# A meterstick (L= 1m) has a mass of m= 0.218 kg. Initially, it hangs from two short strings: one...

## Question:

A meterstick (L= 1m) has a mass of m= 0.218 kg. Initially, it hangs from two short strings: one at the 25 cm mark and the other one at the 75 cm mark.

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 acceleration of the center of mass of the meterstick when it is vertical?

- What is the tension in the string when the meterstick is vertical?

## Circular motion

When an object moves in a circular motion then it experiences a force toward its center known as centripetal force. Due to this force, it experiences a radial acceleration towards its center.

Given

Mass of the meterstick (m) = 0.218 kg

Now, the moment of inertia of the about the pivot

{eq}I = \dfrac{mL^{2}}{12} + m(0.25)^{2} \\ I = \dfrac{0.218*1^{2}}{12} + 0.218*(0.25)^{2} \\ I = 0.03179 \ kg-m^{2} {/eq}

From the energy conservation

{eq}mg*0.25 = 0.5Iw^{2} \\ 0.218*9.81*0.25 = 0.5*0.03179*w^{2} \\ w = 5.8 \ rad/s {/eq}

Now, the acceleration of the center of mass, when it is vertical

{eq}a = w^{2}*0.25 \\ a = 5.8^{2}*0.25 \\ a = 8.41 \ m/s^{2} {/eq}

Now applying Newton's second law

{eq}T_{1} - mg = ma \\ T_{1} = m(g+a) \\ T_{1} = 0.218*(9.81 + 8.41) \\ T_{1} = 3.97 \ N {/eq}

Where

• {eq}T_{1} {/eq} is the tension in the string