What is the final rotational speed of the double rotating disks?

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The discussion revolves around calculating the final rotational speed of two disks, where one disk is initially rotating and the other is stationary. The moment of inertia for both disks is calculated using the formula I = (m*r^2)/2. The initial angular velocity of the first disk is converted from rpm to radians per second for calculations. The conservation of angular momentum is emphasized as the key principle, despite energy not being conserved due to friction. The final rotational speed can be determined by equating the initial and final angular momentum of the system.
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Homework Statement


A disk of mass M1 = 350 g and radius R1 = 10 cm rotates about its symmetry axis at f_initial = 152 rpm. A second disk of mass M2 = 258 g and radius R2 = 5 cm, initially not rotating, is dropped on top of the first. Frictional forces act to bring the two disks to a common rotational speed f_final.
a) What is f_final? Please give your answer in units of rpm, but do not enter the units

Homework Equations


T = I*a
I = (m*r2)/2

The Attempt at a Solution


I found the moment of inertia of M1
I = (.5)(.35)(.12) = .00175 kg*m2
M2
I = (.5)(.258)(.052) = .0003225 kg*m2
I found the angular velocity \omega
152rpm = 304pi rads/min = 5.06pi rads/sec
I don't suppose that I can just consider the mass to have just increased can I, because it specifies friction. I know some numbers, but how to put them together?
 
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You know that energy is not conserved because friction is acting and heat is being generated. However, something else is being conserved. What is it?
 
L = I*w and Li = Lf
Momentum would be conserved because the net torques act internally to the system. I get it, thanks for the hint.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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