PHYS 211 Lecture 21 - Moments of inertia 21-1 Lecture 21 - Moments of inertia Text: similar to Fowles and Cassiday, Chap. 8 As discussed previously, the moment of inertia I f a single mass m executing a circle of radius r about an axis is I = mr 2. F a group of masses, all rotating with the same ω: m 1 m 2 m 3 m 4 2 2 2 2 = m 1 r 1 + m 2 r 2 + m 3 r 3 + m 4 r 4 = m i r i i 2 Let s generalize this expression to a distribution of mass with a position-dependent density (r). If the mass distribution is continuous, then the sum over masses becomes an integral: I = Σ r i 2 m = r 2 dm. As in the previous lecture, dm = (r)d n r where (r) is a position-dependent mass density, describing the mass per unit length, area volume, cresponding to n = 1, 2 3, respectively. We now perfm several sample calculations of the moment of inertia. Circular ring F the axis perpendicular to the plane of the ring, and passing through its centre, this is a simple example. All of the mass elements dm are the same distance R from the axis, so I = r 2 dm = R 2 dm = mr 2. dm R Thin rod We break up the rod into small elements of length dx. The mass per unit length along
PHYS 211 Lecture 21 - Moments of inertia 21-2 the rod is = m/l. Hence, the small element has a mass of dm = dx. axis -L /2 +L /2 The mass of this element is dx I = -L/2 L/2 x 2 dm = -L/2 L/2 x 2 dx = (x 3 /3) -L/2 L/2 = [L 3 /8 - (-L 3 /8)]/3 I = L 3 /12. Substituting = m/l gives I = ml 2 /12. Disk, axis perpendicular to the plane dr R r looking down the axis of rotation The areal density of the disk is = m / (πr 2 ). We break the integral up into concentric rings, each with an area π(r + dr) 2 - πr 2 = 2πr dr, and a cresponding mass element dm given by dm = 2π r dr. The moment of inertia is then I = r 2 dm = r 2 2π r dr = 2π 0 R r 3 dr I = 2π (r 4 /4) 0 R = π R 4 /2. Substituting f the density I = (πr 4 /2) (m / (πr 2 ) I = mr 2 /2. = m / (πr 2 ) gives Solid Sphere
PHYS 211 Lecture 21 - Moments of inertia 21-3 Our solid sphere has a mass m and radius R. One way of approaching the solid sphere is to break it into thin slices, each of which has the moment of inertia calculated in the previous example f a disk. Of course, the moment of each slice depends on its radius r, which in turn depends upon its height z along the axis. Alternate derivation uses spherical polar codinates directly (r is now radial): I = (r sin ) 2 dm = (r sin ) 2 r 2 sin dr d d = r 4 dr sin 3 d d Now: r 4 dr = R 5 /5 sin 3 d = (1 - cos 2 ) sin d = - (1 - cos 2 ) dcos = 2-2/3 = 4/3 d = 2π Whence, I = R 5 /5 4/3 2π = (8π/15) R 5., after substituting f the density = m / (4πR 3 /3), I = m (4πR 3 /3) -1 (8π/15)R 5 --> I = (2/5)mR 2. Other Axes of Rotation Moments of inertia are defined with respect to a particular axis. What happens if the axis of rotation is not one of the simple ones? Two theems are available to help. Perpendicular axis theem This theem applies to planar objects lamina: z z = moment of inertia about the z-axis y x
PHYS 211 Lecture 21 - Moments of inertia 21-4 Evaluating the moment of inertia about the z-axis: I z = (x 2 + y 2 )dm = x 2 dm + y 2 dm. But, I x = y 2 dm if the object is planar [otherwise it is I x = (y 2 + z 2 )dm] so, I z = I x + I y (perpendicular axis theem) Parallel axis theem Consider two axes, one of which goes through the centre-of-mass: II cm cm d = distance between the axes I cm = axis through the centre-of-mass = axis parallel to one through cm Take the two axes to lie along the z-direction. Then the moments involve integrals in the xy plane, and we can simplify them by placing the cm on the x-axis with codinates (d, 0), and introducing relative codinates x *, y * : x, y x *, y * igin cm (d, 0) F any position (x, y), x 2 + y 2 = (d + x*) 2 + y* 2 = d 2 + 2dx* + x* 2 + y* 2 Using the discrete representation, the moments are = (x 2 + y 2 ) dm = d 2 dm + 2d x* dm + (x* 2 + y* 2 ) dm. But x* dm = 0 by the definition of the centre-of-mass position, and I cm = (x* 2 + y* 2 )dm. Thus, = md 2 + I cm Note: This tells us that the axis through the cm has the smallest moments of inertia (in a given direction).
PHYS 211 Lecture 21 - Moments of inertia 21-5 Example (Thin rod) II = cm + Md2 = 1 12 ML 2 + M ( L 2 ) 2 II =? cm = 1 12 ML 2 = 1 ( + 1 12 4 ) ML2 II = 1 3 ML 2 > cm