How Does Dropping Clay Affect a Potter's Wheel's Angular Speed?

[math34]

Homework Statement

A potter's wheel, with rotational inertia 6.40 kg * m^2, is spinning freely at 19.0 rpm. The potter drops a 2.80 kg lump of clay onto the wheel, where it sticks a distance of 47.0 cm from the rotation axis.

Homework Equations

I know I need to use angular momentum here, but I am a little confused..

The Attempt at a Solution

conver to SI units:

19 rpm * (2 pi radians/1 revolution) * ( 1 minute / 60 seconds) = 1.99 radians
radius = 0.47 meters
mass = 2.8 kg

I know I need to use angular momentum: L = Iw

and i think this has to do with conservation of momentum but I need some direction.

so the total angular momentum needs to be the sum of the angular momentum before and after the lump is dropped on the wheel,

w(inital)mv^2 + Iw(final)= Iwfirst term being the angular momentum before the drop so we can replace the moment of inertia by mv^2 since it is a disk

second term being after the lump is dropped on the disk

the sum of these is the total angular momentum, is this on the right approach?
something doesn't seem quite right to me..
Thanks

 

[Redbelly98]

The Attempt at a Solution

conver to SI units:

19 rpm * (2 pi radians/1 revolution) * ( 1 minute / 60 seconds) = 1.99 radians
radius = 0.47 meters
mass = 2.8 kg

I know I need to use angular momentum: L = Iw

and i think this has to do with conservation of momentum but I need some direction.

so the total angular momentum needs to be the sum of the angular momentum before and after the lump is dropped on the wheel,

w(inital)mv^2 + Iw(final)= Iw


first term being the angular momentum before the drop so we can replace the moment of inertia by mv^2 since it is a disk

There are 2 problems with that:
1. They tell you the moment of inertia for the potter's wheel. Just use that value for I, there is no need to calculate it.
2. For future reference: the moment of inertia of a disk is (½)mr². v is irrelevant.

second term being after the lump is dropped on the disk

the sum of these is the total angular momentum, is this on the right approach?
something doesn't seem quite right to me..
Thanks

It sounds like you are adding the initial angular momentum to the final angular momentum, and calling that "the total angular momentum". It doesn't work that way.

Conservation of angular momentum means that it doesn't change between the initial and final situations. So the equation should look more like this:

L_initial = L_final​

Either side of the equation can be called the total angular momentum.

As a start, go ahead and calculate what L_initial is.