A uniform rod of length 1.0 m is bent at: its mid-point to make 90° angle. The distance of the centre of mass from the centre of the rod is

Question

A uniform rod of length 1.0 m is bent at: its mid-point to make 90° 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

Tardigrade Admin · Accepted Answer

When rod is bent at its mid point to make angle

9 0^{\circ}

x_{CM} = \frac{m \times \frac{1}{4} + m \times \frac{1}{4}}{m \times m} = \frac{1}{8} m = 12.5 c m

y_{CM} = \frac{m \times \frac{1}{4} + m \times \frac{1}{4}}{m + m} = \frac{1}{8} m = 12.5 c m

The distance of the cantre of mass from the centre of the rod

r = x^{2} + y^{2}

= (12.5)^{2} + (12.5)^{2}

= 312.5 - 17.7 c m

Q. A uniform rod of length 1.0m is bent at: its mid-point to make 90∘ angle. The distance of the centre of mass from the centre of the rod is

36.1 cm

25.2 cm

17.7 cm

zero

When rod is bent at its mid point to make angle 90∘ xCM​=m×mm×41​+m×41​​=81​m=12.5cm yCM​=m+mm×41​+m×41​​=81​m=12.5cm The distance of the cantre of mass from the centre of the rod r=x2+y2​ =(12.5)2+(12.5)2​ =312.5​−17.7cm

Q. A uniform rod of length $1.0 m$ is bent at: its mid-point to make $9 0^{\circ}$ angle. The distance of the centre of mass from the centre of the rod is