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.

 

[kerimek]

I would like to point out that the homework statement is asking for the speed of cylinder #3 before it hits the ground, not the acceleration. This means that we need to use the equations for rotational kinematics and energy to solve this problem.

First, we can use the conservation of energy to find the speed of cylinder #3 just before it hits the ground. At the starting position, all of the potential energy is stored in cylinder #3, so we can write:

mgh = 0.5*I*ω^2 + 0.5*m*v^2

Where m is the mass of the cylinder, g is the acceleration due to gravity, h is the initial height of cylinder #3, I is the moment of inertia of the cylinder (0.5*m*R^2 for a solid cylinder), ω is the angular velocity, and v is the linear velocity.

We can simplify this equation by noting that the cylinders are identical, so they have the same mass and moment of inertia. We can also assume that they are released from rest, so ω=0. This leaves us with:

mgh = 0.5*m*v^2

Solving for v, we get:

v = √(2gh)

Substituting in the values given in the problem, we get:

v = √(2*9.8*4.7) = 9.5 m/s

Therefore, the speed of cylinder #3 just before it hits the ground is 9.5 m/s.

As for your attempt at a solution, I believe you may have made a mistake in your calculation for the acceleration. Instead of using T=0.5Ma, you should use the equation a= (T-mg)/m. This is because the tension in the string is not equal to the weight of the cylinder, but rather the difference between the weight and the force needed to accelerate the cylinder. Also, you should use the moment of inertia for a solid cylinder, which is 0.5*m*R^2, instead of 0.5Ma. This will give you the correct acceleration of 9.8 m/s^2, which can then be used to find the speed as shown above.

I hope this helps! Let me know if you have any further questions.