Question

A uniform 2.00-kg circular disk of radius 20.0 cm is rotating clockwise about an axis through its center with an angular speed 30.0 revolutions per second. A second uniform 1.50-kg circular disk of radius 15.0 cm that is not rotating is dropped onto the first disk so that the axis of rotation of the first disk passes through the center of the second disk. What is the final angular speed of the two disks when they are rotating together

Answer 1

Answer:

   w = 132.57 rad / s

Explanation:

To solve this exercise we must define a system formed by the two discs, therefore the angular momentum is conserved

initial

         L₀ = I₀ w₀

final

         L_f = I w

how the moment is preserved

         L₀ = L_f

         I₀ w₀ = I w

         

the moment of inertia of disk 1 is

         I₀ = ½ m₁ r₁²

The moment of inertia of the set is

        I = I₀ + I₂

        I = ½ m₁ r₁² + ½ m₂ r₂²

we substitute

       ½ m₁ r₁² w₀ = (½ m₁ r₁² + ½ m₂ r₂²) w

       w = $\frac{ m_1r_1^2 }{ m_1 r_1^2+ m_2r_2^2} \omega_o$

let's reduce the magnitudes to the SI system

        w₀ = 30.0 rev / s (2π rad / 1 rev) = 188.5 rad / s

let's calculate

        w = ( $\frac{ 2 \ 0.20^2}{ 2 \ 0.20^2 + 1.5 \ 0.15^2 }$) 188.5

        w = ( $\frac{0.08}{ 0.11375}$  ) 188.5

        w = 132.57 rad / s