Rotational Motion Question 156
Question: ABC is a uniform wire. If centre of mass of wire lies vertically below point A, then $\frac{BC}{AB}$ Disclose to:
Options:
A)1.85
B) 1.5
C)1.37
D) 3
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Centre of mass
$x _{cm}=\frac{x}{2}\frac{(\rho x)\left( \frac{x}{2} \right)\frac{1}{2}+\rho y\left( \frac{y}{2} \right)}{\rho (x+y)}\Rightarrow \frac{1}{2}+\frac{y}{x}=\frac{y^{2}}{x^{2}}$
$\therefore \frac{BC}{AB}=\frac{y}{2}$
$=\frac{1+\sqrt{3}}{2}=1.37$
