Conservation of Angular Momentum w/ Spring

[kabailey]

Homework Statement

Relaxed spring is sitting on a horizontal surface. A block is attached at one of its ends is kicked with a horizontal velocity, v₁, given to it. The block will move and stretch. Find the distance, x, the spring will stretch. X is in meters.

l₀=1 m
m=1 kg
k=1 N/m
v₁=1 m/s

Homework Equations

F=-kx T=2∏√(m/k) w=√(k/m) w=v/l₀ kx=mrw² KE=(1/2)mwI

The Attempt at a Solution

Plugging in the given and solving for the variables I get w=1 rad/s, T=2∏, l₀=1 m. The equation that I get to relate the initial and final lengths is mw₂(l₀+x)=mwl₀²

I am having hard time understanding this material. I would really appreciate some help. Thanks.

 

[sweet springs]

Hi. Consetvations of angular momentum and energy make sense, do they?

 

[kabailey]

yes it makes sense. similar to the idea of conservation of momentum, P=m*v.I also understand kinetic energy initial will equal kinetic energy final.

 

[sweet springs]

Energy 1/2 mv1^2 = 1/2 mv2^2 + 1/2 k x^2
Angular momentum mv1l0 = mv2(l0+x)
They would make sense to you.
Regards.

 

[kabailey]

Energy 1/2 mv1^2 = 1/2 mv2^2 + 1/2 k x^2
Angular momentum mv1l0 = mv2(l0+x)

Solving for x in the energy equation, I get x=1/v₂

Substituting 1/v₂ in the angular momentum equation, I get v₂=0. When I go back to solve for x, I get x=0.

Is this correct?

 

[sweet springs]

Hi.

You should get quadratic equation from these equation. Do good practice in mathematics.

Regards.

 

[gneill]

Query: Do they want the extension when the spring is 90° to its initial position, or when the spring is at maximum extension?

The reason I ask is there's no guarantee that the angle that the velocity V2 will be perpendicular to the spring when the "true anomaly" (the angle between the initial position and the current position) is 90°. In fact if I were to guess, I'd put that occurrence at 180° from the initial position.

 

[kabailey]

Query: Do they want the extension when the spring is 90° to its initial position, or when the spring is at maximum extension?

The reason I ask is there's no guarantee that the angle that the velocity V2 will be perpendicular to the spring when the "true anomaly" (the angle between the initial position and the current position) is 90°. In fact if I were to guess, I'd put that occurrence at 180° from the initial position.

Sorry I did not notice your post. The extension is at 90 degrees.

 

[kabailey]

Sweet Springs thank you for your help!

The final equation I get is x^3+2(x^2)-2=0, which gives me 0.839m as the final answer

 

[kabailey]

Question regarding this problem, how come I cannot use ma=m(v^2/r)=kx to solve this equation?

 

[kid_123]

Yes I also have this question.. Why can't we use ma=(mV^2)/r=kx

 

[sweet springs]

Hi.

Sweet Springs thank you for your help!
The final equation I get is x^3+2(x^2)-2=0, which gives me 0.839m as the final answer

By deleting v2, I have got the equation
y^4 + 2y^3 + y^2 - 2By - B = 0 where B = mv1^2 / kl0^2 = 1, y=x/lo

Question regarding this problem, how come I cannot use ma=m(v^2/r)=kx to solve this equation?

Interesting. Show me your result please.
Regards.

 

[kabailey]

I did not use the m(v^2/r)=kx. I was curious as to why one could not use this relationship. m(v^2/r) is to be used in an uniform circle and this system would not produce an uniform circle, thus one would yield an incorrect result.

Thank you for all of your help sweet springs.