What is the Total Energy of a System with Given Values of Mass and Length?

[Clever_name]

Homework Statement

see link for a picture of the situation, http://i50.tinypic.com/n5g3z5.jpg

Homework Equations

(1/2)mv^(2) , (1/2)k*l^(2)

The Attempt at a Solution

Total energy of the system at the given point of time is

mv^(2)/2+kl^(2)/2 = 0.6*100/2+100*0.25/2 = 30+12.5 = 42.5 J

At the maximum and minimum values of l, v = 0

=> kl2/2 = 42.5

=> l = √(85/k)

=> l = ±0.922 m

maximum l = 0.922m and minimum l = -0.922m

Not terribly confident in this approach however, any guidance would be appreciated. Thanks!

 

[voko]

Is angular momentum conserved here?

 

[Clever_name]

yes indeed, so this is a conservation of angular momentum problem.

 

[voko]

Is this enough to get on with the solution?

 

[Clever_name]

I'm now reading up on angular momentum, but please if you have some tips to get the ball rolling that would be greatly appreciated.

 

[voko]

One thing you need to think about is the angle of the velocity vector with the string at the min/max positions. You need the angle to compute the angular momentum.

 

[Clever_name]

Ok so far i have

(1/2)k(xf^(2)-xi^(2)) = mv*l*sin(60) , I have taken xf=0.5m, so now i solve for xi.

((-(0.6)(10)sin(60)*2)/100) +0.5
which gives me xi = 0.44m

am i on the right track with this problem? any guidance would be appreciated, thanks!

 

[voko]

I do not understand what that equation means. The right hand side is the initial angular momentum. But what is on the left hand side? Is that the potential energy of the spring? How could you possibly equate energy and angular momentum?

 

[Clever_name]

ok to start since angular momentum is conserved do i start out by writing the following equation?
m*v1*l1*sin(60) = m*v2*l2sin(a)=m*v3*l3sin(b),
l2 = min and l3 = max

 

[Clever_name]

I'm guessing i need to use r-polar coordinates here also?

 

[voko]

Think about what I wrote in #6.

 

[Clever_name]

ok so m*v1*l1*sin(60)-m*v3*l3sin(a) =0

divide through by m*v3*l3,
i get (mv1l1)/(mv3l3)(sin(60)-sin(a)) therefore 'a' must be equal to 60 at max L?

 

[voko]

I do not see why that would be true.

Have you heard about radial and azimuthal components of velocity?

 

[Clever_name]

no i haven't

 

[voko]

You mentioned polar coordinates earlier. Do you know what the polar-coordinates representation of a velocity vector looks like?

 

[Clever_name]

yes indeed, you have a velocity in the theta direction and a velocity in the radial direction where the square root of the two terms squared are then the total magnitude of the velocity

 

[voko]

Very well. Now think about the radial velocity. When the radius is at a minimum or at a maximum, what can you say about the radial velocity?

 

[Clever_name]

the radial velocity will be equal to 0, so v(theta) when l is at a maximum will be the velocity of the mass?

 

[Clever_name]

oh no velocity in the radial direction for both L max/min will be 0

 

[voko]

And if the radial velocity at the max/min radii is zero, what is the overall direction of the velocity? What is its angle with the radius at min/max?

 

[Clever_name]

90 degrees! the direction of the velocity would be perpendicular to the length of the cable L

 

[voko]

Very well. That should simplify the angular momentum equation significantly. Now all you need to do is add an equation for conservation of energy, and solve these two equations.

 

[Clever_name]

so that would be,

(1/2)k(xf^(2)-xi^(2))=(1/2)m(v1^(2)+v2^(2))

 

[voko]

The conservation of energy literally means that the sum of kinetic energy and potential energy in one configuration equals the sum of kinetic and potential energies in another configuration. I do not think that the equation in #23 expresses that.

 

[Clever_name]

ok so the equation should look like this,

(1/2)kxi^(2)+(1/2)mv1^(2)=(1/2)kxf^(2)+(1/2)mv2^(2)

 

[voko]

What are xi and xf?

 

[Clever_name]

lets assume I'm first trying to find the minimum L value so, xi=0.5m and xf=0?

 

[voko]

What do xi and xf mean?

 

[Clever_name]

oh, xf equals the final distance the spring is stretched, and xi is the distance the spring is stretched initially. hmmm i guess cause the cord is inextensible then xi=l1 and xf=l2.

 

[voko]

so what equations do you finally get?

 

[Clever_name]

(1/2)kl1^(2)+(1/2)mv1^(2) = (1/2)kl2^(2)+(1/2)mv2^(2)

where l1 = 0.5,

and

m*v1*l1*sin(60)=m*v2*l2

 

[voko]

Looking good.

 

[Clever_name]

I'm having trouble solving the two equations however,

i get v2=(4.33/l2) then subbing this into the energy equation i get,

42.5= 50l2^(2) + (11.24934/l2^(2))

i do not know how to go about solving that equation.

 

[voko]

Multiply by l2 squared throughout.

 

[Clever_name]

i get 42.5l2^(2)-50l2^(4) still can't see how to solve this.

 

[voko]

I think you have lost one term. Regardless, this is a called a bi-quadratic equation and I am pretty sure you studied it at some stage.

 

[Clever_name]

my bad yes, 42.5l2^(2)-50l2^(4)-11.2493.
I have not studied bi-quadratic equations before though, i'll have a browse around the internet for information

 

[Clever_name]

ok so i have 42.5l2^(2)-50l2^(4)-11.2493.
let l2^(2) =t

so -50t^(2)+42.5t-11.24934 this is a quadratic with roots of -0.212 or 1.062
so i disregard the negative value here because the root of a negative does not exist

so l2 must equal +- 1.03 m

 

[voko]

I do not think this is correct. The equation for t should have two positive roots.

 

[Clever_name]

ok now i think I've made a mistake somewhere because my b^(2)-4ac term is negative

-50t^(2)+42.5t-11.24934=0

when evaluating (1/2)mv1^(2) am i just using the 10m/s given, i think this is my mistake.

 

[Clever_name]

actually no, disregard that statement..

 

[voko]

The equation you got in #33 seems incorrect. Double check everything.

 

[Clever_name]

so i get l2 = +-0.828 or +-0.405, not sure what to do here with these figures however.

 

[voko]

Can the values of l be negative?

 

[Clever_name]

ah!, so l(min) =0.405 and l(max)=0.828?

 

[voko]

This seems correct.

 

[Clever_name]

thank you voko, for the guidance.