(a) Calculate the magnitude of the momentum of the bola at the moment of release

Question

(a) Calculate the magnitude of the momentum of the bola at the moment of release.

(b) Calculate the horizontal speed of the center of mass of the bola after its release.

(c) Calculate the angular momentum of the bola about its center of mass after its release.

(d) Calculate the angular speed of the bola about its center of mass after it has settled into its Y shape.

(e) Calculate the kinetic energy of the bola at the instant of release.

(f) Calculate the kinetic energy of the bola in its stable Y shape.

(g) Explain how the conservation laws apply to the bola as its configuration changes. i.e., why are (e) and (f) different, and what might the source of the difference be?

Explanation / Answer

(a) The speeds of the two stones are v0 for each of the stones at the moment of the release.

The magnitude of momentum = mv0 + mv0 = 2mv0

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(b) Just before the bola is released, the cener of mass is located at a distance d from the stone in hand, where

d = [2m(2l)]/3m = (4/3)l

Now the before the release the angular speed of rotation is .

then v0 = (2l) = 2l

or = v0/2l

So, speed at the distance d from the stone in hand is v = d = (v0/2l)d = [ (v0/2l)d][(4/3)l] = 2v0/3

So the speed of the center of mass before release is v = 2v0/3

The speed of the center of mass after release will remain the same due to absence of any external force acting on the system,

so horizontal speed of the center of mass the release is v = 2v0/3

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(c) Just after the release, the center of mass is located at a distance d = [2m(2l)]/3m = (4/3)l from the stone which was in hand before the release.

Before the release, let the angular momentum about the point P at a distance d = (4/3)l from the stone in hand be L.

Now, the speed of the point P is vp = d =  (v0/2l)[(4/3)l] = 2v0/3

Speed of the two masses at the other end of the string is v2m = v0

So speed of these two masses with respect to point P = vp - v2m =  v0 - 2v0/3 = v0/3,

Now the speed of the stone in hand is zero.

So speed of the stone in hand with respect to to point P = 0 - v0/2 = -2v0/3, negative sign just means that the velocity is in opposite direction to the that of point P.

So the angular momentum about point P before the release is;

L = 2m(speed of these two masses with respect to point P) + m(speed of the stone in hand with respect to point P)

L = 2m(v0/3)l + m(2v0/3)l = 4mv0l/3,

Now this point P (at a distance l from the stone which was in hand before the release) will be the center of mass just after the release. So angular momentum about P (or center of mass after release) before and after release will be the same due to conservation of angular momentum as there are no external torques present.

So, angular momentum of the bola about its center of mass after its release is L = 4mv0l/3

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(d) Let the angular speed be about the center of mass after release.

So, the angular momentum about the center of mass due to the three masses is L = 3ml2

But we saw in (c) that L = 4mv0l/3

So,  3ml2 = 4mv0l/3

or l2 = 4v0l/9

or = 4v0/9l

So the angular speed of the bola about its center of mass will be = 4v0/9l.

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This concludes the answers. Check the answer and let me know if it's correct. If you need anymore clarification or correction I will be happy to oblige....

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