# A meterstick (L= 1m) has a mass of m= 0.258 kg. Initially, it hangs from two short strings: one...

## Question:

A meterstick (L= 1m) has a mass of m= 0.258 kg. Initially, it hangs from two short strings: one at the 25 cm mark and the other one at the 75 cm mark. Now the right string is cut! After the right string is cut, the meterstick swings down to where it is vertical for an instant before it swings back up in the other direction. What is the angular speed when the meterstick is vertical?

## Energy conservation

From the law of energy conservation, we know that the energy converts itself from one form to other but the net energy of the system remains constant. Therefore the at any instant we can equate the total mechanical energy of the system.

Given

Mass of the meterstick (m) = 0.258 kg

It is tied at the end of 25 cm mark

Therefore the moment of inertia about pivot would be

{eq}I =\dfrac{mL^{2}}{12} + m(0.5-0.25)^{2} \\ I = \dfrac{0.258*1^{2}}{12}+ 0.258*0.25^{2} \\ I= 0.0376 \ kg-m^{2} {/eq}

Now from the energy conservation

{eq}E_{1} = E_{2} \\ mgh = 0.5Iw^{2} \\ 0.258*9.81*(0.25) = 0.5*0.0376*w^{2} \\ w= 5.801 \ rad/s {/eq}