Finding the angular velocity of a rotating arm

[Like Tony Stark]

Homework Statement

There's an arm which is $2R$ long and rotates with constant angular velocity on a horizontal plane. There's also a slider block with mass $0.4 kg$. This slider is pushed against a support $A$ by a spring of constant $2 N/m$ and natural length $2R$.
A) Suppose that the force exerted by the support is half the force exerted by the elastic force, find the angular velocity of the system.
B) Find the minimum angular velocity so that the slider doesn't interact with the support.
C) If the angular velocity found previously is doubled, which would be the equilibrium position with respect to the arm?
D) Which would be the difference if the angular velocity is now rotating the other way?

Relevant Equations

Newton's equations

A) The slider experiments three forces, all of them are on the $x$ axis (considering $x$ axis as the axis aligned with the arm): Normal force (exerted by the support), elastic force and centrifugal force, which is $m.(\omega^2 r)$

Elastic force is equal to
$Fe=-k \delta =-2 (R-2R)=2R$ and therefore $N=R$. Then, Newton equations would be
$N-Fe+m (\omega^2 R)=0$. So, solving for $\omega$ I get:
$m.\omega^2 R=R$, and then, replacing the value of the mass we get $\omega = \sqrt{2.5}$

B) For B we solve the same equation but set $N=0$.

C) For C, we solve
$k(R_{eq} -2R)+m\omega^2 R_{eq} =0$, where $R_{eq}$ is the radius for which the system is in equilibrium.

D) The angular velocity affect the centrifugal force, which is $m.(\omega^2 r)$. As $\omega$ is squared, the sign doesn't change anything.
Have I solved the problem correctly? I don't know if my concepts are correct

 

[haruspex]

C) For C, we solve
$k(R_{eq} -2R)+m\omega^2 R_{eq} =0$, where $R_{eq}$ is the radius for which the system is in equilibrium.

Where omega is what value?

 

[BvU]

Your concepts are correct.

Detail: a statement like $R = N$ hurts my eyes: the dimensions have disappeared. Write e.g:$$
N = -{1\over2} k(R-2R) \quad \Rightarrow \\
N + k(R - 2R) = {1\over2} k(R-2R) = {1\over2}kR\quad \Rightarrow\\
N + k(R - 2R) +m (\omega^2 R)=0 \ \ \Leftrightarrow\ \ \omega^2 = {1\over2} {kR\over mR} \quad \Rightarrow \\ \omega = \sqrt {2.5}\ \text {rad/s } $$

 

[Like Tony Stark]

Where omega is what value?

The double of the calculated in B. Sorry I didn't make any difference, but I didn't want to write $\omega_2$ or something like that. I thought it was easy to understand just like that

 

[Like Tony Stark]

Your concepts are correct.

Detail: a statement like $R = N$ hurts my eyes: the dimensions have disappeared. Write e.g:$$
N = -{1\over2} k(R-2R) \quad \Rightarrow \\
N + k(R - 2R) = {1\over2} k(R-2R) = {1\over2}kR\quad \Rightarrow\\
N + k(R - 2R) +m (\omega^2 R)=0 \ \ \Leftrightarrow\ \ \omega^2 = {1\over2} {kR\over mR} \quad \Rightarrow \\ \omega = \sqrt {2.5}\ \text {rad/s } $$

Thanks and sorry! It looks ugly to me too, but it was messy to me to write all the units in the post (I write them when doing the homework, obviously)

 

[BvU]

Practice makes perfect