Ó 2000, W. E. Haisler Mass Moment of Inertia 10

 2000, W. E. Haisler Mass Moment of Inertia 1 Mass Moment of Inertia Examples

The moment of inertia about the a-axis is given by

I  r2 dm  r 2  dV a n  n where rn is the direction normal to the a-axis. y For the circular disk, we have dm  dV

In this case, the integral is much easier to evaluate in polar rn  x coordinates. ua  k z  2000, W. E. Haisler Mass Moment of Inertia 2

Example: Consider a circular cylinder of diameter D, length L, and mass m. z y dA=2rdr dV=dAdz

r x G y L z

Show that Iz (about an axis parallel to the bar's length through G) 2 is given by: Iz = mD /8.

2 GI z   r dm

We can write dm  dV . In cylindrical coordinates, a volume element is given by dV dAdz  rdrd dz  rdr(2  ) L  2  Lrdr.  2000, W. E. Haisler Mass Moment of Inertia 3

The density is given by   m/ V . The volume of the rod is 4m V AL   ( D / 2)2 L. Thus the density is  m/ V  .  D2 L The perpendicular distance from the z-axis to any point of the volume element is r. With this, the mass moment of inertia about the z-axis becomes

2D / 2 2 4m  GI z r n dm   r  2 Lrdr 0  D2 L  D / 2 D / 2 4 8mr3 dr  8 m r 20 2 4 D D 0  mD2 /8 ( cylinder )

2 2 We can also show that GI y  m(3 D / 4  L ) /12 ( cylinder ).  2000, W. E. Haisler Mass Moment of Inertia 4

Long, slender rod of length L z Suppose the diameter D of the cylinder is very small compared to its length L (D= L), i.e., a very slender rod of length L. G y L Using the results for a cylinder and letting D  0, we obtain: x B 2 Long Slender Rod GI z  0 and GI y  mL/12 ( slender rod )

We could also derive GI y in the following way. dm If the rod has a mass of m and length of L, then dz the differential mass is given by dm=(m/L)dz. z Then G y L L/ 22 L / 2 2 2 I z dm  z( m / L ) dz  mL /12 x G y L/ 2   L / 2 B  2000, W. E. Haisler Mass Moment of Inertia 5

Moment of inertia about one end of slender rod: Suppose we wanted the mass moment of z inertia about a y-axis located at point B (the end of the rod). The last integral becomes: dz dm G L L L I z2 dm  z 2( m / L ) dz  mL 2 /3 z B y 0  0 y Notice: The moment of inertia about some x B point other then the center of mass will have a large value. In this case, the moment of inertia about B ( BI y ) is 4 times larger then the value about the center of mass G (GI y).

An easier way to obtain this is by using the parallel axis theorem discussed below.  2000, W. E. Haisler Mass Moment of Inertia 6

Suppose a mass of M is added to each end of the z slender rod.

B M For GI z, we assume that M is a point mass so that the result is the same as for a long slender I  0 rod, i.e., G z . G y L I For G y, we note that each lumped mass is x located a distance d=L/2 from G and has a M moment of inertia about G equal to 2 2 2 GI y r n dm  d M  ( L / 2) M . Since, moment of inertia is additive, we add the additional moment of inertia of the two lumped masses to the previous result to obtain:

I  mL2 /12  ( L / 2) 2 M  ( L / 2) 2 M  mL 2 /12  ML 2 / 2 G y  2000, W. E. Haisler Mass Moment of Inertia 7

Parallel Axis Theorem Consider a plane body with the x-y Cartesian coordinate system located at the center of mass G as shown below. Consider another Cartesian coordinate system (x', y') located at B such that the x'-axis is parallel to the x-axis. The perpendicular distance between the two parallel axes is "d." Assume we know the mass moment of inertia about the x-axis (GI x) and wish to determine the moment of inertia about the parallel x'-axis (Ix'). y’ y y’ = y - d B x’ d x G  2000, W. E. Haisler Mass Moment of Inertia 8

We first note that the moment of intertia about the x-axis through 2 2 the center of gravity is given by GIx  G r n dm   y dm. Recall that rn is normal to the x-axis and is thus equal to y. We wish to obtain the moment of inertia about the parallel x'-axis through point B (i.e., about a parallel axis that is a distance d I ( y ')2 dm from the center of gravity G). This is given by x'  . I We note that y' = y - d so that x' becomes 2 2 Ix' ( y ') dm   ( y  d ) dm y2 dm   2 y ( d ) dm   d 2 dm

The first integral in the above equation is the moment of inertia 2 about the x-axis (through G): GI x   y dm. For the middle integral, the constant 2d may be factored outside the integral so  2000, W. E. Haisler Mass Moment of Inertia 9 that it becomes 2d ( y ) dm  0. This integral is zero since y is measured from the center of mass. For the third integral, d is a constant so the integral becomes md 2. Thus we obtain

2 Ix' G I x  md

This is called the parallel axis theorem or the transfer theorem. It allows one to transfer the moment of inertia about an axis through the center of gravity (the x-axis) to a parallel axis (the x- axis) that is a distance "d" from the center of gravity.

The parallel axis theorem applies to any two parallel axes so long as one of them passes through the center of mass.  2000, W. E. Haisler Mass Moment of Inertia y 10 x Two rigid slender bars and a disk 2 kg disk are used to make the rigid 1 kg O structure pinned at O. The 3 kg 2 m horizontal bar has a length of 3 2 m 1 m m and a mass of 1 kg; the 0.5 m vertical bar has a length of 2 m and a mass of 2 kg. The disk has a diameter of 0.5 m and a mass of 3 kg. Determine the mass moment of inertia about a z-axis through point O. 2 2 2 Disk: GI z  mR/ 2  3 kg (0.5 m ) / 2  0.375 kgm 2 2 OI z  mR/ 2  md thus,  3kg (0.5 m )2 / 2  3 kg (2.25 m ) 2  15.5625 kgm 2 Horizontal Rod: 2 2 2 2 2 OI z  mL/12  md  1 kg (3 m ) /12  1 kg (0.5 m )  1 kgm Vertical Rod: 2 2 2 2 2 OI z  mL/12  md  2 kg (2 m ) /12  2 kg (1 m )  2.667 kgm 2 Total: Summing the three components gives: OI z 19.229 kgm