Question

A uniform rod of length $1.0\, m$ is bent at: its mid-point to make $90^{\circ}$ angle. The distance of the centre of mass from the centre of the rod is

• A. 36.1 cm
• B. 25.2 cm
• C. 17.7 cm
• D. zero

Answer 1

Answer: (C)

When rod is bent at its mid point to make angle $90^{\circ}$
$x_{C M}=\frac{m \times \frac{1}{4}+m \times \frac{1}{4}}{m \times m}=\frac{1}{8} m=12.5 \,cm$
$y_{C M}=\frac{m \times \frac{1}{4}+m \times \frac{1}{4}}{m+m}=\frac{1}{8} m=12.5\, cm$
The distance of the cantre of mass from the centre of the rod
$r=\sqrt{x^{2}+y^{2}} $
$=\sqrt{(12.5)^{2}+(12.5)^{2}}$
$=\sqrt{312.5}-17.7\, cm$