Question

Wire is bent into the structure as shown in the figure, and placed on a table. it consists of two half rings of radius r and two straight parts of length ππ r. the height of com from the table is.​

Answer 1

Answer:

I don't see any figure bro

Answer 2

Height of C.O.M = R(2 + π)/2π

Height:

Given,

Radius of ring = R

Length of straight parts = πR

The figure is given below with all the representations

C.O.M of semi-circular disc is 2R/ π

Now, from center OA= O'B = 2R/ π

C.O.M = $\frac{(\pi R)(2R/\pi ) + 2(\pi R)(\pi R/2) + \pi R(\pi R + 2R/\pi )}{\pi R + 2\pi R + \pi R}$

After eliminating the common terms we'll get

C.O.M = $\frac{4R/\pi + 2\pi R}{4}$

C.O.M = $\frac{R}{\pi y} + \frac{\pi R}{2}$

Hence, the height of COM from the table is $\frac{R(2 + \pi )}{2\pi }$