Question

A uniform disk turns at 4.1 rev/s around a frictionless spindle. A nonrotating rod, of the same mass as the disk and length equal to the disk's diameter, is dropped onto the freely spinning disk, see the figure. They then turn together around the spindle with their centers superposed.

What is the angular velocity of the combination?

State fundamental principle used, objects in the system, and write out equation first

Answer 1

The formula for angular momentum is:
L = Iw
Conservation of angular momentum says that L before = L after:
I₁w₁ = I₂w₂

The moments of inertia are listed on p. 223 in your text, a disk or cylinder through the center is:
I = ¹/₂mr²

And a thin rod through the center is
I = ¹/₁₂ml²
and the rod is 2r long, so the moment of inertia in terms of r is:
I = ¹/₁₂m(2r)²
I = ⁴/₁₂mr² = ¹/₃mr²

So the moment of inertia before the rod is dropped on is just the disk:
I₁ = ¹/₂mr²

And after it the rod is dropped on, it is both moments:
I₂ = ¹/₂mr² +   ¹/₃mr² = ⁵/₆mr²

So plug these into:
I₁w₁ = I₂w₂

(¹/₂mr²)(7.0 rev/s) = (⁵/₆mr²)w₂
(¹/₂)(4.1 rev/s) = (⁵/₆)w₂
w₂ = 2.46 rev/s