9 Kinetics of 3D rigid bodies - rotating frames

Transcription

1 9 Kinetics of 3D rigid bodies - rotating frames 9. Consider the two gears depicted in the figure. The gear B of radius R B is fixed to the ground, while the gear A of mass m A and radius R A turns freely about the axis CE.A constant torque M is applied to the vertical shaft CD. The system is initially at rest. Determine the angular velocity ω CD of the shaft CD after two complete revolutions. Neglect the mass of shaft CD and axle CE and assume that gear A can be approximated by a thin disk. Hint: The centroidal moment of inertia tensor of a thin disk of mass m and radius r is I C = 4 mr2 2 E A R B C M R A D B (9.) 4πM (a ) ω CD = 8 mr2 A mr2 B (b ) ω CD = 2πM 8 mr2 A 4πM (c ) ω CD = 3 4 mr2 A + 8 mr2 B 8πM (d ) ω CD = 2 mr2 A + 4 mr2 B M (e ) ω CD = 3 4 mr2 B Solution If shaft CD, the axle CE, and gear A are considered as a system of connected bodies, only the applied torque M does work. For two complete revolutions of CD, this work is simply given by W 2 = 4πM (9.2) Our choice of reference system is shown in figure 9.. The unit vectors for the reference system xyz are denoted by [ e x, e y, e z ]. Since the gear is initially at rest, its initial kinetic energy is zero. In order to compute 9-

2 9 Kinetics of 3D rigid bodies - rotating frames 9-2 the kinetic energy of the system, we need to relate the angular velocities of the shaft and the gear. The angular velocities components are shown in Figure 9.. Axis of instantaneous rotation e z R A E A H! CD! A C D M R B! CE B e x Figure 9.: Reference system and angular velocity components If we denote the angular velocity of CD by ω CD, then the angular velocity of the gear A is ω A = ω CD + ω CE. The instantaneous axis of rotation for this body is along line CH, because both points C and H on the body (gear) have zero velocity and must therefore lie on this axis. Note that C and H do not belong to the same body, and therefore the proposed reasoning does not strictly apply. However, if we imagine to extend the gear up to the point C, this point will have zero velocity, and therefore the instantaneous axis of rotation can be identified. This requires the components of ω CD and ω CE to be related by the equation ω CD R A = ω CE R B ω CE = R B R A ω CD. (9.3) Thus, the angular velocity of gear A can be expressed as ω A = ω CE e x + ω CD e z = R B R A ω CD e x + ω CD e z. (9.4) The x, y, z axes in the figure are aligned with the principal axes of inertia of the gear, shifted to have their origin at C. Since the point C is a fixed point of rotation, the kinetic energy can be expressed as T 2 = 2 I xxω 2 x + 2 I yyω 2 y + 2 I zzω 2 z. (9.5) Using the parallel-axis theorem, the moments of inertia of the gear about point C are as follows: I xx = 2 mr2 A. (9.6) I yy = I zz = 4 mr2 A + mr 2 B. (9.7) Since ω x = R B R A ω CD, ω y = and ω z = ω CD, eq.(9.5) becomes [ T 2 = 8 mr2 A + 3 ] 4 mr2 B ωcd. 2 (9.8)

3 9 Kinetics of 3D rigid bodies - rotating frames 9-3 We can now use the work-energy principle to relate the change in kinetic energy to the externally applied work. For the gear, we simply have [ T 2 = W 2 + T 8 mr2 A + 3 ] 4 mr2 B ωcd 2 = 4πM + (9.9) thus 4πM ω CD = 8 mr2 A + 3. (9.) m2 4 And therefore the correct answer is answer (a).

