Rotational Kinematics and Energy

[nahanksh]

Homework Statement

https://online-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys211/spring09/homework/10/three_cylinders/6.gif
Three identical, solid, uniform density cylinders, each of mass 17 kg and radius 1.67 m, are mounted on frictionless axles that are attached to brackets of negligible mass. A string connects the brackets of cylinders #1 and #3 and passes without slipping over cylinder #2, whose bracket is attached to the ledge. Cylinder #1 rolls without slipping across the rough ledge as cylinder #3 falls downward.

This system is released from rest from the position shown -- with cylinder #3 at a height of 4.7 m above the ground.

Q) How fast is cylinder #3 moving just before it hits the ground? (v=?)

Homework Equations

The Attempt at a Solution

For Q1, i tried to use Newton's second law.
While doing it,
a= mg-T from #3
RT = I*\alpha
or RT = 0.5*m*R^2*\alpha

And i used alpha as (a/R)

Then i got T=0.5Ma

When substituting this into the first equation, i got 'a' as 17.55 which seems to be wrong..

I think i did something wrong in replacing alpha as (a/R)...
But i can't find it exactly...

Please Could someone help me out here?

 

[AEM]

Homework Statement

https://online-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys211/spring09/homework/10/three_cylinders/6.gif
Three identical, solid, uniform density cylinders, each of mass 17 kg and radius 1.67 m, are mounted on frictionless axles that are attached to brackets of negligible mass. A string connects the brackets of cylinders #1 and #3 and passes without slipping over cylinder #2, whose bracket is attached to the ledge. Cylinder #1 rolls without slipping across the rough ledge as cylinder #3 falls downward.

This system is released from rest from the position shown -- with cylinder #3 at a height of 4.7 m above the ground.

Q) How fast is cylinder #3 moving just before it hits the ground? (v=?)

Homework Equations

The Attempt at a Solution

For Q1, i tried to use Newton's second law.
While doing it,
a= mg-T from #3
RT = I*\alpha
or RT = 0.5*m*R^2*\alpha

And i used alpha as (a/R)

Then i got T=0.5Ma

When substituting this into the first equation, i got 'a' as 17.55 which seems to be wrong..

I think i did something wrong in replacing alpha as (a/R)...
But i can't find it exactly...

Please Could someone help me out here?

It strikes me that you could do this most simply by conservation of energy. The potential energy of the falling cylinder goes into kinetic energy. Don't forget that the KE of cylinder (1) is both KE of translation and KE of rotation and that the linear velocity of the center of mass of cylinders (1) and (3) are equal.