What Forces Act on a Pulley System with Two Masses?

[TSny]

Your original approach was good. You noted that the downward force on m₂ is greater than the downward force on m₁ (except initially when θ = 0 and the forces are equal). Equivalently, you could say that the upward force on m₂ is less than the upward force on m₁. So, the upward acceleration of m₂ is less than the upward acceleration of m₁. Thus m₁ rises more quickly.

 

[Jahnavi]

Your original approach was good. You noted that the downward force on m₂ is greater than the downward force on m₁ (except initially when θ = 0 and the forces are equal). Equivalently, you could say that the upward force on m₂ is less than the upward force on m₁. So, the upward acceleration of m₂ is less than the upward acceleration of m₁. Thus m₁ rises more quickly.

What about this reasoning ?

After a very short time , right mass would be moving horizontally towards right .This would lead to lengthening of right string .Correspondingly left string shortens and in the process left mass moves upwards .Left mass would be higher than right mass .

 

[Jahnavi]

Yes, I enjoyed this one a lot.

Now , I can also say , ME TOO

 

[TSny]

What about this reasoning ?

After a very short time , right mass would be moving horizontally towards right .This would lead to lengthening of right string .Correspondingly left string shortens and in the process left mass moves upwards .Left mass would be higher than right mass .

After a very short but finite time, the right mass would be moving both to the right and upward.
At t = 0, the right mass is only moving to the right. But at this instant the string on the right is not getting longer ($\dot r_2 = 0$). Neither mass is moving upward at t = 0. So, the argument is a little vague. But the argument is suggestive of what's going on.

 

[TSny]

Now , I can also say , ME TOO

That is very good to hear

 

[Jahnavi]

I learned a lot from you TSny . Excellent discussion !

Thanks a lot !

A big Thank You @Chestermiller