Solving a Problem: Forces Acting on P

[greg_rack]

Homework Statement

The smooth particle P is attached to an elastic cord extending from O to P. Due to the slotted arm guide, P moves along an horizontal path($r=0.8sin(\theta)$) with constant angular velocity of $\dot{\theta}=5rad/s$.
The mass of P is 0.08kg, the cord stiffness is k=30N/m and its unstretched length 0.25m.

Find the forces of the guide acting on P for theta=60

Relevant Equations

velocity and acceleration components in polar coordinates, Newton's 2nd law.

Hello guys,

here's a problem which I'm having troubles solving.
It asks for the forces acting on P when $\theta=60^{\circ}$.
I thought for this problem it would have been convenient to consider a polar reference system(r, theta). Drawing the FBD of pin P at a moment in time, we will have 3(?) forces acting on P:
-one($F_S$) along the r direction(pointing towards O caused by the spring-modeled cord, function of its stretch and k;
-one($F_P$) along the positive(direction of movement) theta direction, caused by the push "from backwards" of the guide exerted on P;
-a last normal force, exerted by the circle normal to the path, thus not aligned with the r-theta system defined.

Could you tell me if these are all the forces acting, and thus the equations of motion(FBD+KD) might be built from them?
In case yes, now we could start writing down the EOMs to solve for $F_P$.
$$
\left\{\begin{matrix}
F_S-Ncos\phi=m(\ddot{r}-r\dot{\theta}^2)\\
F_P-Nsin\phi=m(r\ddot{\theta}+2\dot{r}\dot{\theta})
\end{matrix}\right.$$

We may calculate $F_S$ from Hooke's law, but to solve for $F_P$ and N we still need $\phi$.
The procedure to calculate such angle seemed kinda convoluted and thus got me thinking I wasn't on the right path... I was as well confused on the problem's request: do they mean "my" $F_P$ by "forces of the guide acting on P"?

Greg

 

[Lnewqban]

Yes, those are the forces.
There is an isosceles triangle formed among points O, P and center of circle.
Two of its angles should be 90-60 degrees.

 

[kuruman]

$\dots~$we still need ϕ.

Suppose you defined $\phi$ as the angle subtended by the center of the wheel when the particle is at $\theta$ and such that $\phi=0$ when $\theta=0$ (particle at the 6 o'clock position). Note that $\phi=180^{\circ}$ when $\theta=90^{\circ}$ (particle at the 12 o' clock position). Can you deduce a general relation between $\theta$ and $\phi$? After all, the dependence between $\theta$ and $\phi$ can only be linear. If you'd rather measure $\phi$ conventionally, i.e. relative to the 3 o' clock position, then it's a matter of adding (or subtracting) $90^{\circ}$. Draw a good picture and you will see.

 

[greg_rack]

Thank you so much guys, now I see the relation!
I didn't realize that the direction of the normal force, was that of the radius of the circle. That helped me the most to spot $\phi=f(\theta)$.
Too many triangles and scribbles

 

[Lnewqban]

Thank you so much guys, now I see the relation!

http://507movements.com/mm_178.html

 

[greg_rack]

http://507movements.com/mm_178.html

Oh! Now I even see it moving

Thanks @Lnewqban for the super cool website, it's going to prove useful for this course!