Hi, this question was from an archived thread but I have another relevant question that was not answered by any of the posters--(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Hanging Meter Stick tension

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