Three Cylinders (rotational kinematics)

AI Thread Summary
Three identical cylinders are connected in a system where cylinder #1 rolls on a ledge while cylinder #3 falls. The speed of cylinder #3 just before hitting the ground is calculated to be 5.36 m/s. When cylinder #3 is replaced by a sphere, the same height results in a different speed, but the user struggles with the calculations, leading to confusion over energy contributions. The key issue identified is the omission of the translational kinetic energy of cylinders #1 and #3 in the user's calculations. Accurate accounting of all energy forms is essential for solving the problem correctly.
r34racer01
Messages
62
Reaction score
0
showmepl-2.gif


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?


 
Physics news on Phys.org
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:
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top