A thin road of length 3L in bent at a right angle at a distance of 2L on y axis and L on x axis locate the center of mass if L is 1.2 meter
To locate the center of mass of the road, we'll break it down into two straight segments and find the center of mass for each segment. Then, we'll find the center of mass considering the mass distributions of both segments.
Let's denote the two straight segments as Segment A and Segment B. Segment A lies on the y-axis with a length of 2L, and Segment B lies on the x-axis with a length of L.
Segment A:
Since Segment A lies on the y-axis, its center of mass will be at (0, Ly), where "Ly" is the distance of the center of mass from the origin. In this case, "Ly" is half the length of Segment A, which is half of 2L. Therefore, Ly = L.
Segment B:
Since Segment B lies on the x-axis, its center of mass will be at (Lx, 0), where "Lx" is the distance of the center of mass from the origin. In this case, "Lx" is half the length of Segment B, which is half of L. Therefore, Lx = L/2.
To find the overall center of mass, we need to consider the mass distribution of Segment A and Segment B. Since both segments have equal lengths, their masses will also be equal.
The overall center of mass (Cx, Cy) can be found using the following formula:
Cx = (m1 * x1 + m2 * x2) / (m1 + m2)
Cy = (m1 * y1 + m2 * y2) / (m1 + m2)
In this case, m1 and m2 are the masses of Segment A and Segment B respectively, which are equal. Therefore, we can simplify the equation as:
Cx = (Lx * m + Lx * m) / (2m)
Cy = (Ly * m + 0) / (2m)
= Ly/2
Simplifying further, we get:
Cx = (L/2 + L/2) / 2
= L/2
Cy = L/2
Hence, the center of mass of the road is located at (L/2, L/2).
Given that L = 1.2 meters, the center of mass of the road is located at (0.6 meters, 0.6 meters)