What is the solution to the spring and disk rotation problem?

[Arman777]

Homework Statement

Overhead view of a spring lying on a frictionless surface and attached to a pivot at its right end.The spring has a relaxed length of $l_0=1.00m$ and a negligible mass.A small 0.100 kg disk is attached to the free end at left.That disk is then gvena velocity $\vec {v_0}$ of magnitude $11\frac {m} {s}$ perpendicular to the springs lenght.The disk and spring then move around the pivot.
(a)When the streching of the spring reaches its maximum value of $0.100l_0$, what is the speed of disk,
(b)What is the spring constant ?

Homework Equations

[/B]
Energy conservation.

The Attempt at a Solution

[/B]From Newtonian equations we know that $fx=\frac {mv^2} {r}$ so from that we can obtain $1.1k=v^2$

Then I wrote energy conservation

İnitally it has a speed $11\frac {m} {s}$ and later time it will have some velocity v' and there's also spring potantial energy

Here I stucked,Do I have to add rotational energy of spring-disk system ? I think I should but I got wrong result.
The equation will be $\frac 1 2m(v_0)^2=\frac 1 2m(v')^2+\frac 1 2kx^2$
here
$x=0.1m$
or $\frac 1 2m(v_0)^2=\frac 1 2mI(w)^2+\frac 1 2kx^2$ but its same as upper equation.

Any help ?

 

[kuruman]

When you write F = kx for the magnitude of the spring force, x is not the length of the spring. It is the extra stretch (or compression) of the spring.
Because you have a point mass, you can either write the rotational energy or the translational energy. As you have seen, they give the same result.

 

[Arman777]

When you write F = kx for the magnitude of the spring force, x is not the length of the spring. It is the extra stretch (or compression) of the spring.
Because you have a point mass, you can either write the rotational energy or the translational energy. As you have seen, they give the same result.

I know I wrote k0.1=0.1v^2/1.1

 

[kuruman]

Yes, I see now that you canceled the 0.1 for the mass m with the 0.1 for the displacement x.

When you say you got the wrong result, what result did you get and how? Do you know what the "correct" result is?

 

[Arman777]

$1.1k=(v')^2$
$\frac 1 2m(v_0)^2=\frac 1 2m(v')^2+\frac 1 2kx^2$here
$x=0.1m$

so we can cancel $\frac 1 2$ then
$0.1kg (11\frac {m} {s})^2=0.1kg(v')^2+k(0.1m)^2$
$12.1J=0.11k+0.01k$
$12.1J=0.12k$
$k=100.83 \frac {N} {m}$

answer is $k=210\frac {N} {m}$

 

[kuruman]

You conserved mechanical energy and that gave you one equation. Is there anything else that is also conserved? That is where you get the second equation you need.

 

[Arman777]

You conserved mechanical energy and that gave you one equation. Is there anything else that is also conserved? That is where you get the second equation you need.

Angular momentum will be conserved cause $τ=0$ so $L_f=L_i$
$1.1mv'=1.m(v_0)$

from that $v'=10\frac {m} {s}$ and $k=210\frac {N} {m}$

but still didnt understand why $kx=\frac {mv^2} {r}$ is not correct ?

 

[kuruman]

but still didnt understand why $kx=\frac {mv^2} {r}$ is not correct ?

Because it does not guarantee you that there is no radial component of the velocity. The best way to tackle problems of 2d motion involving central forces (forces that depend on r only) where angular momentum is conserved is to write the mechanical energy as
$$ME = \frac{L^2}{2mr^2}+\frac{1}{2}mv_r^2+V(r)$$ where $L$ is the conserved angular momentum, here $L=m v_0 l_0$, $r=l_0+x$ and $V(r)=\frac{1}{2}k x^2$. The first term in the equation is the rotational energy of the mass and is also known as "the centrifugal potential."

The condition for maximum extension of the spring is obtained by solving the ME conservation equation for $v_r$, and then setting $v_r = 0$. This results in $$\frac{k}{m}=\frac{v_0^2(2l_0+x)}{x(l_0+x)^2}.$$

 

[Arman777]

Because it does not guarantee you that there is no radial component of the velocity. The best way to tackle problems of 2d motion involving central forces (forces that depend on r only) where angular momentum is conserved is to write the mechanical energy as
$$ME = \frac{L^2}{2mr^2}+\frac{1}{2}mv_r^2+V(r)$$ where $L$ is the conserved angular momentum, here $L=m v_0 l_0$, $r=l_0+x$ and $V(r)=\frac{1}{2}k x^2$. The first term in the equation is the rotational energy of the mass and is also known as "the centrifugal potential."

The condition for maximum extension of the spring is obtained by solving the ME conservation equation for $v_r$, and then setting $v_r = 0$. This results in $$\frac{k}{m}=\frac{v_0^2(2l_0+x)}{x(l_0+x)^2}.$$

I see I guess,Is that means if there's radial veloctiy the force will be different then $kx=\frac {mv^2} {r}$ ?

 

[kuruman]

see I guess,Is that means if there's radial veloctiy the force will be different then $kx=\frac {mv^2} {r} ?$

When you say "the force" which force do you mean? There is only the force, the spring force, which provides both a centripetal acceleration and a radial acceleration. In that case, Newton's second law in the radial direction is $$m\frac{d^2x}{dt^2}=\frac{mv_{\theta}^2}{l_0+x}-kx$$
Here, the left side of the equation, mass times radial acceleration, is zero instantaneously at maximum extension (or compression) of the spring. When the spring goes from zero to maximum extension, it does so in a way that conserves angular momentum. The mass goes around while executing radial oscillations. It is not in a circular orbit, which is what your initial solution assumes.

 

[Arman777]

When you say "the force" which force do you mean?

I tried to meant force equation or the moton equation.Ok.Now I understand the whole idea.Thanks a lot :)

 

[Arman777]

Or wait I hate this but

Here, the left side of the equation, mass times radial acceleration, is zero instantaneously at maximum extension (or compression) of the spring.

It gives my equation when left side is zero ? sorry I am tired but I have to ask

 

[Arman777]

Nevermind maybe I understand later.thanks again

 

[Arman777]

Or wait its not a general equaiton ($kx=\frac {mv^2} {r})$its true for instantenously that's why we can't apply ?

 

[kuruman]

Or wait its not a general equaiton $(kx=\frac {mv^2} {r})$ its true for instantenously that's why we can't apply ?

Yes.

 

[Arman777]

Yes.

Finally,I understand no more questions.Thanks again :)