Three Cylinders (rotational kinematics)

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Homework Help Overview

The problem involves three identical cylinders in a rotational kinematics scenario, focusing on the motion of the cylinders as one falls and the others roll. The original poster is attempting to determine the speed of cylinder #3 just before it hits the ground after replacing it with a sphere.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to apply energy conservation principles to find the speed of cylinder #3 after replacing it with a sphere. They express confusion regarding the inclusion of energy contributions from other cylinders.

Discussion Status

Participants are engaging in a back-and-forth discussion, with some offering guidance on missing energy components in the calculations. There is an ongoing exploration of the energy equations involved, and multiple interpretations of the problem setup are being considered.

Contextual Notes

Participants note the importance of including the energy of cylinder #2 and the translational kinetic energy of the other masses in the calculations. The original poster's calculations have been challenged, indicating potential misunderstandings in the application of energy conservation.

r34racer01
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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?


 
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r34racer01 said:
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?

Hi r34racer01! :smile:

You've left out the energy of cylinder #2.
 
tiny-tim said:
Hi r34racer01! :smile:

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

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?
 
sorry!

Hi r34racer01! :smile:

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

(have an omega: ω and a squared: ² :smile:)
r34racer01 said:
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?

You've left out the translational KE of masses 1 and 3. :smile:
 

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