A long thin solid rod lies along the positive x-axis. One end is at x = 1.68 m a
ID: 1399010 • Letter: A
Question
A long thin solid rod lies along the positive x-axis. One end is at x = 1.68 m and the other at x = 3.16 m. The linear mass density is = ax2 + b, where is measured in kg/m, and the constants have the following values: a = 2.20 kg/m3, b = 3.41 kg/m. Determine the total mass of the rod.
Calculate the x-coordinate of the center of the mass for this rod.
*I've seen other posts with this question but in each of those the unites for the a and b values, respectivly, were kg/m^4 and kg/m^2, while mine are kg/m^3 and kg/m. I'm not sure why this makes a difference but none of those solutions were working for mine. please help!
Explanation / Answer
total mass of the rod = m = integral ( dx ) from 1.68 to 3.16
integral ( 2.2x^2 + 3.41 dx ) from 1.68 to 3.16
= ( 2.2 x^3/ 3 + 3.41 x ) from 1.68 to 3.16
= ( 2.2 *3.16^3/ 3 + 3.41 * 3.16) - ( 2.2 * 1.68^3/ 3 + 3.41 * 1.68) = 24.709 kg
X coordinate of centre of mass = integral ( x dm ) / m
= integral ( 2.2 x^3 + 3.41 x^2 ) / m
= (1/24.709)*( 2.2 x^4/ 4 + 3.41 x^2/2 ) from 1.68 to 3.16
= (1/24.709)*( 2.2 x^4/ 4 + 3.41 x^2/2 ) - (1/24.709)*( 2.2 x^4/ 4 + 3.41 x^2/2 )
= (1/24.709)*(2.2 * 3.16^4/ 4 + 3.41 * 3.16^2/2 ) - (1/24.709)*( 2.2 *1.68^4/ 4 + 3.41 *1.68^2/2 )
= 2.536 m