A uniform rectangular bar with cross-section 66 mm x 54 mm and of length 1.7 m has mass 16kg. It pivots about an axis prependicular to the wider face. The axis is equidistant from the edges and 170 mm from one end. What is the relevant second moment of mass?

Question

A uniform rectangular bar with cross-section 66 mm x 54 mm and of length 1.7 m has mass 16kg. It pivots about an axis prependicular to the wider face. The axis is equidistant from the edges and 170 mm from one end. What is the relevant second moment of mass?

Answer 1

The "wider face" is the pair of 66 mm x 1.7 m parallel faces.

First, get the moment of inertia about an axis passing through the centroid of that face. Call it Icm.

Then, use the parallel axis theorem, which is explained at
http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html

Answer 2

For the moment of inertia about an axis through the center of mass, see the "cuboid" formula at

http://en.wikipedia.org/wiki/List_of_moments_of_inertia

Answer 3

To find the relevant second moment of mass for the uniform rectangular bar, we need to calculate the moment of inertia for rotation around its axis of rotation (perpendicular to the wider face). The formula for the moment of inertia of a rectangular bar with respect to an axis passing through its center of mass and perpendicular to its cross-section is:

I = (1/12) * m * (h^2 + b^2)

Where:
- I is the moment of inertia
- m is the mass of the bar
- h is the height (longer side) of the cross-section
- b is the breadth (shorter side) of the cross-section

Given:
- The mass of the bar, m = 16 kg
- The height of the cross-section, h = 66 mm = 0.066 m
- The breadth of the cross-section, b = 54 mm = 0.054 m
- The distance of the axis of rotation from one end, x = 170 mm = 0.17 m

First, we need to calculate the distance from the axis of rotation to the center of mass of the bar. Since the axis is equidistant from the edges, it is located at the midpoint of the longer side. Therefore, the distance d from the axis of rotation to the center of mass is half the height of the cross-section:

d = h/2

Next, we can calculate the relevant second moment of mass (often called the moment of inertia) using the formula:

I = m * (d^2 + x^2)

Substituting the known values:

I = 16 kg * ((h/2)^2 + x^2)

I = 16 kg * ((0.066/2)^2 + 0.17^2)

I = 16 kg * ((0.033)^2 + 0.029)^2)

I = 16 kg * (0.001089 + 0.000841)

I = 16 kg * 0.001930

I = 0.0309 kg * m^2

Therefore, the relevant second moment of mass (moment of inertia) for the given uniform rectangular bar is 0.0309 kg * m^2.