# Second Newton's law in rotation with pulley.

## Homework Statement

The rope doesn't slide on the pulley, it's mass is negligible and it doesn't stretch .The pulley's moment of inertia is 0.00300 kg * m^2 with a radius of 10 cm. As the first mass goes down the rope's friction generates a moment of force of 0.150 N*m opposed to the angular speed of the pulley. Initially , both masses are motionless and m1 is at a height of 1.5m(h = 1.5m).
m1 = 1.5 kg
m2 = 0.8 kg

https://gyazo.com/074104f2af4ac27811136c21b4e1bcfb

## Homework Equations

τ = r*F
τresultant = Σ(τ) = I* αz
Fresultant = ma
αz = aθ / r
-ay1 = ay2

## The Attempt at a Solution

T1 = -m1ay1 + m1g
T2 = m2ay2 + m2g
-T1 + T2 - 1.5 = I*ay1
replacing the values in the last equation with the first two equations and isolating a results 3.69 m/s2

the real solution is a1 = -2.06 m/s2

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The equations used by you are valid for massless pulley. As it is given that pulley has MI, which means it has mass. So please use energy conservation to solve this problemM1 loses gravitational energy m2 gains and both m1, m2 gain translational KE and pulley gains rotational KE = 0.5*I*(v/r)^2. If magnitude of acceleartion is ay then angular acceleration of pulley = ay/r. Please give it a try.

If you wish to apply, then Tr = torque = I*(ay//r)

If you wish to apply, then Tr = torque = I*(ay//r)
yes ,but what about the friction ?

haruspex