Torque about an accelerating point

[xkcda]

The total force acting on the pulley is zero so:
F=mg+T1+T2 (1)Analyzing the torque and angular acceleration about the actual axis of rotation, the axle of the pulley, gives:
τnet=T1R−T2R=Iα (2)If we analyze about point P, the right edge of the pulley where T1 is applied, we get:
τnet=(F−mg)R−T2×2R=(I+mR2)α WRONG(3)Using Equation 1 to eliminate F−mg from Equation 3 gives:
τnet=T1R−T2R=(I+mR2)α WRONG(4)The net torque in Equations 2 and 4 is the same, but the moments of inertia are different so the angular accelerations are also different. Note that if we think of point P as attached to the right-hand string, if T1 ≠ T2 then it is accelerating.My question is if we think P as point fixed in space and not attached to to the right-hand string, then what will be the equation of torque about point P?

In the case of instantaneous axis of rotation, we say that the center of rotation is a point in space and does not undergo radial acceleration. So we think of it as an inertial frame of reference. So can't we analyze the torque about point P thinking that it not subject to any kind of acceleration.In that case we should get the actual torque about point P.

 

[Lnewqban]

Welcome, @xkcda !

The fact that I and m are given, suggests that T1 and T2 are not equal and that there is a unique angular acceleration of the pulley.

I believe that equation 4 is incorrect because the moment of inertia is incorrect.
Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#cmi

 

[kuruman]

Is there a question to be answered or a task to be performed that relates to the diagram? This is not a forum for guessing games.

 

[haruspex]

Torque about an accelerating point​

τnet=(F−mg)R−T2×2R=(I+mR2)α

Your title is the clue. If you take torques about an accelerating point which is not the mass centre then you need to consider the consequences of using a non-inertial frame.
As you noted, the mass centre is not accelerating here, so although you took torques about some other point the angular acceleration is still about the mass centre: I, not $I+mR^2$.
Or, if you take a non-rotating frame moving with point P of the wheel as your reference frame, making P stationary, then you must introduce a virtual force to compensate for your non-inertial frame. On, the wheel, that will be $mR\alpha$, upwards.
$\tau_{net}=(F−mg)R−T_2\cdot 2R+mR^2\alpha=(I+mR^2)\alpha$.
If you take a rotating frame moving with P, the reasoning may be different again, but will lead to the same equation.

 

[xkcda]

Your title is the clue. If you take torques about an accelerating point which is not the mass centre then you need to consider the consequences of using a non-inertial frame.
As you noted, the mass centre is not accelerating here, so although you took torques about some other point the angular acceleration is still about the mass centre: I, not $I+mR^2$.
Or, if you take a non-rotating frame moving with point P of the wheel as your reference frame, making P stationary, then you must introduce a virtual force to compensate for your non-inertial frame. On, the wheel, that will be $mR\alpha$, upwards.
$\tau_{net}=(F−mg)R−T_2\cdot 2R+mR^2\alpha=(I+mR^2)\alpha$.
If you take a rotating frame moving with P, the reasoning may be different again, but will lead to the same equation.

Can you describe me the details about that pseudo force please?Did you apply $mR\alpha$ along P or along the center of mass?

 

[haruspex]

Can you describe me the details about that pseudo force please?Did you apply $mR\alpha$ along P or along the center of mass?

The "inertial force" is taken to act on each element of mass directly, in proportion to its mass. So here it would be taken to act on the mass centre.
https://en.wikipedia.org/wiki/Fictitious_force