How Do You Calculate Moment of Inertia for Connected Masses?

[Jaklynn429]

Homework Statement

http://i241.photobucket.com/albums/ff4/alg5045/p13-17.gif
The four masses shown in the figure below are connected by massless, rigid rods.
(a) Find the coordinates of the center of mass if MA = 130 g and MB = MC = MD = 260 g.
(b) Find the moment of inertia about an axis that passes through mass A and is perpendicular to the page.

Homework Equations

I=mA(rA^2)+mB(rB^2)...

The Attempt at a Solution

I got part A as being .057 m for both x and y.
now wouldn't i use:
(.130*0)+(.260*.05^2)+(.260*.0707^2)+(.260*.05^2)
?? Its telling me my answer is wrong?!

 

[alphysicist]

Hi Jaklynn429,

Homework Statement

http://i241.photobucket.com/albums/ff4/alg5045/p13-17.gif
The four masses shown in the figure below are connected by massless, rigid rods.
(a) Find the coordinates of the center of mass if MA = 130 g and MB = MC = MD = 260 g.
(b) Find the moment of inertia about an axis that passes through mass A and is perpendicular to the page.

Homework Equations

I=mA(rA^2)+mB(rB^2)...

The Attempt at a Solution

I got part A as being .057 m for both x and y.
now wouldn't i use:
(.130*0)+(.260*.05^2)+(.260*.0707^2)+(.260*.05^2)

In the formula, r is the distance from each particle to the axis. The first term here is correct, because the 0.130kg particle is 0m from the axis. The other three terms seem to be saying that there are two 0.260kg particles that are 5cm from the axis, and one 0.26kg particle is is 7.07cm from the axis. Do you see what these last three terms need to be?

 

[baba89]

Hello,

Thank you for sharing your work on this problem. Moment of inertia is a property of an object that describes its resistance to rotational motion. It is calculated by summing the products of each mass and its distance from the axis of rotation squared. In this problem, the axis of rotation is through mass A and perpendicular to the page.

For part (a), you have correctly found the coordinates of the center of mass to be (0.057, 0.057) meters. This means that the center of mass is located at the intersection of the two diagonals of the square formed by the masses.

For part (b), you are correct in using the equation I=m(r^2) to find the moment of inertia. However, the distances (r) you are using for each mass are not correct. The distance for mass A should be 0 meters, since it is on the axis of rotation. For mass B, the distance should be the length of the diagonal of the square, which is 0.0707 meters. For masses C and D, the distance should be the length of one side of the square, which is 0.05 meters.

Using this information, the moment of inertia would be calculated as: I = (0.130*0) + (0.260*0.0707^2) + (0.260*0.05^2) + (0.260*0.05^2) = 0.000923 kg*m^2.

I hope this helps clarify the problem for you. Keep up the good work in your studies!