Thank you, JimB, Steve.
Steve, you still remember a lot of physics!
Method 1: Straightforward Force/Torque Analysis
Let the rocket force be represented by Ft=30N and the tension in the rope by Fr, which we will need to calculate. The pulley inertia is I=20 kg m^2 and the mass is m=2 kg, suspended in a gravitational field with g=10 m/s/s.
The torque equation becomes α=dω/dt = (R Ft - R Fr) / I , which leaves only the rope tension as an unknown to be calculated/represented.
The rope tension is caused by the force from the mass, and the mass has two components of force that cause rope tension, gravitational force (mg), and inertial force m dv/dt. Hence, Fr=m g+m dv/dt, where v is the linear velocity of the mass.
However, linear velocity of the mass v can be related to the rotational acceleration of the pulley by v=R ω, and since R does not change in time then dv/dt = R dw/dt=Rα.
We can now substitute this into the original torque equations and do some algebra to find that α=( R Ft - R m g )/ ( I + m R^2)
In our case we get α = ( (1) 30 - (1) 2 (10) ) / ( 20 + 2 (1)^2 )=10/22=0.4545 rad/s/s
This answer is lower than the 0.5 rad/s/s , as you predicted.
I would go with the first method because it seems more natural and straightforward.
Let's do another case. This time the pulley is already rotating with angular velocity, ω, of 2π rad/s. There is no angular acceleration which means torque is zero and if we further assume that there is no friction involved then the rocket isn't really doing any work; it's basically in off mode.
The setup looks
like this. But this time as I mentioned above the pulley was already rotating at a constant angular velocity of 2π rad/s which translates to linear velocity of 2π m/s when the rope which has 2 kg mass suspended to it and resting on ground gets somehow attached to it. The vertical height between the pulley and ground could be approximated to be 20 meters.
Our purpose is to lift the mass to some height above the ground and we also want to keep things as calm as they could be; I mean no jerky motion.
We can assume that the rocket has some kind of a sensor which regulates the angular velocity at 2π rad/s. It turns the rocket on if the velocity is under 2π rad/s and off if it goes above. Further, we can also assume that the sensor makes rocket to exert different amount of force as required.
As soon as the rope gets attached to the pulley, the angular velocity of pulley will go down for an infinitesimal amount of time and during that time both pulley and mass would have same speed.
Then, instantly, the rocket turns on and starts exerting force. The rocket aims to get the system to 2π rad/s within 1 second. Therefore the system needs the acceleration of (2π-5. 9924)/1s=0.291 rad/s/s for one second. This time we need to know rocket force and I believe that the formula derived by you can be used, α=( R Ft - R m g )/ ( I + m R^2). After rearranging Ft={α/(R(I+mR^2))}+mg={0.291/(20+2)}+2(10)=20.013 N. Once the system has reached 2π rad/s angular velocity then the rocket just has to exert 20 N force to counterbalance the clockwise torque to 2 kg mass. Do you agree with everything I have said?
As to why this is true, that is a harder question from a scientific point of view. Ultimately, we never get to the final "why" in physics, but we can always try to probe deeper. In this case, we could consider that the see-saw is not really an ideal rigid object. It is composed of atoms. There are atomic forces (electrical in nature) that keep the board together as one piece and provides forces and torques within the system. We could try to analyze every atom and do a free body diagram for each and every one, but this is of course impossible for a person to do and would take a computer an impractical amount of time to complete. But, in principle we could try to imagine how all these electrostatic forces are interacting to create a nearly rigid board that transmits forces and torques to the people involved in having fun on a see-saw.
I would say that JimB's reply was regular and typical answer to the question. I was more interested in 'why'. I'm happy that you understood my query and I do agree that such a simple question might be really hard to answer. It's like asking that why the inertia exist in the first place.
This question came to my mind while writing about the case when the pulley rotating at constant angular velocity gets connected to the rope. The pulley possessed angular moment and momentum is always conserved; linear and angular momentums work independent of each. When the pulley gets joined/connected to rope and mass starts get lifted, some of angular momentum gets transformed into linear momentum. I don't see how angular momentum is conserved in this case.
Let's try a simple case. Suppose a bicycle is being driven at a constant speed which means constant angular velocity and hence angular momentum. Once the brakes are applied, where does angular momentum disappear? Thank you.
Best regards
PG
Some good links about linear and angular momentums:
1:
https://ccrma.stanford.edu/~jos/pasp/Relation_Angular_Linear_Momentum.html
2:
https://www.physicsforums.com/threa...nation-of-angular-and-linear-momentum.473302/
3:
https://imechanica.org/node/8288