How Fast Does the Collar Travel Before Hitting Stop B?

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The discussion revolves around calculating the speed of a collar just before it hits a stop after being released from rest. The collar, weighing 5 lbs, travels along a smooth guide and involves both gravitational and elastic potential energy. Participants emphasize using energy conservation principles to determine the initial and final stretches of the spring, which is crucial for calculating the elastic potential energy. There are challenges with unit conversions and ensuring the correct application of formulas, particularly when transitioning from gravitational to elastic potential energy. Ultimately, the correct speed of the collar is derived through careful calculations, with a final value of approximately 15.71 ft/sec.
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Homework Statement



Ok sorry for the horrible drawing i freehanded it on paint.

The 5lb collar is released from rest at A and travels along the smooth guide. Determine the speed of the collar just before it strikes the stop at B. The spring has an unstretched length of 12in. k=2 lb/in

please note that from point C to point A it is a 1/4 circle of radius 12in
From C to B, it is horizontal.


Homework Equations





The Attempt at a Solution



I could do this problem if it was traveling from just C to B but the curve has me lost.

I attempted to find my Y and i got y=(144-x^2)^1/2+10
I solved that for x and got x=(144-(y-10)^2)^1/2
I did this to find the hypotenus or the s length of the spring. would it be correct for my s to be (x^2+y^2)^1/2. i feel like I am way off hlep
 

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Not sure what you are trying to do. What are y, x, and s?

Hint: Use energy conservation. Find the initial and final stretch of the spring.
 
s is the length of spring. the initial stretch is 22in, and the final is 12in, but how does that help
 
joemama69 said:
s is the length of spring. the initial stretch is 22in, and the final is 12in, but how does that help
It will allow you to figure out the elastic potential energy stored in the spring. Be sure to take the unstretched length of the spring into account.
 
I attempted to find my Y and i got y=(144-x^2)^1/2+10
I solved that for x and got x=(144-(y-10)^2)^1/2


am i correct is saying that my s would the hypotenus using the x and y above.

I would then plug it into V(e) = .5ks^2 for elastic potential energy

I believe I also have to use V(g) = Wy

My main problem is that my y value is a peacewise function. From point B-C it is a different function than from C-A

Note the Picture attached above
 
joemama69 said:
I attempted to find my Y and i got y=(144-x^2)^1/2+10
I solved that for x and got x=(144-(y-10)^2)^1/2
Sorry, but I still have no idea what you are trying to calculate here.


am i correct is saying that my s would the hypotenus using the x and y above.
No. All you need to find is the initial and final stretch of the spring from its unstretched length--the curved section is irrelevant. Your previous post had all the data needed to find the spring PE.

I would then plug it into V(e) = .5ks^2 for elastic potential energy
Yes, once you find the amount that the spring is stretched, that's how you find the elastic PE.

I believe I also have to use V(g) = Wy
If you mean gravitational PE, then yes, you definitely need that.

My main problem is that my y value is a peacewise function. From point B-C it is a different function than from C-A
So what? Energy is conserved. All you care about it is the total energy (elastic, gravitational, and kinetic) at points A and B.
 
Ok I think i got it, i thought it would be much more complicated, my book is confusing

ok here's what i did,

V(A) = V(e) - V(g) = .5ks^2 - Wy = -3292 k=2lb/in s=10in W=5lb*32.2 ft/s^2 <-- should this be in/s^2 since k is in lb/in... & y=22in

T(A) = .5mv^2 = (5/2)v^2

And V(C) & T(C) = 0 because the spring is no longer stretcher and the y = o

so i got v = 36.29 in/s

Are my units correct.
 
Last edited by a moderator:
joemama69 said:
ok here's what i did,

V(A) = V(e) - V(g) = .5ks^2 - Wy = -3292
Why the minus sign?
k=2lb/in s=10in W=5lb*32.2 ft/s^2 <-- should this be in/s^2 since k is in lb/in... & y=22in
I would put all quantities into standard units, distances in feet. (k = 2lb/in = 24lb/ft.) Note that 5 lbs is the object's weight, not its mass, so no need to multiply by g when finding the gravitational PE. (What's the object's mass?)

T(A) = .5mv^2 = (5/2)v^2
The collar starts from rest at point A.

And V(C) & T(C) = 0 because the spring is no longer stretcher and the y = o
The elastic and gravitational PEs are both zero at point B (not point C, at least according to your diagram). The KE is not zero.
 
  • #10
I subtracted Wy because of a similar example in my book but i guess i read into it wrong

ok so my new mass is 2.26 slug

ok so my T(a) = 0, but my T(B) is now 1.134v^2

plug and chug and i got a velocity of 11.98 ft/sec, that just seems wrong, too fast

note the new pic
 

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  • #11
joemama69 said:
I subtracted Wy because of a similar example in my book but i guess i read into it wrong

ok so my new mass is 2.26 slug
How did you get that?

As a sanity check for the speed, how fast would something be moving if it were simply dropped from that same height? (That speed must be less than the speed of the collar in this problem, since there's no spring pulling it down.)
 
  • #12
1 pound = 0.0310809502 slug

5lb = .1554045... oops calc error

ok so i got at point A
V(g)=9.157366, V(e)=9.96, T=0

Point B
V(g)=0, V(e)=0, T=.77v^2

v=4.96 ft/sec I hope
 
  • #13
joemama69 said:
1 pound = 0.0310809502 slug

5lb = .1554045... oops calc error
That's better.

ok so i got at point A
V(g)=9.157366, V(e)=9.96, T=0
Your V(g) is close to my value: mgh = (5)*(22/12) = 9.167
Show your calculation for V(e).

Point B
V(g)=0, V(e)=0, T=.77v^2
Show your calculation for T.

v=4.96 ft/sec
Did you do the sanity check I suggested?
 
  • #14
Point A

Vg = .554*32.2*1.83 = 9.157366<-- Different than urs Doc
Ve = .5*24*.83 = 9.96

Point B

T = .5*1.554*v^2 = .777v^2
 
  • #15
oops Ve = .5 *24*.83^2 = 8.2668

v = 4.74
 
  • #16
joemama69 said:
Point A

Vg = .554*32.2*1.83 = 9.157366<-- Different than urs Doc
You must have a typo in here, since the left and right sides don't match.
Since mg is given as 5 lb, mgh = 5*(22/12) = 9.167
Since mg = 5; m = 5/g = 5/32.2 = 0.1553
Ve = .5*24*.83 = 9.96
The 0.83 (really 0.8333...) should be squared.

Point B

T = .5*1.554*v^2 = .777v^2
Wrong mass.
 
  • #17
Sorry for the typo, i meant 1.554

I thought that the mass was 1.554 slug, and 5lb was the force
 
  • #18
joemama69 said:
I thought that the mass was 1.554 slug, and 5lb was the force
You're off by a factor of ten. The mass is 0.1554 slugs. (See my previous post.)
 
  • #19
man i keep making stupid mistakes

V(g) = .1554*32.2*1.83333333333 = 9.1738
V(e) = .5*24*.8333333333333333^2 = 10
T(B) = .5*.1554*v^2 = .0777v^2

Therefor v = 15.71 ft/sec
 
  • #20
joemama69 said:
V(e) = .5*24*.8333333333333333^2 = 10
Redo this one.
 

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