# A mass attached to a string rotates, string meets a nail halfway thus the rotation continues

I have a ball attached to a masless string that rotates, and the string meets a nail midway thus the string with the mass now rotates around the nail with half the radius it rotated before, lets call that radius $L$, and $X$ is equal to $L/2$

I have angular velocity just a moment before the string meets the nail, when mass is at P, lets call that velocity $w_{1}$ and I need to find the Tension of the rope a moment before the rope meets the nail and a moment after it.

$T=mg+Lw_{1}^{2}m$

Now I can use conservation of angular momentum here(fix me if I'm wrong here) before and after meeting the nail and say:

$I_{i}w_{1} = I_{f}w_{2} \Rightarrow mL^{2}w_{1} = m(L/2)^{2}w_{2}$

and from that I see $w_{2} = 4w_{1}$ Is there something wrong here? Because this result says the velocity instantly changed once meeting the nail. to sum up my question, because it is requested to find the tension an instant after meeting the nail should I use:

1. $T = mg +\frac{L}{2}w_{1}^{2}m$

or

1. $T = mg +\frac{L}{2}w_{2}^{2}m$ ?

One argues the angular velocity has instantly changed and the other argues the velocity hasnt yet changed just a moment after meeting the nail because its not accelration which can instantly change thus we will still calculate the tension with $w_{1}$. I think number 2 is right with $w_{2}$, but most of my peers tell me I'm wrong, which is right in your opinion?

• Almost completely irrelevant; but if X = L/2 then this will actually smack into the nail that the string comes from at A
– JMac
Feb 2, 2017 at 19:24

Pirx is correct on his answer that the velocity does not change. The reason is that angular momentum should be conserved because the force acts on the center of rotation and does not create torque. Angular momentum should be computed in the same coordinate system, let us choose the origin at B. Then $mv_1x=mv_2x$ and so $v_1=v_2$ (same if you use conservation of energy). Then you can conclude that $w_2=2w_1$ (same as if you use conservation of energy). That is why your result is wrong, you used conservation of angular momentum but each side of the equal sign was computed using a different origin.
And yes, you should use $w_2$ because what changes instantaneously is not the angular speed, but your choice of system of coordinates. Even before the collision, the angular speed measured from A ($v/L$) is different than the angular speed as measured from B ($v/X)$ which does not change during the collision). What changes is the angular speed around A as compared around B.
• Thanks ! I suspected it has something to do with me changinc cordinates. But if we are using the same cordinates, I'm not sure how I find $w_{2} = 2w_{1}$. If we calculate them around the same origin, they have the same radius relative to origin( distance from origin) and that concludes the angular velocities are the same from the conservation around let's say point A $mL^{2}w_{1} = mL^{2}w_{2}$, wheere does the radius change comes in play ? Feb 2, 2017 at 19:22
• yes, if you use either A or B you will find both angular speeds equal (for A: $w_1^A=w_2^A$). But you want to relate the angular speed measured on reference frame A to that measured on ref. frame B. If you use conservation of momentum and obtain v1=v2, then you can replace $Lw_1^A=Xw_2^B$ to get the angular speeds in the two frames.
• Oh ok thanks! And I also think me calculating the moment of inertia $I$ after hitting the nail with the same pot of refrence is really wrong because the radius isn't the same for each point so $I=mr^{2}$ is wrong to say to one of the cases.I would need to do an intergral to actually calculate that I assume. Feb 3, 2017 at 8:56