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Thanks to all those who responded to my little problem. An
interesting point is that not everyone got the right answer!! but I am not going
to publish who did and who did not. Here is my interpretation.
(a) The answer to the first part of the question is quite straight
forward. Since angular momentum remains constant (no external
torques),then the ratio of the final and initial kinetic energies is
given by:
I1 w1 = I2 w2
1.5 I1 = 5 I2 or I1 = 3.33 I2
(1/2 I1 w1^2) / (1/2 I2 w2^2) (ratio of initial to final KE)
after substitution the ratio is 1:3.33
(b) To increase the energy of the system, work must be done. The only
force that you can apply is the centripetal one and you can clearly
see that it must do positive work since the masses move inward.
Here is my attempt at the explanation. I hope that you can all
follow. If not, I will send you a hard copy done in an equation editor.
Imagine that you model the system as a point mass located at a
distance r from the centre axle of the turntable (this distance could
actually be the radius of gyration). If you let the velocity of this mass
be v and the centripetal force equal F the the work done by
this force is,
Fs = (mv^2)/r dr
By using the work-energy principle, we can say that the change in
energy is equal to the work done. That is
Fs = (mv^2)/r dr = dE = d(1/2 m v^2)
-(mv^2)/r dr = m v dv (I have included the minus sign because r is
decreasing)
-dr/r v = dv
0 = (v + r) dv/dr
0 = d/dr(vr) - which says that the rate of change of the vr (which
is effectively the moment of the linear momentum) is zero.
Thus, the whole thing works out nicely. The increase in energy can be
accounted for and it turns out that the angular momentum (moment of the linear momentum) remains constant.
Cheers and best wishes.
Rob
_____________________
Robert Neal, PhD
Department of Human Movement Studies
The University of Queensland
QLD, AUSTRALIA
ph 61 7 365 6240
FAX 61 7 365 6877
EMAIL NEAL@HMS01.HMS.UQ.OZ.AU
_____________________________
interesting point is that not everyone got the right answer!! but I am not going
to publish who did and who did not. Here is my interpretation.
(a) The answer to the first part of the question is quite straight
forward. Since angular momentum remains constant (no external
torques),then the ratio of the final and initial kinetic energies is
given by:
I1 w1 = I2 w2
1.5 I1 = 5 I2 or I1 = 3.33 I2
(1/2 I1 w1^2) / (1/2 I2 w2^2) (ratio of initial to final KE)
after substitution the ratio is 1:3.33
(b) To increase the energy of the system, work must be done. The only
force that you can apply is the centripetal one and you can clearly
see that it must do positive work since the masses move inward.
Here is my attempt at the explanation. I hope that you can all
follow. If not, I will send you a hard copy done in an equation editor.
Imagine that you model the system as a point mass located at a
distance r from the centre axle of the turntable (this distance could
actually be the radius of gyration). If you let the velocity of this mass
be v and the centripetal force equal F the the work done by
this force is,
Fs = (mv^2)/r dr
By using the work-energy principle, we can say that the change in
energy is equal to the work done. That is
Fs = (mv^2)/r dr = dE = d(1/2 m v^2)
-(mv^2)/r dr = m v dv (I have included the minus sign because r is
decreasing)
-dr/r v = dv
0 = (v + r) dv/dr
0 = d/dr(vr) - which says that the rate of change of the vr (which
is effectively the moment of the linear momentum) is zero.
Thus, the whole thing works out nicely. The increase in energy can be
accounted for and it turns out that the angular momentum (moment of the linear momentum) remains constant.
Cheers and best wishes.
Rob
_____________________
Robert Neal, PhD
Department of Human Movement Studies
The University of Queensland
QLD, AUSTRALIA
ph 61 7 365 6240
FAX 61 7 365 6877
EMAIL NEAL@HMS01.HMS.UQ.OZ.AU
_____________________________