Angular speed and intertia problem

[coey]

I'm kinda having trouble figuring out this problem...i already know that i have to use conservation of angular momentum Iiwi=Ifwf but I'm having trouble calculating the key part, the second moment of inertia. I'm pretty sure that im' supposed to be using this form of the calculation of inertia, I=Mr^2... but i still can't seem to solve it... like when i use it to test whether the numbers work out for the first moment of inertia 2.92 = 2.92*2 * 1.09^2, these numbers don't work out.



A student sits on a freely rotating stool holding two weights, each of mass 2.92 kg. When his arms are extended horizontally, the weights are 1.09 m from the axis of rotation and he rotates with an angular speed of 0.746 rad/s.

The moment of inertia of the student plus stool is 2.92 kg · m2 and is assumed to be constant. The student pulls the weights in horizontally to a position 0.307 m from the rotation axis.


Question: Find the new angular speed and find the kinetic energy before and after he pulls the weight inward.

 

[ZapperZ]

I'm kinda having trouble figuring out this problem...i already know that i have to use conservation of angular momentum Iiwi=Ifwf but I'm having trouble calculating the key part, the second moment of inertia. I'm pretty sure that im' supposed to be using this form of the calculation of inertia, I=Mr^2... but i still can't seem to solve it... like when i use it to test whether the numbers work out for the first moment of inertia 2.92 = 2.92*2 * 1.09^2, these numbers don't work out.



A student sits on a freely rotating stool holding two weights, each of mass 2.92 kg. When his arms are extended horizontally, the weights are 1.09 m from the axis of rotation and he rotates with an angular speed of 0.746 rad/s.

The moment of inertia of the student plus stool is 2.92 kg · m2 and is assumed to be constant. The student pulls the weights in horizontally to a position 0.307 m from the rotation axis.


Question: Find the new angular speed and find the kinetic energy before and after he pulls the weight inward.

A number of issues here:

1. Never, EVER solve a physics problem by plugging in numbers at the very beginning. This is a big no-no and will create hell in your attempts at solving physics problems later on. Assign symbols to all the relevant quantities and solve the problem algebraically first. This also allows whoever is looking at your work to follow what you are doing clearer.

2. What is "The moment of inertia of the student plus stool is 2.92 kg · m2..."? m2? In any case, I will assume that I, the moment of inertia of the student+stool is known (at least to you).

3. There are TWO moment of inertias here: (i) moment of inertia of student+stool, which is I, which doesn't change, and (ii) moment of inertia of the two masses, which does change. Call it I1 and I2 for before and after.

Then your conservation equation should look like

(I+I1)w1 = (I+I2)w2

You want to solve for w2. You know I, you can calculate I1 and I2 (you were given the radius of revolution for each case), and you were given w1. Thus, w2 can be solved.

From this, the rotational KE should be baby algebra.

Zz.

 

[thepatientmental]

Hi there, it seems like you are struggling with a problem involving angular speed and inertia. To solve this problem, you are correct in using the conservation of angular momentum equation, Iiwi=Ifwf. However, you are also correct in saying that you need to calculate the second moment of inertia (I) in order to use this equation.

The formula for calculating the moment of inertia of a point mass is I=Mr^2, where M is the mass and r is the distance from the axis of rotation. However, in this problem, we are dealing with two point masses (the weights) at different distances from the axis of rotation. In order to calculate the moment of inertia for this system, we need to use the parallel axis theorem, which states that I=Icm+Md^2, where Icm is the moment of inertia about the center of mass and d is the distance between the center of mass and the axis of rotation.

In this problem, the student plus stool system can be treated as a point mass located at the center of mass, which is in the middle of the stool. Therefore, the moment of inertia for the student plus stool system is simply I=Mr^2, where M is the total mass of the student and stool and r is the distance from the axis of rotation to the center of mass.

To calculate the moment of inertia for the system with the weights pulled in, we need to use the parallel axis theorem. The distance between the center of mass and the new position of the weights is 0.307 m, so the moment of inertia for the system with the weights pulled in is I=Icm+Md^2=Mr^2+(2*2.92 kg)*(0.307 m)^2=2.92 kg · m^2 + 0.898 kg · m^2=3.818 kg · m^2.

Now, using the conservation of angular momentum equation, we can solve for the new angular speed. Plugging in the initial and final moments of inertia and the initial angular speed, we get:

Iiwi=Ifwf
(2.92 kg · m2)(0.746 rad/s)=(3.818 kg · m2)(wf)
wf=(2.92 kg · m2)(0.746 rad/s)/(3.818 kg · m2)=0.572 rad/s

To calculate the kinetic energy before and after the weights are pulled in