Motion of a point mass (circular motion)

AI Thread Summary
The discussion focuses on the motion of a point mass constrained by a thread being pulled through a hole, with initial conditions and equations of motion established in polar coordinates. Participants clarify the separation of forces in the radial and angular directions, emphasizing the need to account for the changing radius as the thread is pulled. A misunderstanding arises regarding the acceleration of the radius, which is clarified by noting that while the thread is pulled at a constant velocity, the radius is indeed changing. The conversation highlights the importance of formulating a differential equation for the angular position as a function of time, using the known rate of change of the radius. Overall, the thread emphasizes the application of Newton's laws to derive the equations of motion for this system.
Carson Birth
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Homework Statement


A point mass, sliding over an even, horizontal plane is bounded on an inextensible, massless thread. During the motion, the thread is pulled by a force F with constant velocity ##v_{o} ## through a hole O. In the beginning (##t_{o} ## = 0) r(##t_{o} ##)=b holds (the length of the thread on the plane is b at the beginning then will get shorter due to the force F as it is pulled through). The initial velocity of the point mass perpendicular to the thread is ##v_{1} ## in ##\phi ## direction, and the angle ##\phi ##(##t_{o} ##) = ##\phi_{o} ## = 0Formulate the system's equation of motion and the equation of constraint forces in polar coordinates. Apply Newton's Law. Also the polar coordinate system is attached to the masspoint with r pointing away from the hole and ##\phi ## pointing toward the trajectory.

Homework Equations


##\overrightarrow{a} ## = (##\ddot{r} ## - r##\dot{\phi}^2 ##) in r direction + (r##\ddot{\phi} ##+2##\dot{r} ####\dot{\phi} ##) in ##\phi ## direction

The Attempt at a Solution


I separated the forces into there respected directions:
##\ddot{\phi} ## direction: mr##\ddot{\phi} ## + m2##\dot{r}####\dot{\phi} ## = 0
r direction: F + mr##\dot{\phi}^2 ## = 0

Now I am pretty sure there isn't any constraint forces, since there is no N force effecting the mass point.

So now I need to create an equation of motion, and from my understanding I need to create one equation. Is it as simple as just solving for ##\dot{\phi} ## and plugging it into the other equation? I had a similar equation where I solved the ##\phi ## direction equation as a differential but it didnt have the 2##\dot{r}####\dot{\phi} ## term with it, it was a gravity force making it very simple to solve. I know this isn't for people to solve my homework so I am just looking for advice on how to get it all set up for the further sub-questions. Any advice would be appreciated :D
 
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Carson Birth said:

Homework Statement


A point mass, sliding over an even, horizontal plane is bounded on an inextensible, massless thread. During the motion, the thread is pulled by a force F with constant velocity ##v_{o} ## through a hole O. In the beginning (##t_{o} ## = 0) r(##t_{o} ##)=b holds (the length of the thread on the plane is b at the beginning then will get shorter due to the force F as it is pulled through). The initial velocity of the point mass perpendicular to the thread is ##v_{1} ## in ##\phi ## direction, and the angle ##\phi ##(##t_{o} ##) = ##\phi_{o} ## = 0Formulate the system's equation of motion and the equation of constraint forces in polar coordinates. Apply Newton's Law. Also the polar coordinate system is attached to the masspoint with r pointing away from the hole and ##\phi ## pointing toward the trajectory.

Homework Equations


##\overrightarrow{a} ## = (##\ddot{r} ## - r##\dot{\phi}^2 ##) in r direction + (r##\ddot{\phi} ##+2##\dot{r} ####\dot{\phi} ##) in ##\phi ## direction

The Attempt at a Solution


I separated the forces into there respected directions:
##\ddot{\phi} ## direction: mr##\ddot{\phi} ## + m2##\dot{r}####\dot{\phi} ## = 0
r direction: F + mr##\dot{\phi}^2 ## = 0
r is changing. Why did you ignored ##\ddot r##?
 
ehild said:
r is changing. Why did you ignored ##\ddot r##?
My thought was since the thread is being pulled by a force with constant velocity, that it wouldn't be accelerating.
 
Carson Birth said:
My thought was since the thread is being pulled by a force with constant velocity, that it wouldn't be accelerating.
You are right, I misread it as "constant force". Sorry.
So you do not know F, but you know that ##\dot r ## is constant. Go ahead. Solve the first equation.
 
Last edited:
ehild said:
You are right, I misread it as "constant force". Sorry.
So you do not know F, but you know that ##\dot r ## is constant. Go ahead. Solve the first equation.
When you say solve the first equation, what do you mean? Would I solve the first equation as a differential equation or you mean solve for a variable then put it into the first equation? :D
 
Carson Birth said:
When you say solve the first equation, what do you mean? Would I solve the first equation as a differential equation or you mean solve for a variable then put it into the first equation? :D

It is a differential equation for Φ as function of time. You know ##\dot r## hence also r(t).
 
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