Question

A rod of length L is placed along the x-axis between x=0 and x=L. The linear mass density is lambda such that lambda=a+bx. Find the mass of the rod.

Answer 1

linear mass density of the rod is given by

\lambda = \alpha + \beta xλ=α+βx

now we have

\frac{dm}{dx} = \alpha + \beta x

dx

dm

=α+βx

integrate both sides

\int dm = \int \alpha + \beta x dx∫dm=∫α+βxdx

M = \alpha L+\frac{ \beta L^2}{2}M=αL+

2

βL

2

so above is the total mass of rod

now by the formula of center of mass

r_{cm} = \frac{1}{M}\int xdmr

cm

=

M

1

∫xdm

r_{cm} = \frac{1}{M} \int x*(\alpha + \beta x) dxr

cm

=

M

1

∫x∗(α+βx)dx

r_{cm} = \frac{1}{M} (\alpha*\frac{L^2}{2} + \beta*\frac{L^3}{3})r

cm

=

M

1

(α∗

2

L

2

+β∗

3

L

3

)

now plug in value of mass of rod

r_{cm} = \frac{(\alpha*\frac{L^2}{2} + \beta*\frac{L^3}{3})}{( \alpha L+\frac{ \beta L^2}{2})}r

cm

=

(αL+

2

βL

2

)

(α∗

2

L

2

+β∗

3

L

3

)

so above is the position of center of mass of rod