# Homework Help: Three Cylinders (rotational kinematics!)

1. Nov 4, 2008

### r34racer01

Three identical, solid, uniform density cylinders, each of mass 16 kg and radius 1 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 4.4 m above the ground.
a) How fast is cylinder #3 moving just before it hits the ground?
v = 5.36

b) What is the rotational speed of cylinder #2 at the time in (a)?
w = 5.36

c) Cylinder #3 is now replaced by a sphere of the same mass. How fast is it moving just before it hits the ground?
v'= 5.36

d) Finally, cylinder #1 is replaced by a sphere of the same mass and radius. How fast is cylinder #3 moving just before it hits the ground?
v'' = ????

This is where I'm having problems. This should be exactly the same as pt. A except we have a sphere so c=2/5. So I did
mgh = 2/5mv^2
(16)(9.81)(4.4) = 2/5(16)v^2
and I got 10.388 = v'' but apparently that's wrong, anyone know what's going on?

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 4, 2008

### tiny-tim

Hi r34racer01!

You've left out the energy of cylinder #2.

3. Nov 4, 2008

### r34racer01

Ok now I tried 690.624 = (1/2)*I(sphere)*w^2 + (1/2)*I(cylinder)*w^2
690.624 = (1/2)((2/5)mr^2)(v^2/r^2) + (1/2)((1/2)mr^2)(v^2/r^2)
690.624 = (0.2)mv^2 + (0.25)mv^2
690.624 = 3.2v^2 + 4v^2
9.79388 = v
But that's still wrong. What am I doing wrong?

4. Nov 7, 2008

### tiny-tim

sorry!

Hi r34racer01!

I'm sorry I've taken so long to reply

(have an omega: ω and a squared: ² )
You've left out the translational KE of masses 1 and 3.