PHySICS!!!!?

yep: PHySICS!!!!?
A particular horizontal turntable can be modeled as a uniform disk with a mass of 200 g and a
radius of 20 cm that rotates without friction about a vertical axis passing through its center. The
initial angular speed of the turntable is 2.4 rad/s. A ball of clay, with a mass of 80 g, is dropped
from a height of 35 cm above the turntable. It hits the turntable at a distance of 10 cm from the
center, and sticks where it hits so that the clay and the turntable rotate together at a new angular
speed. Assuming the turntable is firmly supported by its axle so it remains horizontal at all times,
find the final angular speed of the turntable-clay system.
PLEASE EXPLAIN HOW TO DO IT!!

Answers and Views:

Answer by Ness
1.6 rad/s

Answer by FJSL
This is a conservation of angular momentum question.

Wbefore = angular speed before clay
Wafter = angular speed after clay
Jbefore = angular momentum before clay
Jafter = angular momentum after clay
Id = moment of inertia of disc
Ic = moment of inertia of clay
Rd = radius of disc
Rc = radius of disc
Md = mass of disc
Mc = mass of clay

Assuming no frictional losses anywhere in the system and a point mass of clay:

Id = (1/2)(Md)(Rd)^2 (from definition of moment of intertia of a solid disc)

Id = (1/2)(0.2 kg)(0.2 mt)^2 (converting to kilograms and meters because I can’t think in any other units)

Id = 0.004 kg-mt^2

Ic = (Mc)(Rc)^2 (from definition of moment of inertia of a point mass about a center point)

Ic = (0.08 kg)(0.10 mt)^2

Ic = 0.0008 kg-mt^2

Jbefore = (Id)(Wbefore) (from definition of angular momentum)

Jafter = (Id + Ic)(Wafter) (moments of inertia are added)

Jbefore = Jafter (from definition of conservation of angular momentum)

Substituting, rearanging and solving for Wafter gives:

Wafter = (Id)(Wbefore)/(Id + lc)

Wafter = (0.004 kg-mt^2)(2.4 rad/sec)/(0.004 kg-mt^2 + 0.0008 kg-mt^2)

Wafter = angular speed after clay = 2.0 radians/sec