omega = 1.37

Not the right answer

I also used the equation L = r x mv (v = r\omega) for both and I got the same answer. any insight?

 

[RobL14]

Your initial L is correct as far as I can tell. It is your final L that is the problem.

The new moment of inertia is not simply (m+M)r². Treat the person as a point particle, which will have the same final angular velocity as the disk, whose moment of inertia is the same. You treat them separately, because the disk and the person each have angular momentum about the same rotation axis. What you've done there is treated it as if the disk has simply become more massive, which is not the case (unless the person's body splattered and spread uniformly across the disk, but that'd be pretty grotesque).

 

[elsternj]

having trouble finding the moment of inertia once the parachutist lands then. I = \summr²

1120 = (140)(2)² + (65)(2)²\omega

and the answer still comes out to be 1.37. Wouldn't this be treating it as a 2 particle system?

 

[elsternj]

actually i just tried mr^2 / 2 (disk) + mr^2 (parachutist) and got the right answer.

this the right way?