Question

Find the total mass of the given rod and the center of mass.

1. The length of a rod is 6 m and the linear density of the rod at a point x meters from one end is (2x + 3) kg/m.

2. The length of a rod is 9 in. and the linear density of the rod at a point x inches from one end is (4x + 1) slugs/in.

3. The length of a rod is 12 cm, and the measure of the linear density at a point is a linear function of the measure of the distance from the left end of the rod. The linear density at the left end is 3 g/cm and at the right end is 4 g/cm.

4. The measure of the linear density at any point of a rod 6 m long varies directly as the distance from the point to an external point in the line of the rod and 4 m from an end, where the density is 3 kg/m.

5. The measure of the linear density at a point of a rod varies directly as the third power of the measure of the distance of the point from one end. The length of the rod is 4 ft and the linear density is 2 slugs/ft at the center.

Answer 1

1.

mass:

$m=\int^6_0(2x+3)dx=(x^2+3x)|^6_0=54$ kg

center of mass:

$\overline{x}=M_y/m$

$M_y=\int x\rho(x)dx=\int^6_0x(2x+3)dx=(2x^3/3+3x^2/2)|^6_0=198$

$\overline{x}=198/54=3.67$ m

2.

mass:

$m=\int^9_0(4x+1)dx=(2x^2+x)|^9_0=171$ slugs

center of mass:

$\overline{x}=M_y/m$

$M_y=\int x\rho(x)dx=\int^9_0x(4x+1)dx=(4x^3/3+x^2/2)|^9_0=1012.5$

$\overline{x}=1012.5/171=5.92$ inches

3.

the linear density of the rod:

$(3+x/12)$ g/cm

mass:

$m=\int^{12}_0(3+x/12)dx=(x^2/24+3x)|^{12}_0=42$ g

center of mass:

$\overline{x}=M_y/m$

$M_y=\int x\rho(x)dx=\int^{12}_0x(3+x/12)dx=(x^3/36+3x^2/2)|^{12}_0=264$

$\overline{x}=264/42=6.29$ cm

4.

the linear density of the rod:

$\frac{3}{4}(10-x)$ kg/m

mass:

$m=\int^{6}_0\frac{3}{4}(10-x)dx=(15x/2-3x^2/8)|^{6}_0=31.5$ kg

center of mass:

$\overline{x}=M_y/m$

$M_y=\int x\rho(x)dx=\int^{6}_0\frac{3}{4}x(10-x)dx=(15x^2/4-x^3/4)|^{6}_0=81$

$\overline{x}=81/31.5=2.57$ m

5.

the linear density of the rod:

$(x^3/4)$ slugs/ft

mass:

$m=\int^{4}_0(x^3/4)dx=(x^4/16)|^{4}_0=16$ slugs

center of mass:

$\overline{x}=M_y/m$

$M_y=\int x\rho(x)dx=\int^{4}_0x(x^3/4)dx=(x^5/20)|^{4}_0=51.2$

$\overline{x}=51.2/16=3.20$ ft