# Homework Help: Conservation of Angular Momentum w/ Spring

1. Apr 29, 2012

### kabailey

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

2. Relevant equations

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

3. 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.

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2. Apr 29, 2012

### sweet springs

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

3. Apr 29, 2012

### 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.

4. Apr 29, 2012

### 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.

5. Apr 29, 2012

### kabailey

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???

Last edited: Apr 29, 2012
6. Apr 30, 2012

### sweet springs

Hi.

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

Regards.

7. Apr 30, 2012

### Staff: Mentor

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.

8. May 7, 2012

### kabailey

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

9. May 7, 2012

### 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

10. May 7, 2012

### kabailey

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

11. May 8, 2012

### kid_123

Yes I also have this question.. Why cant we use ma=(mV^2)/r=kx

12. May 9, 2012

### sweet springs

Hi.
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