Rotational Motion on an Axis Problem: Helpppp?

1. Oct 28, 2006

cheechnchong

Rotational Motion on an Axis Problem: Helpppp???

Problem: Two thin rectangular sheets (0.20 m x 0.40 m) are identical. In the first sheet the axis of rotation lies along the 0.20-m side, and in the second it lies along the 0.40-m side. The same torque is applied to each sheet. The first sheet, starting from rest, reaches its final angular velocity in 8.0 s. How Long does it take for the second sheet, starting from rest, to reach teh same angular velocity?

My Approach: I Didn't setup any equations just yet...kinda confused about how to come up with the TIME!!!!?? I drew the 2 identical rectangles on different axes (on on the 0.40-m side and the other on the 0.20-m side). From the given information, I know that both have equal torque and mass; however, radius im not sure about since they're on different axis (should i use the moment of inertia I=mr^2 here? to figure the radius). Ok once i have the radius, WHERE do i plug it in to find TIME!!!!!???

I think I know how to solve it, but it's just a few things i cannot figure from this problem. Thanks!

2. Oct 28, 2006

Office_Shredder

Staff Emeritus
You should use I=mr2 to figure out what the moment of inertia is (well, something like that). Remember that T=I*a where T is torque, I moment of inertia and a angular acceleration. So figure out what their relative angular accelerations are, and from there you should be able to figure out how long it takes one to reach a certain angular velocity in relation with the other

3. Oct 28, 2006

cheechnchong

How can i find the Torque without a GIVEN acceleration? and how can i utilize the angular velocities given at REST (w initial = 0 m/s). lol it's a hellva problem.

4. Oct 28, 2006

rsk

I think, as office shredder says, it's about relative acceleration, Intertia and time. Doing it that way gave me 2s. Am I close?

5. Oct 28, 2006

cheechnchong

YOU ARE RIGHT ON THE MONEY!!! how did you do it--in steps? oh and is the I = 1/3mr^2 (since it's a rectangle)?

Last edited: Oct 28, 2006
6. Oct 28, 2006

cheechnchong

^^how in the world did you calculate that answer??? my first question is...where did you get the first acceleration in order to find the torque shared by the two rectangles? because if i can get the torque i can set the other one equal to it and FIND the 2nd acceleration. From that acceleration i just dont know what else to do....???

7. Oct 28, 2006

Office_Shredder

Staff Emeritus
Ok, here we go. Let's start by getting the moment of inertia of each one.

I=m*r2 Clearly m is the same for both. Now, you should realize that by r in this problem, r is the distance from the axis of rotation to the center of mass of the object. So r is either .1, or .2 depending on the orientation. So if it rotates through the .4 meter side, I.4=m*.12, and if it rotates around the .2 meter side, I.2=m*.22

The important point is that I.4/I.2 = 1/4 (do you see how?)

To if the two torques are equivalent, we have T=Ia for both. So I.4*a.4 = I.2*a.2

So a.2/a.4 = 1/4

Assume both are rotating up to a final angular velocity V We know the general formula for the angular velocity given angular acceleration and time is:

v= a*t

if the initial angular momentum is zero. So we have V=a.4*t.4=a.2*t.2

Or

4 = t.2/t.4

But we know t.2 = 8, so t.4 = 8/4 = 2

8. Oct 28, 2006

cheechnchong

^^beautiful connections...i knew it some kinda tricky question! i woulda stayed up nights trying to figure this one out. Now i see that you carried this out as a proportional problem; therefore, next time i'll know what to look for. BIG THANKS!!!