Acceleration and tension in a system

[furor celtica]

Homework Statement

A tape is wound round two smooth cylinders as shown. The higher cylinder is fixed, but the lower cylinder sits in a loop formed by the tape.
Square mass= 5kg
Lower cylinder=12kg
The system is held at rest in the position shown, with the tape taut, and then released. Find the acceleration of (A) the lower cylinder and (B) the 5 kg mass.

Homework Equations

OK so first I took 2T-120=12a and 50-T=5(a’), (a and (a’) being the respective accelerations obviously) but I couldn’t figure out right away what relationship between a and a’. After I took 2a=a’ and got the right result (0.625 and 1.25 ms^2 respectively).

The Attempt at a Solution

But although it might seem fairly evident to most of you, I was wondering how it could be proven that 2a=a’ in systems like this.
Thanks and excuse the rubbish Paint diagram.

 

[kuruman]

But although it might seem fairly evident to most of you, I was wondering how it could be proven that 2a=a’ in systems like this.

This is why a' = 2a is evident to (most of) us. There are three pieces of rope to consider in your diagram, left, middle and right. When the smaller mass goes up by distance h, the right rope loses length h which gets added to the combined length of middle+left rope. Clearly each length increases by h/2 so the larger mass goes down by distance h/2.

Therefore

In the same time that the smaller mass is displaced by h the larger mass is displaced by h/2.

How do you think the velocities and accelerations of the two masses ought to be related?

 

[furor celtica]

thanks

 

[rajput635]

can we use F=ma istead of using height concept
for F=ma

since mass in case of 12kg is double so acceleration will b half

or a'=2a