- #1

mattmannmf

- 172

- 0

A uniform meter stick is hung at its center from a thin wire. It is twisted and oscillates with a period of 5 s. The meter stick is then sawed off to a length of 0.76 m, rebalanced at its center, and set into oscillation.

what is its new period

I= 1/12 mL^2

equation: T= 2 pi * sqrt (I/k)

where T= 5

L= 1

I solved for K in terms of m (mass)...where i got k=7.599 m

then i replugged it into the same equation but where

T= unknown (trying to find)

L=.76

the mass cancels out and my answer gives me 3.8... which is wrong when i checked.

The only thing i can see is that I am using the wrong "I" for using the same equation 2nd time when i replugged it in. Should i be using the parallel axis theorem where

I= 1/12 mL^2 + Md^2

(since its not revolving around its center of mass)

L=1

d= .76

when i tried that i got my answer to be 6.28...wrong again

what is its new period

I= 1/12 mL^2

equation: T= 2 pi * sqrt (I/k)

where T= 5

L= 1

I solved for K in terms of m (mass)...where i got k=7.599 m

then i replugged it into the same equation but where

T= unknown (trying to find)

L=.76

the mass cancels out and my answer gives me 3.8... which is wrong when i checked.

The only thing i can see is that I am using the wrong "I" for using the same equation 2nd time when i replugged it in. Should i be using the parallel axis theorem where

I= 1/12 mL^2 + Md^2

(since its not revolving around its center of mass)

L=1

d= .76

when i tried that i got my answer to be 6.28...wrong again