Hanging Meter Stick

1. Dec 12, 2006

franniemeow07

Hi, this question was from an archived thread but I have another relevant question that was not answered by any of the posters--

P.S. The answer to c is NOT mg.

1. The problem statement, all variables and given/known data

A meter stick (L = 1 m) with mass M = 0.409 kg is supported by two strings, one at the 0 cm mark and the other at the 90 cm mark.

c) Now the string at 0 cm is cut. What is the tension in the remaining string immediately thereafter?

I got parts a and b figured out, but I can't get c...Can I please get some help...

2. Relevant equations

When the left string is cut, the stick pivots about the 90cm mark. Take torques about that point:
(1) mgx = I α (Where "x" is the distance from the pivot to the center of mass of the stick; I is the rotational inertia of the stick about the pivot.)
Note that the acceleration of the c.m. is related to α by a = αx. Thus you can solve for "a".

Now consider the forces acting on the stick:
(2) mg - T = ma
This will give you T, the tension in the remaining string.

3. The attempt at a solution

What formula for I should I use, I = (1/3)ML^2, or I = (1/12)ML^2?
Also, should L be equal to "x", the distance from the pivot to the center of mass, or should it be 1meter?

2. Dec 12, 2006

OlderDan

It would be helpful if you posted the whole question, even if you do know how to do part of it. I will assume the stick is horizontal before the string is cut. If it's not, that could affect the outcome.

As for the moment if inertia, it is neither of the choices you listed. You will need to use the parallel axis theorem to find the appropriate I.

http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html

3. Dec 12, 2006

franniemeow07

I'm sorry, you assumed right, the stick is horizontal before the string is cut. I guess I forgot to post the diagram :tongue:

THANK YOU SO MUCH, you are awesome! I figured out the I, it's just the Icm+Mx^2, and yes, I do use x as r, in calculating a = alpha / r. YAYYY :rofl: