Physics, 30.12.2020 03:50 singhmanny3526

Consider a rigid 3-mass system ( with origin at the leftmost mass 1 kg) which can rotate about an axis perpendicular to the system. The mass are separated by rods of length 5m, so that the entire length is 10m. Find the x-coordinate of the center of the mass for the three-mass system with respect to the origin. Treat mass as particles. Answer in unit of m. now consider a rotation axis perpendicular to the system and passing through the point Xo at distance 3.8 m from the leftmost mass 1kg. find the moment of inertia of the 3-mass system about the new axis. Answer in unit of kg. m^2

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Answer from: journeyburks07

The moment of inertia of the 3-mass system about the new axis is 54.32 kgm/s².

We have three masses each of mass = 1kg such that they are in line with mass m at origin, m at 5m and m at 10m

(a) The center of mass:

$X=\frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3} \\
\\
X =\frac{ 1*0+1*5+1*10}{1+1+1}\\
 \\
X = 5m$

Hence the center of mass of the system is at x = 5m.

(b) The moment of inertia about the axis passing through x = 3.8m

from the parallel axis theorem:

$I = I_{cm} + Md^2$

where, $I_{cm}$ is the moment of inertial along an axis passing through the center of mass of the system, M is the total mass of the system and d is the distance of the given axis from center of mass.

M = 3kg

d = 5 - 3.8 = 1.2m

$I_{cm}=1*5^2+1*0+1*5^2\\\\
I_{cm}=50 kgm/s^2$

Md² = 3×(1.2)²

Md² = 4.32 kgm/s²

I = 50 + 4.32

I = 54.32 kgm/s² is the moment of inertial about the given axis.

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Answer from: Perdikaris

1) x_{cm} = 5 m , 2) I = 168.32 kg m²

Explanation:

1) An important concept of center of mass is

$x_{cm} = \frac{1}{M} \sum x_{i} m_{i}$

where M is the total mass of the system

Let's apply this equation to our case, suppose that all masses are equal and are worth 1 kg

$x_{cm}$ = ⅓ (1 0 + 1 5 + 1 10)

x_{cm} = 5 m

2) In this para indicates that there is an axis of rotation at the point xo = 3.8 m and they ask to calculate the moment inertia.

Let's use the parallel axes theorem

I = I_{cm} + M D

where I_{cm} is the moment of inertia with respect to the center of mass, D the distance between the two axes of rotation and M the total mass of the system

Let's look for the moment of inertia of the center of mass

$I_{cm}$ = 1 0 + 1 5² + 1 10²

I_{cm} = 125 kg m²

the total moment of inertia is

I = 125 + 3 3.8²

I = 168.32 kg m²

Answer from: Quest

33: i think it’s b

Answer from: Quest

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