Circular wire with ball and spring problem

[Victoria_235]

Homework Statement

Here is the problem.
A small ball of mass m moves without friction attached to a cwire.
The ball moves under the action of gravity and a spring of constant k whose other end is fixed in the point A, with OA = R / 2.
The spring is tension free when the spring is R/2
to. Calculate the potential energy of the mobile as a function of the angle φ.
B. Calculate the speed at which the mobile must be released in position with free tension of the spring so that the ball can go around.

The system is on the horizontal plane.
Solve also in case it is in the vertical plane.

Homework Equations

The Attempt at a Solution

[/B]
So, I have been trying to solve it...

Firstly, I have considered Newton's equations.

On k direction, perpendicular to the problem plane, is gravity, g and the reaction force of the wire, as the ball is moving in OXY plane, the resultant should be 0.
N_k - m*g = 0

On the other hand, we have the OXY forces.

F_s (spring force) and the reaction of the wire in the radial direction (N_r)

so:

F_s + N_r = m*a, where m is the ball mass, and a is the aceleration

The energy balance, I think is this way:

Kinetic energy -> K_e = 1/2m(v_i)², where v_i is the initial speed.
Potential energy -> If I consider the system in the horizontal plane with 0 high, it should be 0.
Spring energy -> S_e= 1/2k*x^2, where x is the distance to any point in the wire to A.

But I have no idea to continue it...

Could anybody help me? :)

 

[I like Serena]

Hi Victoria, welcome to PF,

The spring energy is indeed of the form 1/2 k s^2, where s is the length of the spring beyond the length at which it is at rest.
Can we find that length?
We need it as a function of the angle φ, which is the angle between OP and the X-axis.
How about the length AP?
Can we calculate it as function of φ?

Suppose we express the x-coordinate of P as a function of φ.
That should have a cosine in it shouldn't it?
What could it be?

 

[Victoria_235]

Hello, thank you very much for your response.
We could give "AP" as function of R, where R is the radio. With law of cosines, we have that AP = R\sqrt \frac{5}{4} - cos\alpha , where alpha is the angle between OA and AP.
This way, we could give potential energy as function of angle.
But, what I really don't know how to solve is the condition of initial speed to turn full circle arround.
Any clue about it?

Thank you. :)

 

[I like Serena]

Hello, thank you very much for your response.
We could give "AP" as function of R, where R is the radio. With law of cosines, we have that AP = R\sqrt \frac{5}{4} - cos\alpha , where alpha is the angle between OA and AP.
This way, we could give potential energy as function of angle.
But, what I really don't know how to solve is the condition of initial speed to turn full circle arround.
Any clue about it?

Thank you. :)

To turn full circle, we need sufficient speed to 'reach' the hardest point.
The hardest point is where the spring is extended the most.
At this point the speed has to be 'just' positive.
Let's set the speed in this 'hardest' point to zero, meaning that the kinetic energy is zero.
Since we have conservation of energy, the initial spring energy plus the initial kinetic energy has to be at least the spring energy of the 'hardest' point.

Btw, the formula for AP doesn't look quite right.

 

[Victoria_235]

I have put AP distant as function of radio and radio/2. Because of the cosine theorem, $c^2 = a² + b^2 - 2abcos\phi$, I wrote $L^2 = (R/2)^2 + R² - (2R^2)/2 cos\phi$ --> then, $R \sqrt{ ((\frac{5}{4})- cos)\phi)}$, where L = AP.

So, analyzing the energy in the points A and B, (see the new draw, I have put the potential energy origin in B, and the maximum in A, beeing the circle in vertical position)
Energy in A:
$$Ep_a=2mgR$$
$$Ec_a = \frac{1}{2} m(v_a)²$$
$$Ee_a= \frac{1}{2}K (R/2)^2 = 1/8 kR²$$
Where, Ep is potential energy, Ec kinetic, and Ee the elastic one.

Energy in B:
$$Ep_b= 0$$
$$Ec_b= \frac{1}{2}m(v_b)^2$$
$$Ee_b =\frac{1}{2} K(R+(R/2)^2) $$

Doing $Ep_a + Ec_a + Ee_a = Ep_b + Ec_b + Ee_b$, and $v_b= 0$

I obtain that initial speed in A should be more than $\sqrt {\frac{2KR²}{m}-4Rg}$

where, v_a is the initial speed, v_b speed in B, m the mass, K spring constant.

 

[Victoria_235]

Sorry, may be not too much clear in the last image, might here is better.

 

[I like Serena]

Nice drawings! :)

Let's reorder the parentheses a bit.
It should be $AP=R\sqrt{\frac 54 - \cos(\phi)}$.

It is given that the spring is at rest if its length is R/2. Shouldn't we subtract that from AP before calculating the elastic energy?

Btw, doesn't potential energy include both gravitational energy and elastic energy?