Homework Help: Rotational Kinetic Energy

1. Apr 23, 2006

M98Ranger

QUESTION: "A student on a piano stool rotates freely with an angular speed of 29.5rps. The student holds a 1.25 kg mass in each outstretched arm, .759m from the axis of rotation. The combined moment of Inertia of the student and the stool (ignoring the two masses) is 5.430kg.m^2 (a value that remains constant).

a.) As the student's arm are pull inward the angular speed is 3.54 rps. Determine the distance the masses are from the axis of rotation at this time (consider the masses to be modeled a mass-points).<I KNOW THE ENGLISH IS CHOPPED UP, BUT THIS IS HOW IT WAS WRITTEN ON THE PAPER.>

b.)Calculate KEfinal\KEINITIAL

MY THOUGHT PROCESS: This seems to me that the answers lie in utilizing the work-energy theorem which says that the net work done by external forces in a rotating rigid object about a fixed axis equals the change in the object's rotational energy. ie

(.5*IW^2)final-.5*IW^2)initial=Sum of Work
Where I=moment of inertia of an extended rigid object and W=instantaneous angular speed

To get radius (r) we know that W*r=tangential Velocity
we also know that I=m*r^2

MY questions:
-I don't understand how you can "ignore the two masses" in the calculation of I to get 5.430kg.m^2.
-How do I find the I of the whole system if the 2 masses are ignored
-Do I need to find the tangential velocity value of angular speed in order to find the R?
-If so, how would I go about doing it?

Thanks for reading this. Hopefully somebody can make sense of all this.

2. Apr 23, 2006

lightgrav

The PERSON's I is 5.43 [kgm^2] ... the TOTAL I is I_person + I_masses .
I_masses will depend on their distance from the rotation axis.

the person is on a "piano stool" so that the EXTERNAL torque is small.
That is, Angular momentum is conserved while the masses move inward.

Find the initial angular momentum, then the final angular speed ...
KE = 1/2 L omega ... (just like KE = 1/2 p v for linear motion).

3. Apr 23, 2006

M98Ranger

You said find the initial angular momentum and then find the final angular speed, but I thought that it gives you the final angular speed of 3.54rps. Maybe I am totally confused though.

here is what I did, but my answer is all messed up. I got 13.73 meters as my radius for my masses. Here goes....
Ip+Imasses= Itotal(initial)--->5.430[kg.m^2]+2*(1.25kg)*(.759m)^2=6.8702[kg.m^2] then, (IF X = the final I of the masses as they are brought in)
(.5*LW)initial=(.5*LW)final--->
(.5)*(6.87[kg.m^2])*(29.5rps)^2=(.5)(5.430+X[kg.m^2])(3.54rps)^2-->
X=471.667[kg.m^2] then, (IF 2mr^2 = X)
2*m*r^2=471.667[kg.m^2] | m=1.25kg
r=13.736meters........Pretty LONG arms, so where did I go wrong.
I am not understanding what you mean obviously when you said "find the initial angular momentum, then the final angular speed"

Thanks for the explanation thus far. It helped me tremendously.

Last edited: Apr 23, 2006
4. Apr 23, 2006

nrqed

Wow, he/she must be dizzy!!
there is something fishy. The number of rotation per second should *increase* as the mass are pulled inward.
there is no work done (no extrenal force acts on the system assuming that the friction force on the stool is negligibe.

Just calculate the initial K energy (using I_total = I_stool+student + I_masses) (the I of the two masses is just m* distance from the axis of rotation). This is a number you know.

Now, the final KE will contain one unknown: the distance at which the masses are. You can solve for this

5. Apr 23, 2006

nrqed

looks like you did it right (normally, omega should be put in radians per second not left in rps...here by luck it does not matter since it cancels out..but watch out for this in general).

The reason you get a larger value is that the values given in the initial question don't make sense. It would make much more sense if the initial value was 2.95 rps, not 29.5.

6. Apr 23, 2006

M98Ranger

EXACTLY WHAT I WAS THINKING.... now I feel better that somebody else was thinking that it didn't make sense. I wish my teacher would proof homework before he gives it out to students.