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

_____________________________