Solving a Problem: Forces Acting on P

  • Thread starter Thread starter greg_rack
  • Start date Start date
  • Tags Tags
    Forces
AI Thread Summary
The discussion revolves around analyzing the forces acting on point P at an angle of 60 degrees, specifically focusing on three forces: the spring force (F_S), the force from the guide (F_P), and the normal force. The user seeks confirmation on whether these are the correct forces to consider and how to derive the equations of motion from them. A key point raised is the need to determine the angle φ, which relates to the geometry of the system, particularly the isosceles triangle formed by points O, P, and the circle's center. The conversation highlights the importance of understanding the relationship between angles θ and φ to solve for the forces accurately. Overall, the user gains clarity on the problem and finds useful resources for further study.
greg_rack
Gold Member
Messages
361
Reaction score
79
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.
Screenshot 2021-11-22 174214.jpg
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
 
Physics news on Phys.org
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.
 
greg_rack said:
##\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.
 
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 o_O
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top