Question

Disk A, with a mass of 2.0 kg and a radius of 70 cm , rotates clockwise about a frictionless vertical axle at 50 rev/s . Disk B, also 2.0 kg but with a radius of 50 cm , rotates counterclockwise about that same axle, but at a greater height than disk A, at 50 rev/s . Disk B slides down the axle until it lands on top of disk A, after which they rotate together. After the collision, what is their common angular speed (in rev/s) and in which direction do they rotate?

Answer 1

w = - 197.5 rad / s

The negative sign indicates that the rotations are clockwise

Step-by-step explanation:

To solve this exercise, let's use the concept of conservation of the angular number.

We create a system formed by the two discs, in this case the forces last the shock are internal

initial instant .. just before shock

L₀ = I₀ w₀ + I₁ w₁

instnte final. Right after crash

L_f = (I₀ + I1) w

angular momentum is conserved

I₀ w₀ + I₁ w₁ = (I₀ + I₁) w

w = I₀ w₀ + I₁ w₁ / Io + I1

The moment of inertia of a disk with an axis passing through its thermometric center

I₀ = ½ m² r₀²

I₁ = ½ m₁ r₁²

we substitute

I₀ = ½ 2.0 0.70²

I₀ = 0.49 kg m

I₁ = ½ 2.0 0.5²

I₁ = 0.25

₁

let's reduce the magnitudes the SI system

w₀ = -50 rev / (2pi rad / 1rev) = -314.15 rad / s

w₁ = 70 rev (2pi rad / 1rev) = 439.82 rad / s

we will assume that the counterclockwise turns are positive

w = -0.49 314.15 + 0.25 439.82 / (0.49 + 0.25)

w = (- 4.696 + 1.0995) 102) / 0.74

w = -197.75 + 0.25

w = - 197.5 rad / s

The negative sign indicates that the rotations are clockwise