Question

A uniform piece of wire, 20 cm long, is bent in a right angle in the center to give it an L-shape. How far from the bend is the center of mass of the bent wire?

Answer 1

The center of mass of the bent wire is approximately 11.18 cm from the bend.

Step-by-step explanation:

In order to find the center of mass of the bent wire, we need to divide it into two segments: the horizontal segment and the vertical segment. The length of each segment is half of the total length of the wire, which is 20 cm, so each segment is 10 cm long.

The center of mass of the horizontal segment is located exactly at its middle point, which is 5 cm from the corner. The center of mass of the vertical segment is also located at its middle point, which is 10 cm from the corner. Since the horizontal and vertical segments are orthogonal, the distance from the bend to the center of mass of the bent wire is the hypotenuse of a right triangle with legs of length 5 cm and 10 cm. Using the Pythagorean theorem, we can calculate the distance:

d = sqrt(5^2 + 10^2) = sqrt(25 + 100) = sqrt(125) = 11.18 cm

Therefore, the center of mass of the bent wire is approximately 11.18 cm from the bend.

Answer 2

Answer:TL;DR: 3.535 cm

Step-by-step explanation:

Xcm = ΣxMoments/ΣMasses = (10*0 + 10*5)/(10+10) = 50/20 = 2.5 cm

by symmetry,

Ycm = 2.5 cm

The distance D from the point Xcm,Ycm to the origin is D = √(2.5²+2.5²) = 3.535 cm