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Conservation of angular momentum 
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#1
Jan2113, 09:53 PM

P: 248

I have 2 questions about the situations below
Let vcp be the initial tangential velocity required for the upper body for that, when the low body is released, the upper body describes a circular trajectory I  In situation A, if we leave upper body with an initial tangential velocity (different from vcp), we have F2 ≠ F1 II  If the initial tangential velocity = 0, F2=F1 III  if initial tangential velocity is vcp, F2 = F1 IV  In situation B if we consider the pulley to have no mass we have F2=F1 a) Why do we have F2 ≠ F1 in case I and F2 = F1 in cases II and III ? b) What would happen (the shape of the movement) if in situation A we had a pulley with mass=0 in the central hole? **Note that the white circle in A has nothing to do with the trajectory of the upper body, it's just a circle **There's no friction between any surfaces (only between the string and the pulley in case B, otherwise the string would slide in the pulley, not rotating it. But the pulley has no mass, so no energy is lost) **Try to explain both qualitative and quantitatively 


#2
Jan2113, 10:04 PM

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F1 is always equal to F2 in magnitude. The only difference between these situations is the accelerations of the masses. If the tangental velocity of the mass on the table is less than vcp, then the lower mass accelerates downwards, reducing tension in the string. Over a period of time, the upper mass will increase speed, and the system may reach a point where the lower mass accelerates upwards. Assuming no friction or damping factors, depending on the initial conditions, the system could oscillate between accelerating the lower mass downwards and upwards.



#3
Jan2213, 12:24 AM

P: 248

I don't think so. Let m = 1kg be the mas of the upper body, M=2kg the mas of the low body, R = 0.4m the initial radius and v0 = √0.8 the initial velocity (tangential) of the upper body (I got this numbers from an exercise).
We have conservation of angular momentum in the axis that passes trough the hole and is perpendicular to the table. So v0.R = vt.r (vt is the tangential velocity and r is radius at that moment) Let vy be the "centripetal" velocity of the upper body (velocity that point to the hole =dr/dt). vy is equal to the velocity of the low body Let ay be d^{2}r/dt^{2}. ay is the acceleration of the low body If a = F1/m, we have ay = ( avt^{2}/r)/2 (I) (look at the picture below) Conservation of energy: MgR + mv0^{2}/2 = Mgr + m.vt^{2}/2 + (M+m).vy^{2}/2 Substituting vt = v0.R/r, substituting values, deriving in relation to time and dividing by vy we get ay = 2/3 (10(0.4/r)^{3}) So in the initial moment ay = 6m/s² Now let's suppose by absurd that the forces were equal F1=F2 = F In the initial moment: MgF = M.ay But by I we get ay = F/21 So 20F = F2 > F = 11N So ay = 4.5m/s², absurd. So F1 != F2 


#4
Jan2213, 06:40 AM

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Conservation of angular momentum
The acceleration of the lower mass equals the radial component of acceleration of the upper mass: d^{2}y/dt^{2} = d^{2}r/dt^{2} The tension in the string related to the lower mass: m (g + d^{2}y/dt^{2}) = m (g + ay) where ay is the acceleration of the lower mass (ay is negative if lower mass is accelerating downwards). Note that if the velocity of the upper mass is not vcp, then it's path is not circular, and the centripetal force, being perpendicular to the path of the upper mass, does not point directly to the hole in the table. I don't know how to reuse an attachment, so here is an image link to an attachment I have from a previous post. The short inward lines are perpendicular to the path of the mass, and would correspond to the direction of the centripetal component of acceleration of the mass. 


#5
Jan2213, 09:05 AM

P: 248

But note that my solution is NOT that, as well as the ay I've mentioned is NOT centripetal, it's RADIAL (look at the demonstration) The radial acceleration is resultant acceleration, since the resultant force is radial, so ay is the resultant acceleration on the upper mass. But the centripetal acceleration is not the resultant acceleration, since we have a tangential acceleration too that summed to the previous one, has the radius direction. Note that this does not violates the principle of resultant force in the string being zero (if it was not zero the acceleration would tend ∞). When the string touches a point, this point makes another force in the string that can cancel the difference in tensions Some examples: 1) The black dot is the hole of the case A If F1 = 4N and F2 = 3N and if tg(θ) = 3/4, we have that F cancels the difference in tensions and the resultant force in the string is zero. Even with different tensions. 2) If the pulley in case B had mass, we would still have a difference of tensions making a torque in the pulley, rotating it. For this to be possible, the pulley have to have static friction with the string, otherwise it would slide. The friction and the normal forces act like the F force in the example above, canceling the difference of tension inn the string. 


#6
Jan2213, 10:05 AM

P: 963

In the original post it was stated that: "**There's no friction between any surfaces" (only between the string and the pulley in case B, otherwise the string would slide in the pulley, not rotating it. But the pulley has no mass, so no energy is lost)"
Together with an assumption that the string is massless, this entails that the tension on the string is the same from end to end. [Note that the parenthetical is wasted. If there's no friction between string and pulley, the pulley still works just fine. The pulley does not rotate. But it does not need to. That's precisely the situation for a string running over a frictionfree corner] The example of a 3 4 5 triangle fails the equaltension criterion. There has to be friction for the tension to be 3 on the one leg and 4 on the other. If we ignore the "no friction" clause from the original post then the problem is ambiguous and cannot be solved. 


