# Homework Help: Dynamics: Max horizontal speed of a collar with three restricting springs

1. Feb 25, 2012

### jaredogden

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

A 1.2-kg collar C may slide without friction along a horizontal rod.
It is attached to three springs, each of constant k 5 400 N/m and
150-mm undeformed length. Knowing that the collar is released
from rest in the position shown, determine the maximum speed it
will reach in the ensuing motion.

The position shown shows the collar C on a horizontal rod. There are three springs attached to it at the same point on the collar. Spring 1 is attached from the collar straight down to a pin at 150 mm below the collar (undeformed). Spring 2 is attached to another pin connection 150 mm below the collar and 150 mm to the left of the collar (45-45-90 triangle). And finally Spring 3 is attached 150 mm below the collar and 300 mm to the left of it.

2. Relevant equations

E = KE + PE + U
PE of a spring = 1/2kΔx2
Ei = Ef

3. The attempt at a solution

First find the deformed length of the springs.
Spring 3: l = √((.15m)2 + (.3m)2)
l = 0.3354 m
Δx3 = 0.3354m - 0.15m
Δx3 = 0.1854m

Spring 2: l = √((.15m)2 + (.15m)2)
l = 0.21213 m
Δx2 = 0.21213m - 0.15m
Δx2 = 0.06213m

Spring 1: Undeformed Δx = 0m

now Ei = PE (from the deformed springs, no KE or U)
Ei = 1/2k1(Δx1)1 + 1/2k2(Δx2)2 + 1/2k3(Δx3)3
Ei = 1/2(400N/m)(0m)2 + 1/2(400N/m)(0.06213m)2 + 1/2(400N/m)(0.1854m)2
Ei = 7.6459J

From the Law of Conservation of Energy Ei = Ef and velocity will be max when there is no spring potential energy restricting the collar and only KE.

Ef = 1/2mv2
7.6459J = 1/2(1.2kg)v2
3.57 m/s = v

That is the answer I got however the answer in the book is v = 3.19 m/s
I'm guessing one of the springs will still no matter what restrict some movement? I'm not sure what else to do on this one.

2. Feb 25, 2012

### tiny-tim

hi jaredogden!
nope

velocity will be max when spring potential energy is min (not zero), won't it? …

and they've set up the question so that it's fairly obvious where the min is, so just calculate the PE there

3. Feb 25, 2012

### jaredogden

That was my last question, I guess there will always be some sort of spring potential acting on it, but it is at its least when the collar has moved 150 mm to the left and the outer springs are barely deformed while the middle one isn't at all.

I added in a 2(1/2)(400N/m)(0.06213m)2 component which would account for the two outer springs.

I got the right answer after that! Thanks so much for the little push in the right direction!