Non-Uniform Motion in a circle

[myheadhurts]

I am a bit confused by this question(see attached image). This has been posted before but im not any further forward. My problems are:

 

 

P1 - has this got the following forces acting on it: Weight, Tension, and Natural reaction force?

 

and thus does the equation of motion end up as:

 

mr''= -|T|er - (mgcos a)er - (mgsin a)eo + |N|er

 

where er and eo are the radial and tangential components respectively

 

Then is it a case of gathering the similar terms and resolving in the er direction to get:

 

ml0'^2 = mg cos a - |T|

 

 

P2 - is this just experiencing the tension of the string and its own weight?

[swansont]

Natural reaction force? I've not heard the term — I'd call it the Normal force, i.e. normal to the surface.

 

From just a quick glance — do you have an equation for the tangential acceleration?

[myheadhurts]

is there not a natural reaction force due to the particle being in contact with surface of the cylinder?

[swansont]

is there not a natural reaction force due to the particle being in contact with surface of the cylinder?

 

That's what is often called the normal force, at least in my experience. Calling it a reaction force begs the question, reaction to what? The tension is a reaction force of one particle on the other, via the connection. The weight is a reaction force of the gravitational force exerted by each body on the earth.

[Pete]

is there not a natural reaction force due to the particle being in contact with surface of the cylinder?

To provide an answer you need to define the term natural reaction force since I too have never heard of this term.

 

Pete

[salaw]

This is a cool problem! I'm not sure I follow the OP's comments on it, but the problem itself is pretty nifty. Note the salient points:

• The cylinder is fixed -- it does not rotate.
  
• It is not stated explicitly, but the implication of the smooth cylinder is that weight P1 slides frictionlessly on it.
  
• The initial motion of P1 is circular ... however ....
  
• The normal force keeps P1 from entering the cylinder, but the only thing holding it on that surface is its weight. At some point, as P1 slides up and over the cylinder, it will break away from the surface of the cylinder, which leads to some interesting questions:
   
   
  
  1. The breakaway point is presumably somewhere past the top of the cylinder -- but where, exactly?
      
      
     
  2. After breakaway, P1 will be following a roughly parabolic path. Will that path intercept the surface of the cylinder? I.e., will it strike the cylinder again, after breaking away?
      
      
     
  3. What is the shape of the path followed by P1 after it breaks away from the surface, and what is the shape of the path followed by P2 after P1 breaks away from the surface? Can they be described with elementary functions?
      
     

 

Unfortunately I haven't got time to pursue this tonight, but if nobody else posts a solution I'll fiddle with it a bit this weekend. (Not that I'm at all sure I can solve it -- the first question I listed looks straightforward, but the second and third look hard!)

 

*********************************************

The foregoing text was written Friday night; the following additional analysis was appended the next morning:

 

I played around with this one some more this morning. Sure enough, finding the point where P1 breaks away from the cylinder is straightforward.

 

If we set the potential energy when [math]\theta[/math]=0 to 0, then we have

 

[math](1)\ V = mgR(\sin\theta - 2\theta)[/math]

 

where g is the acceleration due to gravity. The kinetic energy is

 

[math](2)\ T = {1 \over 2} m [R^2\dot \theta^2 + 2R^2 \dot \theta^2]

= {3 \over 2} mR^2\dot\theta^2

[/math]

 

and the Lagrangian is

 

[math](3)\ L = mR [{3 \over 2}R\dot\theta^2 + 2g\theta - g\sin\theta][/math]

 

Taking the usual partials,

 

[math]

(4)\ {{\partial L} \over {\partial \theta}} = gmR [2 - \cos\theta]

[/math]

 

[math]

(5)\ {{d} \over {dt}} {{\partial L} \over {\partial \dot\theta}} = 3mR^2\ddot\theta

[/math]

 

and setting [math]{{\partial L}\over{\partial\theta}}={d\over{dt}}{{\partial L}\over{\partial \dot\theta}}[/math], we find the equation of motion up to the moment when P1 breaks away from the cylinder:

 

[math]

(6)\ 3R\ddot\theta = g(2 - \cos\theta)

[/math]

 

If we can solve that, then we'll have the angle and velocity as a function of time. Unfortunately, I can't. But as it happens, to find the angle at which P1 breaks away, we don't need to solve (6) anyway; we just need the velocity as a function of the angle, and we can find that just from the total energy. The system is conservative, and we start with T=V=0, so we can equate T with -V, which gives us exactly what we want:

 

[math]

(7)\ {3 \over 2}mR^2\dot\theta^2 = g [2mR\theta - mR\sin\theta]

[/math]

or

[math]

(8)\ \dot\theta^2 = g [{4 \over {3R}} \theta - {2 \over {3R}} \sin\theta]

[/math]

 

And now we need the centrifugal "force". As usual I've forgotten the expression for that, so we'll take a moment to rederive it:

 

Lemma -- Centrifugal "Force":

For a free particle of mass M, in polar coordinates (r,theta), we have:

 

[math]T = {1 \over 2}r^2\dot\theta^2M + {1 \over 2}M\dot r^2

[/math]

 

[math]

V = 0

[/math]

 

All we're interested in is the radial acceleration, so we look at the equations in r:

 

[math]

{{\partial L}\over{\partial r}} = Mr\dot\theta^2

[/math]

 

[math]

{d \over {dt}}{{\partial L}\over{\partial\dot r}} = M\ddot r

[/math]

 

Equating those we obtain the centrifugal acceleration:

[math]\ddot r = r \dot\theta^2[/math]

 

At breakaway, the central force must equal the (fictitious) centrifugal "force". Plugging in the mass of P1 times the expression for the centrifugal acceleration in polar coordinates, to get the centrifugal "force", and comparing it with (8) above, we find that it's equal to:

 

[math]

(9)\ mR\dot\theta^2 = mg [{4 \over 3}\theta - {2 \over 3}\sin\theta]

[/math]

 

The central force acting on P1 must be equal to that at the moment of breakaway. The central force is due to gravity, and is

 

[math]

(10)\ F_c = mg\sin\theta

[/math]

 

Equating (9) and (10) we find

 

[math]

(11)\ \sin\theta = {4 \over 3}\theta - {2 \over 3}\sin\theta

[/math]

 

or

 

[math]

(12)\ \sin\theta = {4 \over 5}\theta

[/math]

 

A little fiddling with a calculator indicates the breakaway point is at about 1.1311 radians, which is about 65 degrees. That's well before the top of the cylinder!

 

Note that the string from P1 is still wrapped partly around the cylinder at that point, so P1 is being pulled toward the cylinder by the force on the string. Consequently, it may very well hit the cylinder again before it finally falls free; we can't be sure from what I've written here. Unfortunately, answering that question, and finding the shapes of the paths of P1 and P2, are both well beyond my limited differential equation skills; to go from here I'd have to use a numerical simulation and the amusement value of such a thing wouldn't pay for the time it would take .

Edited September 13, 2008 by salaw
multiple post merged

[swansont]

One should realize that when a homework-like question is posted, the goal is to help the poster find the answer him/herself, rather than do the work for them.