Question

The 7-kg collar has a velocity of 6 m/s to the right when it is at A. It then travels down along the smooth guide shown. The spring has an unstretched length of 100 mm and B is located just before the end of the curved portion of the rod. Determine the speed of the collar when it reaches point B.

Answer 1

I know this problem would probably be better suited for say, Physics or Engineering, but nobody ever checks those sections.

Answer 2

i'm not sure,
but it looks like the spring force is acting as the centripital force, curve A-B is circular, since the length of the spring isn't changing, centripetal force is constant, thus tangential velocity is constant. so at B, velocity should be 6m/s in the downward direction...?

Answer 3

no, it looks like the length of the spring is changing :p

Answer 4

Yeah, that's what I figured. I'm supposed to use the conservation of energy equation; KE_1 + V_1 = KE_2 + V_2;

I figured for the potential, there are only two forces that are doing work; the weight of the collar and the force of the spring. I've tried this twice so far, and I only get so many attempts.

Answer 5

ohh...nevermind what I said, the spring doesn't even pass through the center of the circluar arc xD

Answer 6

Well, it kind of does, it's more that where it passes through between A and B is irrelevant (maybe? Still not totally sure).

Answer 7

If anyone who's reading this wants to chime in, please feel free!

Answer 8

@ganeshie8 and @irishboy, I've been told maybe one of you could help me, so I'm tagging you here in the hopes that perhaps one of you will see this later after the fact.

Answer 9

Well, at least one of them worked.

Answer 10

Have you tried setting up the newtonian way, showing all the forces

Answer 11

Maybe that would be a good first step

Answer 12

@IrishBoy123 @Michele_Laino @ganeshie8 I think they would like this question, I'm sure there is a good set up for the lagrangian method as well

Answer 13

I've drawn the Free Body and Kinetic diagrams. TWICE. Because I've attempted this problem twice so far (I only get six, I think, so I'm not in a rush to burn through them all).

Answer 14

Oh actually conservation of energy should work nicely

Answer 15

That's what I thought too, but I keep coming up dry as far as answers go.

Answer 16

try solving :

\(x_i = \sqrt{0.2^2+0.2^2} - 0.1\)

\(x_f = 0.4-0.1\)

\(7*g*0.2 + 1/2*7*6^2 + k*x_i = 1/2mv^2 + k*x_f\)

Answer 17

I tried solving and got approximately 6.09 m/s. Mastering says it's still wrong, but the hint system implies that I'm only off by a small amount due to a rounding error or used wrong number of sig figs (it asks for three, so I don't think it can be a sig fig issue).

Answer 18

I'm getting \(v = 6.184 m/s\)

Answer 19

http://www.wolframalpha.com/input/?i=solve+7*9.8*0.2+%2B+1%2F2*7*6%5E2+%2B+k*x+%3D+1%2F2*7*v%5E2+%2B+k*y%2C+x%3D%5Csqrt%7B0.2%5E2%2B0.2%5E2%7D+-+0.1%2C+y+%3D+0.4-0.1%2C+k%3D50

Answer 20

do we need to use any particular for `g` ?

Answer 21

It accepted 6.18, though the answer was 6.29. Weird. I don't get what you did that was different than what I did though, except for maybe the initial force of the spring; how come you only use the unstretched length? And why was the stretch on both sides of the equation not squared? And why was the force of the spring equation on the left side not multiplied by 1/2?

Answer 22

Ahh so many questions!
I'll try to answer them in multiple replies...

first of all, what do you know about Hookes law ?

Answer 23

Hooke's Law is spring constant times the stretch squared, right? The stretch being the displacement of the stretched length and the unstretched length.

Answer 24

Look it up again

Answer 25

F = -kx. I thought you were supposed to integrate it, though.

Answer 26

Omg. you're absolutely correct,
\(W = \int_0^{x_1} -kx \,dx = -\dfrac{1}{2}kx^2 \)

Answer 27

http://www.wolframalpha.com/input/?i=solve+7*9.8*0.2+%2B+1%2F2*7*6%5E2+%2B+k*x%5E2%2F2+%3D+1%2F2*7*v%5E2+%2B+k*y%5E2%2F2%2C+x%3D%5Csqrt%7B0.2%5E2%2B0.2%5E2%7D+-+0.1%2C+y+%3D+0.4-0.1%2C+k%3D50

Answer 28

My mistake, it should be 1/2 kx^2, not just kx

Answer 29

So the part I was getting wrong wasn't the Hooke's law bit; it was something else. Now that I think about it, there WAS one other thing you did different that I didn't do, and that's probably what screwed me up. You multiplied the weight by 0.2. Why?

Answer 30

are you refering to PE at point A ?

Answer 31

Yeah.

Answer 32

0.2 is the difference in height between A and B

Answer 33

PE = m*g*h

Answer 34

Oh crap, I realized it just as you said it. You set the datum at B, and measure the height from there. Wow, do I feel stupid right about now.

Answer 35

And with Hooke's Law, we assume that the spring is unstretched at point A, I'm guessing?

Answer 36

The spring is streched at both A and B because the length is greater than 100mm at both those points. You're correct if you're saying that the difference in this potential energy of spring is what contributes to the change in kinetic energy between the points A and B.

Answer 37

we're given that 100mm is the equilibrium length of the spring..

Answer 38

Aye, that's correct.

Answer 39

this has a completely analytical solution with its own DE and exact solution.......which i qm minded to latex as it's interesting. it might even give thr right answer :p

there is really only 1 relevant coordinate in this, \(\theta\) , which is the angle the spring makes with the horizontal. the geometry means that the angle at the centre of the circle is \(2 \theta\) which makes things easy-ish.