A uniform metre rule has a mass of 145g and its centre of mass is at the 50 cm mark from end A. A mass of 300g is hung at end A. Where should the pivot be placed for the rule to balance? Give the posi

A uniform metre rule has a mass of $145 \text{g}$ and a length of $100 \text{cm}$. Its center of mass is at the $50 \text{cm}$ mark from end A. A mass of $300 \text{g}$ is hung at end A ($0 \text{cm}$ mark). Let the string be positioned at a distance $x$ from end A. For the rule to balance, the sum of clockwise moments about the pivot (string) must equal the sum of anticlockwise moments.

Mass of rule $M_r = 145 \text{g}$.
Mass at end A $M_A = 300 \text{g}$.
Length of rule $L = 100 \text{cm}$.
Position of rule's center of mass from A $= 50 \text{cm}$.

Taking moments about the pivot (string at position $x$ from A):
Anticlockwise moment due to mass at A: $M_A g \times x$
Clockwise moment due to rule's mass: $M_r g \times (50 \text{cm} - x)$

Equating moments:

This is the position from end A. The question asks for the position from end B.
Position from end B $= 100 \text{cm} - x$
Position from end B $= 100 \text{cm} - 16.292 \text{cm} = 83.708 \text{cm}$.

The position from end B is $\boxed{\text{83.71 cm}}$.

Given:
Mass $m = 56 \text{g} = 0.056 \text{kg}$.
Radius $r = 0.8 \text{m}$.
Frequency $f =