#8
Jan2213, 10:58 AM

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P: 7,133

For a massless string, the tension is the same everywhere on the string and at the points of attacment to both masses. The tension can be stated as a function of acceleration of the lower mass, which I'll call M in this post:
M (g + d^{2}y/dt^{2}) = M (g + ay) where ay is negative if M is accelerating downwards. The tension in the string is the net (and only) force on the upper mass, which I will call m in this post. The net acceleration of the upper mass = tension / m. The path of the upper mass will depend on it's initial velocity. I'm not sure how to solve for a dynamic situation. For a circular path, the tension in the string is relative to 1/r^3. For example, since angular momentum is conserved, 1/2 the radius requires 2 times the speed, and this results in 8 times the tension. 


#9
Jan2213, 11:59 AM

P: 963

This notion may be difficult to apply to the case of a sharp corner, so consider a rounded corner that can be made as sharp as desired. If there is no tangential force then the tension in the string at any point along the curve must be the same as the tension at the next point. So the tension along the whole string must be constant 


#10
Jan2213, 12:01 PM

P: 248

As different tensions in the same string are such an uncommon thing, I posted here to see if my calculations were right. I want to know if I was right in my post and if I don't I want to know WHERE my calculations are wrong and WHY the tensions have to be equal (in specific why there has to be friction as rcgldl said) 


#11
Jan2213, 01:20 PM

P: 248

First I want to say that I'm not doubting any one of you guys can be right, I just want an explanation for something I cannot understand You said that if there was no tangential forces than the tensions should br the same. Here is my counterargument: I) The atoms in the string are together because attraction forces are acting between consecutive atoms (tension) II) The string can be thought like a spring or an elastic where k→∞ III) If we elongate the string (increase the distance between atoms), the force between the consecutive atoms increase Now look at the picture below: The big pink ball is the pulley and the black forces are the normal forces acting on the atoms. If we had different tensions then the atoms would be in different distances. In this case we have α≠θ, and it does not counter any of the 3 first statements (the trianggle is not isosceles so the tensions are changing. Tensions on the right are bigger than tensions on the left .That's plausible since the distances are increasing too.) So if I was not wrong in any consideration, the tensions could change. **** But you can see how microscopic explanation is difficult. Note that all I want want is an explanation for why I'm wrong. In the second post I did, I didn't assume anytime the tensions were different, I just got to it. So all I want is that you guys find a mistake in my SECOND post. If there are mistakes there, then I can see that I'm wrong. 


#12
Jan2213, 02:30 PM

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It appears to me that in your diagram and also in the next to last line of your derivation you used the symbol ##a## where you should have used ##a_y## (or ##\ddot{r}##). But I don't follow your derivation, so I'm not sure. It is not hard to derive the correct expression for ##\ddot{r}## using polar coordinates. See here for a derivation. The result is that the total radial component of acceleration is ##a_r = \ddot{r}r\dot{\theta}^2 = \ddot{r}v_t^2/r##. So, ##\ddot{r} = a_r v_t^2/r##. In your problem, ##a_r## is what I believe you are calling ##a##; that is, ##a_r = F/m = a##. Anyway, if you use the correct expression ##\ddot{r} = a_r v_t^2/r= F/m v_t^2/r##, I think you will find that the tension is the same for both masses. 


#13
Jan2213, 03:04 PM

P: 963

If the string is massless and there is no friction, then the the net tangential force on any small section of the string must be zero. Otherwise Newton's second law would say that the acceleration of that small section would be infinite.
So the tension on the string is constant end to end. 


#14
Jan2213, 03:29 PM

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P: 7,133

ac = centripetal acceleration ar = ay v0 = sqrt(.8) r0 = .4 m ac = .08/.4 = 2 F1 = 1kg (ac + ar) = 1kg (2 + 6) = 8 N 


#15
Jan2213, 03:31 PM

P: 248

[]'s John 


#16
Jan2213, 04:18 PM

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P: 7,175

Just a comment on the "tension in the string" questions:
In your case B, if the pulley is massless the string tension is equal at both ends. If the pulley has mass, the tensions may be different. As you say, the friction between the string and the pulley gives the tangential force that accelerates the pulley, and changes the string tension along its length. For a detailed analysis of this, see http://en.wikipedia.org/wiki/Capstan_equation. If you assume the string does not slip, you can model this in two simpler ways: (1) Apply Newton's 2nd law to the pulley, to get an equation connecting the change in the string tension with the angular acceleration of the pulley. (2) Use conservation of energy, and include the KE of the rotating pulley. In general, you could consider situation A to be the same, with different tension in the two parts of the string. But for the problem in your OP, you don't have enough information to solve that problem. When the string moves over the "rough" edge of the hole, it does work because of the "friction" force, but you are not given any information to find how much work is done, or what the difference is tensions is. So the only assumption you can make to solve the problem is that there is no friction at the hole, and the tension in both ends of hte string is the same. And when you do the analysis correctly assuming no friction, all the results are consistent, as you already figured out! 


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