Question

A light hollow tube of 2.00 cm diameter and 1.0 m length is filled with tiny beads of different density. The resulting density distribution is linear, with the left end having a density of 1.8 g/cm3, and the right end having a density of 9.6 g/cm3. How far from the left end will be the center of mass (give answer in cm)?

Answer 1

Answer:

The centre of mass will be at a distance of 0.614 m from the left end of the light hollow tube.

Explanation:

Given:

Diameter of the hollow pipe= 2 cm

Length of the hollow tube=1 m

Density at the left$=1.8\times10^3\ kg/ m^3$

Density at the right end$=9.6\times 10^3 kg/ m^3$

Let the density of the tube varies as

$\rho=a+bx$

putting the end points of density as follows we have

$1.8\times10^3=a+b\times0$

$9.6\times10^3=a+b\times 1$

we get a=1.8$\times10^3$ and$b=7.8\times10^3$

$\rho=(1.8+7.8x)\times 10^3$

where  

• a and b are constants.
• x is the distance from left end.

We know that the x coordinate  centre of the mass  is given by

$X_{cm}=\dfrac{\int dm\ x}{\int dm}$

where dm is the mass of an element given by

$dm=\pi R^2(1.8+7.8x)\times 10^3dx$

Now we have

$X_{cm}=\dfrac{\int (\pi R^2(1.8+7.8x)\times10^3x\ dx)}{\int (\int \pi R^2(1.8+7.8x) dx}\times 10^3\\X_{cm}=\dfrac{\ 0.9x^2+2.6x^3}{1.8x+3.9x^2}\\\\X_{cm}=\dfrac{\ 0.9(1^2-0^2)+2.6(1^3-0^3)}{1.8(1-0)+3.9(1^2-0^2)}\\\\X_{cm}=0.614\ \rm m$

Answer 2

Answer:

The centre of mass will be at a distance of 0.614 m from the left end of the light hollow tube.