Need help setting up the problem.

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AI Thread Summary
A user seeks assistance with a physics problem involving a 6kg collar sliding on a frictionless pole described by a parabolic equation. The problem requires applying conservation of energy principles, considering kinetic, gravitational potential, and elastic potential energy. After initial calculations, the user arrives at a velocity of approximately 11.93 m/s at x=2, which is confirmed as correct by another participant. Suggestions for simplifying calculations are provided, emphasizing the efficiency of directly calculating changes in potential energy. The discussion highlights the importance of detailed calculations for accuracy in solving physics problems.
qball1982
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Hello,
I was looking to get some help with setting up this problem so I may solve it. For some reason I just don't see what to do to work towards a solution.

Homework Statement


A 6kg collar is allowed to slide over a frictionless pole whose height above the ground obeys the parabolic equation y=8-(1/2)x^2. Attached to the collar is a k=30N/m spring. The spring is 1 m when unstretched and connected in a way that the spring will always start at the origin. If the collar started from rest at (0,8), how fast will it be traveling at x=2? Hint: Don't forget gravity!Any help to point me in the right direction would be fantastic. Thank you.
 
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welcome to pf!

hello qball1982! welcome to pf! :smile:

(try using the X2 button just above the Reply box :wink:)

hint: try conservation of energy …

kinetic energy + gravitational potential energy + elastic potential energy = constant …

what do you get? :smile:​
 
Excellent! Thank you tiny-tim!

I worked this out using the formula: K1+U1=K2+U2
where U= mgy+1/2kx2

Crunching the numbers I get :
0+1205.4=K2+778.05
427.35=K2
427.35=1/2(6)v2

then finally: v=11.935m/s

Does this seem like the right path?
 
hi qball1982! :smile:

(just got up :zzz:)

yes that's the correct method :smile:

(but we need to see your detailed calculations if you want us to check the actual figures … for example, did you use √(x2 + y2) in the spring PE?)
 
Hello tiny-tim!

I suppose exact calculations would help. Sorry about that. :wink:

Spring displacement calculations:
x1 = 8-1 = 7 meters
x2 = √(22+62) -1 = 5.32 meters
And of course the -1 in each is due to the springs unstretched length of 1 meter.

Spring Uel Calculations:
Uel1= 1/2(30N/m)(7m)2 = 735N/m
Uel2= 1/2(30N/m)(5.32m)2 = 425.25N/m

Gravitational Potential Energy Calculations:
Ug1= (6kg)(9.8m/s2)(8m)=470.4
Ug2= (6kg)(9.8m/s2)(6m)= 352.8

To solve:
K1+Uel1[/SUB+Ug1=K2+Uel2+Ug2

0+ 735+ 470.4 =K2+ 425.25 + 352.8
1205.2 = K2 + 778.05
427.15 =K2
427.15 = 1/2(m)(v2)
427.15 = 3(v2)
142.38 = (v2)
v=√(142.38)
v = 11.93 m/s

I think it looks right but I am also wrong quite a bit too. :smile:

Thanks again for all the help!
 
yes that looks ok :smile:

just one thing, you could save time and make it look neater if instead of doing all this …
qball1982 said:
Uel1= 1/2(30N/m)(7m)2 = 735N/m
Uel2= 1/2(30N/m)(5.32m)2 = 425.25N/m

Ug1= (6kg)(9.8m/s2)(8m)=470.4
Ug2= (6kg)(9.8m/s2)(6m)= 352.8

you just work out Uel1 - Uel2 = 1/2(30N/m){(7m)2 - (5.32m)2} :wink:

(and the same for Ug)
 
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