How to solve mass-spring system when affected by torque in a pulley?

[bolzano95]

Homework Statement

Write a differential equation for a mass-spring system.

Relevant Equations

F=ma

 

[haruspex]

There is no friction between the rope and pulley.

You don't mean that. A frictionless pulley means there is no friction at the axle.
If there were no friction between pulley and rope the pulley would not turn; the rope would just slide over it.

A few mistakes towards the end.
Torque = moment of inertia times.. what?
A factor of R seems to have disappeared somewhere.

 

[bolzano95]

You don't mean that. A frictionless pulley means there is no friction at the axle.
If there were no friction between pulley and rope the pulley would not turn; the rope would just slide over it.

A few mistakes towards the end.
Torque = moment of inertia times.. what?
A factor of R seems to have disappeared somewhere.

Sorry, I meant the rope does not slip on the pulley.

I fixed the last equations and double checked the R-s. But I'm still confused by what you mean with
Torque = moment of inertia times ... what?
The torque in this case is given as a product of C and angular velocity, where C is a factor of dampening.

Is this correct?

 

[haruspex]

The torque in this case is given as a product of C and angular velocity

Torque has dimension $ML^2T^{-2}$. Moment of inertia is $ML^2$ and angular velocity is $T^{-1}$. Multiplying those last two gives $ML^2T^{-1}$, not torque.

C is a factor of dampening.

There is no damping here (much less dampening - all is dry). The forces are all conservative.

 

[bolzano95]

I checked the torque and I agree there is something wrong, but unfortunately I have to use the given formula (it's mandatory). But I think that in the coefficient C are hidden necessary units.

I suppose that the goal of this problem is to write a homogeneous linear differential formula, but what I get is a nonhomogeneous one. So I just wanted to check if my solving process is correct.

 

[haruspex]

I have to use the given formula (it's mandatory)

Either you were given the wrong formula or you have misunderstood.
The equation is $\tau=I\alpha$, torque equals moment of inertia times angular acceleration.

 

[Lnewqban]

The statement “... there is a torque in the axis or rotation” suggests to me that the shown diagram does not correspond with the text of this problem.

 

[haruspex]

The statement “... there is a torque in the axis or rotation” suggests to me that the shown diagram does not correspond with the text of this problem.

Good point.
@bolzano95, does that statement correspond to a part of the problem statement that you have left out? I.e. your $-C\omega$ term is correct but your error is that you left out the $-I\alpha$ term from the torque difference?

I remain doubtful of that because axial friction should be $-C\frac{\omega}{|\omega|}(T_1+T_2)$.

Edit: added tension factor above.