Does friction torque in speeding up = speeding down?

[Anak Soleh]

Homework Statement

'm trying to calculate moment of inertia of a pulley connected to a shaft mounted on bearing by using free fall method. A load (270 g) is tied to a string and the string is wrapped around pulley.
The load is dropped from 1,6 m height then i measure rotation speed of pulley when the load is going down, plot the data and get acceleration from the slope (α up). I also measure rotation speed after the load is already on the ground until it is stopped and get the deceleration (α down).

Relevant Equations

Στ=Iα

On speeding up:

τload - τfriction = Iαup

On speeding down:

τfriction = Iαdown

If i substitute τfriction from speeding down to speeding up equation, i get moment of inertia:

I = (τload)/(αup+αdown)

But, is this allowed? Does friction torque in speeding up is equal to friction torque in speeding down?

Additional info: the speeding up range from 0 to 38 rad/s and the speeding down from 38 rad/s to 0 rad/s.

 

[haruspex]

Does friction torque in speeding up is equal to torque in speedingdown?

It's a reasonable assumption.

 

[Anak Soleh]

It's a reasonable assumption.

Thanks. That's assuring.