Rotation Moment of Inertia and Applications

Transcription

1 Rotation Moment of Inertia and Applications Lana Sheridan De Anza College Nov 20, 2016

2 Last time net torque Newton s second law for rotation moments of inertia calculating moments of inertia

3 Overview calculating moments of inertia the parallel axis theorem applications of moments of inertia Atwood machine with massive pulley (?)

4 5 Calculating m/v sometimes is Moment referred to as ofvolumetric Inertiamass ofdensity a Uniform Rod ss per unit volume. Often we use other ways of expresswhen (Example dealing with 10.7) a sheet of uniform thickness t, we can ty s 5 rt, which represents mass per unit area. Finally, when a rod of Moment uniform of cross-sectional inertia of a uniform area A, thin we sometimes rod of length use L and mass M L 5 ra, about which anis axis the mass perpendicular unit length. to the rod (the y axis) and passing through its center of mass. gth L and mass M (Fig. and passing through its y y dx re (Example 10.7) iform rigid rod of length L. moment of inertia about the is is less than that about the y The latter axis is examined in ple O L x x the definition of moment of inertia in Equation As

5 Calculating Moment of Inertia of a Uniform Rod Rod is uniform: let λ = M L be the mass per unit length (density). I y = r 2 dm

6 Calculating Moment of Inertia of a Uniform Rod Rod is uniform: let λ = M L be the mass per unit length (density). I y = = r 2 dm L/2 L/2 (x ) 2 λ dx

7 Calculating Moment of Inertia of a Uniform Rod Rod is uniform: let λ = M L be the mass per unit length (density). I y = r 2 dm = L/2 L/2 [ (x ) 3 = λ 3 = M L = 1 12 ML2 (x ) 2 λ dx ] L/2 L/2 [ ] L L3 24

8 when dealing with a sheet of uniform thickness t, we can ty s 5 rt, which represents mass per unit area. Finally, when a rod of Moment uniform ofcross-sectional inertia a uniform area A, thin we sometimes rod of length use L and mass M L 5 ra, which is the mass per unit length. about an axis perpendicular to the rod (the y axis) and passing through its center of mass. Calculating Moment of Inertia of a Uniform Rod gth L and mass M (Fig. and passing through its y y dx re (Example 10.7) iform rigid rod of length L. moment of inertia about the is is less than that about the y The latter axis is examined in ple O L x x the definition of moment of inertia in Equation As g the integrand to a single variable. ss dm equal to the mass per unit length I y l = 1 multiplied 12 ML2 by dx9. M

9 A Trick: The Parallel Axis Theorem Suppose you know the moment of inertia of an object about an axis through its center of mass. Let that axis be the z-axis. (The coordinates of the center of mass are (0, 0, 0).)

10 A Trick: The Parallel Axis Theorem Suppose you know the moment of inertia of an object about an axis through its center of mass. Let that axis be the z-axis. (The coordinates of the center of mass are (0, 0, 0).) Suppose you want to know the moment of inertia about a different axis, parallel to the first one. Let the (x, y) coordinates of this new axis z be (a, b).

11 A Trick: The Parallel Axis Theorem Suppose you know the moment of inertia of an object about an axis through its center of mass. Let that axis be the z-axis. (The coordinates of the center of mass are (0, 0, 0).) Suppose you want to know the moment of inertia about a different axis, parallel to the first one. Let the (x, y) coordinates of this new axis z be (a, b). Then we must integrate: I z = (r ) 2 dm where r is the distance of the mass increment dm from the new axis, z.

12 The Parallel Axis Theorem The result we will get. For an axis through the center of mass and any parallel axis through some other point: I = I CM + M D 2 where D = a 2 + b 2 is the distance from from the axis through the center of mass to the new axis.

13 The Parallel Axis Theorem Distance formula: (r ) 2 = (x a) 2 + (y b) 2.

14 The Parallel Axis Theorem Distance formula: (r ) 2 = (x a) 2 + (y b) 2. I z = = = (r ) 2 dm (x a) 2 + (y b) 2 dm x 2 dm 2a x dm + a 2 dm + y 2 dm 2b y dm + b 2 dm

15 The Parallel Axis Theorem But: x CM = 1 M x dm = 0 x dm = 0. Also for y.

16 The Parallel Axis Theorem But: x CM = 1 M x dm = 0 x dm = 0. Also for y. I z = x 2 dm 2a x dm + a 2 dm + y 2 dm 2b y dm + b 2 dm

17 The Parallel Axis Theorem But: x CM = 1 M x dm = 0 x dm = 0. Also for y. I z = x 2 dm 2a 0 x dm + a 2 dm 0 + y 2 dm 2b y dm + b 2 dm

