Question

Suppose a metal rod is lying horizontally on the ground. The two ends of the rod are positioned at x = 0,5 meters respectively. The mass density of the rod at position is given, in grams/meter, by the function 10 8(x) = 1+2 Evaluate the center of mass of the rod.

Answer 1

Solution 5m X Ve X=0 :5 given, the mass density of the rod at position x is given by 10 (r) - Itx Find; The center of mass of rod Step - I, The rod has non-uniform density, So Center of mass of rod is given by Xcm Mix + M₂ X2 + mn Хcom (xam -11) sdm Step - I, Value of dm dm mass density x dx 8(t) dx dm =

Step. I Now the value of dm put into equation (1) Fx:(1).dz s slxl.dx Xcm 10 :' 8(x) IX 10 Xcm frul dx IX 10 dx ☺ -10Gede or los de 10 Step- TV find the Value of som dx Itx - ** 2 +1 -1 dx X+1 so xt/ xtl x4) de -) dx - xtl x In Ixtil - (3)

Also find the value of I have dx Let u= xt1 du dx Re write then I du - The integral of with respect to u is In (lui) u So In liui) Replace U = Ill dic. In (1x+1) 19) tr Step 5, putting values from ean (9) and (3) into le) Hen we get Xcm 10 [ x-In(x+11] 11 0 I un Ix+11) - [x - dnlxH17 [an/x+11 11 como 7 en litl1 to x=5 then we get Step - 7 tacking limit's from x=0

Xcm Y 5 - ( un liten 1 un lotil 5 ای با مو) - In 11451 I onolitol uniw*+] u 5 In 161 5 - [: 20161 = 1.79175946923 1.79175996983 Xcm - 2.79 m So, Center of mass of rod Xem is 2.79 meters