Finding the angular velocity when the mass hits the floor. Two hanging masses ar

Question

Finding the angular velocity when the mass hits the floor.

Two hanging masses are suspended from a massive disk by a rope, wound aroung the disk. The moment of inertia of a disk obeys the equation I = (1/2)M(R^2). The hanging masses have m1 = 0.14 kg, m2 = 0.21 kg. the disk has a mass M = 0.86 kg and radius R= 0.25 meter.

I only need help finging the angular velocithy of the disk, when mass two hits the floor in part b!

Part b says, starting from rest, m2 descendes 0.2 meter to the floor. How fast is it moving when it reaches the floor? I got 0.493 m/s

I was thinking angular velocity =

(1/R)(sqr rt.( (2(m2*g*d - m1*g*d))/(m1+m2 + (1/2)*Mpulley))

if I use that would the D(distance for m2 be 0.2 meters and for m1 be -0.2 meters)?

Explanation / Answer

For m1: T1 - m1g = m1a

For m2: m2g - T2 = m2a

For pulley: T2 - T1 = I = (MR2/2)(a/R) = MRa/2

Adding first two equations, we get,

T1 - T2 = (m1 + m2)a +(m1 - m2)g

So,

MRa/2 + (m1 + m2)a + (m1 - m2)g = 0

=> a = (m2 - m1)g / (MR/2 + m1 + m2) = 1.5 m/s2

Time taken to reach ground, t = (2s/a)1/2 = (2 * 0.2 / 1.5)1/2 = 0.52 s

Angular velocity of pulley, = t =at/R = 1.5 * 0.52 / 0.25 = 3.1 rad/s2

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