18 The Parallel Axis Theorem But: x CM = 1 M x dm = 0 x dm = 0. Also for y. I z = = x 2 dm 2a 0 x dm + a 2 dm 0 + y 2 dm 2b y dm + b 2 dm (x 2 + y 2 ) dm + (a 2 + b 2 ) dm

19 The Parallel Axis Theorem But: x CM = 1 M x dm = 0 x dm = 0. Also for y. I z = = x 2 dm 2a 0 x dm + a 2 dm 0 + y 2 dm 2b y dm + b 2 dm (x 2 + y 2 ) dm + (a 2 + b 2 ) dm = I z + M D 2 where D = a 2 + b 2 is the distance from from the origin to the new axis.

20 The Parallel Axis Theorem I = I CM + M D 2 where D = a 2 + b 2 is the distance from from the origin to the new axis.

21 Question R Rectangular plate 1 I CM M(a 2 b 2 ) 12 We found that the moment of inertia of a uniform rod about its midpoint was I = 1 12 ML2. b a What is the moment of inertia of the same rod about the and axis through an endpoint of the rod? L Long, thin rod with rotation axis through end 1 I 3 ML2 L (A) 1 3 ML2 (B) 1 2 ML2 (C) 7 12 ML2 R (D) ML2 Thin spherical shell 2 2 I CM MR 3 R

22 Question R Rectangular plate 1 I CM M(a 2 b 2 ) 12 We found that the moment of inertia of a uniform rod about its midpoint was I = 1 12 ML2. b a What is the moment of inertia of the same rod about the and axis through an endpoint of the rod? L (A) 1 3 ML2 (B) 1 2 ML2 (C) 7 12 ML2 R (D) ML2 Long, thin rod with rotation axis through end 1 I 3 ML2 Thin spherical shell 2 2 I CM MR 3 L R

23 Calculating Moment of Inertia of a Uniform Cylinder Calculation (Example of Moments 10.8) of Inertia 309 A uniform solid cylinder has a radius R, mass M, and length L. Calculate its moment of inertia about its central axis (the z axis). z dr r R L Figure (Example 10.8) Calculating I about the z axis for a uniform solid cylinder.

24 Calculating Moment of Inertia of a Uniform Cylinder We will again need to evaluate I = r 2 dm. Let ρ = M V, so that I = ρ r 2 dv.

25 Calculating Moment of Inertia of a Uniform Cylinder We will again need to evaluate I = r 2 dm. Let ρ = M V, so that I = ρ r 2 dv. We will do this by summing up cylindrical shells. What is the surface area of a cylinder of height L (without the circular ends)?

26 Calculating Moment of Inertia of a Uniform Cylinder We will again need to evaluate I = r 2 dm. Let ρ = M V, so that I = ρ r 2 dv. We will do this by summing up cylindrical shells. What is the surface area of a cylinder of height L (without the circular ends)? A = 2πrL Then the volume of a cylindrical shell of thickness dr is dv = 2πrL dr (The volume of the entire cylinder, V = πr 2 L.)

27 Calculating Moment of Inertia of a Uniform Cylinder I z = R 0 I z = 2πρL r 2 (ρ2πrl dr) R 0 r 3 dr

28 Calculating Moment of Inertia of a Uniform Cylinder I z = R 0 I z = 2πρL r 2 (ρ2πrl dr) R 0 r 3 dr [ ] r 4 R = 2πρL 4 0 [ ] R 4 = 2πρL 4 = 1 2 πl ( M πr 2 L = 1 2 MR2 ) R 4

29 with Different Geometries Hoop or thin cylindrical shell 2 I CM MR R Hollow cylinder 1 I CM M(R R 2 2 ) R 1 R 2 Solid cylinder or disk 1 2 I CM MR 2 R Rectangular plate 1 I CM M(a 2 b 2 ) 12 b a Long, thin rod with rotation axis through center 1 I CM ML 2 12 L Long, thin rod with rotation axis through end 1 I 3 ML2 L Solid sphere 2 I CM MR 2 5 R Thin spherical shell 2 2 I CM MR 3 R

30 Applying Moments of Inertia Now that we can find moments of inertia of various objects, we can use them to calculate angular accelerations from torques, and vice versa. We can solve for the motion of systems with rotating parts. τ = Iα

31 Summary moments of inertia Atwood machine with massive pulley Next test Monday, Nov 27. (Uncollected) Homework Serway & Jewett, PREV: Ch 10, onward from page 288. Probs: 27, 29, 35, 39, 43 PREV: Look at examples 10.6.