You are sitting on a stool that is free to rotate without friction, and you are

Question

You are sitting on a stool that is free to rotate without friction, and you are holding a 2.00kg weight in each hand (throughout this problem, ignore friction and all forces external to the "you +weight" system). For the sake of simplicity, also ignore your own moment of inertia (pretend your moment of inertia is zero and that the moment of inertia of the you + weights system is determined entirely by the weights). You and your weights are rotating with an angular velocity of 3.00 radians per second, and you are holding the weights with arms outstretched so that each is 80.0 cm from the axis of rotation (the distance from the axis to the center of mass of each weight).

+ rotating anticlockwise

Then youpull your arms in so that the weights are 20.0 cm from the axis of rotation.

a) What happens and why?

b) What is your angular velocity after you pull your arms in?

c) Keeping in mind what is and is not conserved as you pullin your arms, what are the initial and final kinetic energies of the rotating system?

d) Let the initial angular momentum of the system be L, the mass of one of the weights be M, and the distance from the axis of rotation to the blocks be R. (Note that if the angular momentum of the system is L, then the angular momentum of one of the blocks is L/2!) Then show that the force required to pull one of the blocks in at constant speed is equal to : F = L^2/(4MR^3)

e) Now show that the magnitude of the work required to pull the blocks in from a distance of 80.0cm to a distance of 20.0 cm really is equal to the difference in kinetic energy that you calculated in part c. That is, even if you couldn't derive it, use the formula for force given above and calculate the work done as you pull in the weights.

DO NOT SIMPLY USE CONSERVATION OF ENERGY TO ASSERT THAT THE WORK IS EQUAL TO THE CHANGE IN KINETIC ENERGY! You are being asked to calculate the work done (using force as a function of distance) to show that it really is equal to the change in kinetic energy! There is an integral involved, but it isn't difficult.

Note :Don't spend too much time worrying about a minus sign! If you get the correct magnitude but the wrong overall sign, that's pretty good. I specifically said to show that the magnitude is correct since there is a tricky minus sign involved.

Explanation / Answer

a) when we pull arms moment of inertia of holding bodies decreses.

so, angualr velocity increases to keep angular moemntum constant.

b) I1 = 2*m*r1^2

= 2*2*0.8^2

= 2.56 kg.m^2

I2 = 2*m*r2^2

= 2*2*0.2^2

= 0.16 kg.m^2

w1 = 3 rad/s

w2 = ?

Apply conservation of nagular momentum,

L1 = L2

I1*w1 = I2*w2

==> w2 = (I1/I2)*w1

= (2.56/0.16)*3

= 48 rad/s

c) angualr momentum is conserved but not kinetic enrgy.

Ki = 0.5*I1*w1^2 = 0.5*2.56*3^2 = 11.52 J

Kf = 0.5*I2*w2^2 = 0.5*0.16*48^2 = 184.32 J

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