Masses, pulley, friction and rotation

[LCSphysicist]

Homework Statement

Two masses of 9 kg and 1 kg, hang from the ends of a string that passes around a pulley of mass 40kg and radius 0.5m (I=MR²/2) as shown in the diagram. The system is released form resst and the 9kg mass drops, starting te pulley rotating

If the coefficient of friction between the string and the pulley is 0.2 what is the least number of turns that the string must make around the pulley to prevent slipping?

Relevant Equations

All i cound find is:
T2-M2g = m2*a2
T1-M1g = -m1*a2
(T2 + f - T1) = M*a2/2
T2 = T1*e^(μ*θ)

In summarize, i have four equations and five incognits.
T2,T1,theta,a2,f

Need to find one more equation, but i don't know how

 

[haruspex]

(T2 + f - T1) = M*a2/2

Please explain this equation. In particular, be precise about the subsystem it applies to.

 

[LCSphysicist]

Please explain this equation. In particular, be precise about the subsystem it applies to.

That equation says about the torque applied on the pulley, i adopt the system coordinates ordinary xyz with z pointing vertically.
assuming the friction force in the direction of my drawn, we have:

(T2*R + f*R - T1*R) = I*alpha = 0.5*M*R²*(a2)/R
We can cut R
T2 + f - T1 = 0.5M*a2

 

[etotheipi]

Friction in pulley questions can be a sightly slippery concept (you're welcome for the pun, I take payment in cash). This paper has a good rundown on the effect of friction at a pulley with a massless string. Section 4 is about torques in this context, which appears to be useful for your question.

The tensions themselves do not act on the pulley, only the normal reactions and frictional forces between the string and the pulley. Of these, only the frictional forces produce a torque about the axle. It turns out that the torque of all these frictional forces happens to equal the torque of the difference in the tensions on either side of the pulley.

As a "trick", you can also get the exact same result by considering the pulley + the section of string in contact with the pulley as one system, with $T_1$ and $T_2$ as the only external forces. This system would have the same moment of inertia as just the pulley, since the string is massless.

 

[haruspex]

That equation says about the torque applied on the pulley, i adopt the system coordinates ordinary xyz with z pointing vertically.
assuming the friction force in the direction of my drawn, we have:

(T2*R + f*R - T1*R) = I*alpha = 0.5*M*R²*(a2)/R
We can cut R
T2 + f - T1 = 0.5M*a2

Let's distinguish three things here: the pulley, the section of rope in contact with the pulley and the vertical sections of rope.
The tensions in the vertical sections act on the contacting section; friction acts between the contacting section and the pulley. In other words, your f is not additional to T2-T1, it is the same thing.
Simplest is to think of the pulley and the contacting section of rope as a single entity.

 

[LCSphysicist]

Wow from my country kkk NIvaldo Lemos has a book too.

I think i get it, so,
T2-M2g = m2*a2
T1-M1g = -m1*a2
(T2 + f - T1) = M*a2/2
T2 = T1*e^(μ*θ)

Become just

T2-M2g = m2*a2
T1-M1g = -m1*a2
(T2 - T1) = M*a2/2
T2 = T1*e^(μ*θ)

right?
:)

 

[LCSphysicist]

it is just a little complicated to realize that the really important tension is only at the ends, is it a consequence of the rope not having mass?

 

[etotheipi]

I believe your adjusted equations are okay, though I'm falling asleep right now so I'll need to check them again in the morning .

To understand why it works, you can either think of it in the way @haruspex mentioned (i.e. pulley + rope in contact with pulley as a system), or you can take a more complicated approach of integrating up the forces on the pulley due to lots of small elements of string. The torque relation is perhaps not an obvious result at first

 

[haruspex]

Wow from my country kkk NIvaldo Lemos has a book too.

I think i get it, so,
T2-M2g = m2*a2
T1-M1g = -m1*a2
(T2 + f - T1) = M*a2/2
T2 = T1*e^(μ*θ)

Become just

T2-M2g = m2*a2
T1-M1g = -m1*a2
(T2 - T1) = M*a2/2
T2 = T1*e^(μ*θ)

right?
:)

Yes.