Two disks of the same thickness and same material are attached to a shaft as sho

Question

Two disks of the same thickness and same material are attached to a shaft as shown. The 8 lb disk A has a radius of 3 in, and disk B has a radius of 4.5 in(weight no given). Knowing that a counter-clockwise couple M of magnitude 20 lb*in is applied to disk A when the system is at rest, determine the time required for the angular velocity of the system to reach 960 rpm

I've seen the same question asked but I am curious if you can solve with Priciple of Impulse and Momentum. Any help is apreciated

Explanation / Answer

Mass of disk A = 8 lb

let mass of disk B be m

As disk A and disk B are of same material , so will have same density .

mass / volume = constant for both disks

volume = area of cross-section * thickness . they both have same thickness.

area of cross-section of disk a = π*32

area of crossection of disk b = π*4.52

so 8 /(9Ï€) = m/(Ï€*4.52 )

mass of disk B , m = 4*4.5 = 18 lb

Total mass moment of inertia of system about shaft = mass moment of inertia of disk A + disk B about the shaft ( I = mr2 /2)

I = Ia + Ib = 8*32 /2 + 18*4.52 /2 = 218.25 lb-in2

Moment applied = mass moment of inertia * angular acceleration

M = I α

α = M/I = 20 / 218.25 = 0.091638 rad/sec2

α = dω /dt

as starting from rest,

ω = ωo + α*t

ωo = 0 as it starts from rest

ω = 2*π*960/60 = 100.53 rad/sec

time ,t = ω / α = 100.53 / 0.091638 = 1097.03 secs = 18.283 minutes

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