4 9 Kinetics of 3D rigid bodies - rotating frames A slender bar of mass m and length L is attached to a vertical shaft rotating with constant angular velocity Ω. The slender bar is allowed to pivot about an an axis perpendicular to the shaft passing through point B. What is the equation of motion of the bar? Neglect the radius of the shaft. Hint: the centroidal moment of inertia of a slender bar of mass m and length l is I C = 2 ml2 (9.). θ + Ω 2 sin θ cos θ + 3g L sin θ =. 2. θ Ω 2 sin θ cos θ + mgl sin θ =. 3. θ Ω 2 sin θ cos θ + 3g sin θ =. 2L 4. θ Ω 2 sin θ cos θ + 5g cos θ =. 2L 5. θ + Ω 2 sin θ sin θ + 5g sin θ =. 2L Solution Since the rotation about the vertical axis is prescribed the system needs only one degree of freedom to describe its motion. We select as DOF the rotation θ of the bar about its hinge. Writing the angular momentum principle with respect to B allows to neglect the presence of the reaction forces at the hinge. However, there action moments provided by the hinge in the two constrained rotations need to be taken into account. If the angular momentum principle is written with respect to the inertial frame e x, e y, e z a time dependent inertial tensor will arise. It is more convenient to use a coordinate frame b, b 2, b 3 attached to the bar, with origin at B, and aligned with the principal axis of the bar. Free-Body Diagram and Coordinate system where the three components of the reaction force at the hinge F, F 2, F 3 and the two moment components M and M 2 are highlighted, together with the gravity force. We need to write Ḣ B = I B ω = M ext B. (9.2) By expressing all the terms with respect to the moving coordinate frame. angular velocity of the bar and angular velocity of the shaft are: The total ω = Ω cos θb + Ω sin θb 2 + θb 3, Ω = Ω cos θb + Ω sin θb 2. (9.3)

5 9 Kinetics of 3D rigid bodies - rotating frames 9-5 respectively. The external moment about point O is given by, M ext B = mg L 2 sin θb 3 + M b + M 2 b 2. (9.4) The moment of inertia tensor I B about point B can be readily obtained by applying the Steiner s theorem: I B = 3 ml2 (9.5) 3 ml2 note the zero entry in the first diagonal position due to the fact that the bar has negligible thickness. We can now write the angular momentum as Ω cos θ H B = I B ω = 3 ml2 Ω sin θ = 3 3 ml2 θ ml2 Ω sin θb ml2 θb3 (9.6) The principle of angular momentum requires ḢO. since this vector is now expressed w.r.t. the moving frame, we need to use the formula Ḣ B = ḢB + ω H B, (9.7) which expresses the time derivative of the vector H B with respect to a moving frame. the vector ω is the angular velocity vector of the moving frame. The term ω H B results to be: b b 2 b 3 ω H B = Ω cos θ Ω sin θ θ 3 ml2 Ω sin θ 3 ml2 θ = 3 ml2 θω sin θ 3 ml2 θω sin θ 3 ml2 θω cos θ (9.8) 3 ml2 Ω 2 sin θ cos θ and therefore, the angular momentum principle writes: = M (9.9) 3 ml2 θω cos θ + 3 ml2 θω cos θ = M2. (9.2) 3 ml2 θ 3 ml2 Ω 2 sin θ cos θ = mg L sin θ. 2 (9.2)

6 9 Kinetics of 3D rigid bodies - rotating frames 9-6 (9.2) does not contain any reaction moment, and therefore governs the system dynamics entirely. Simplifying, we get: θ Ω 2 sin θ cos θ + 3g sin θ =. (9.22) 2L Note that, if we pose Ω =, (9.22) becomes the eq. of motion of a simple pendulum. Also, once the motion θ(t) is determined by solving (9.2), the reaction moments are given by (9.9) and(9.2). Note that M is zero, because the bar has negligible inertia w.r.t. its own axis. Analysis of multiple choice. Incorrect. 2. Dimensionally incorrect. 3. Correct. 4. Incorrect. 5. Incorrect.

7 9 Kinetics of 3D rigid bodies - rotating frames A thin rectangular plate of mass M and sides a and b rotates around its diagonal with a constant angular velocity ω. Neglect the action of gravity. The principal moment of inertia tensor I C of a thin rectangle of mass M and sides a and b is I C = a 2 2 M b 2 (a 2 + b 2 ) bearing a b bearing! (9.23). What is the magnitude of the force F in the bearings? j (a ) F = 2 Mω2 (a 2 b 2 ab ) (a 2 + b 2 ) 3/2 (b ) F = 6 ω2 (a 2 b 2 ab ) (a 2 b 2 ) 3/2 (c ) F = 2 Mω2 (b 2 a 2 ab ) (a 2 + b 2 ) /2 (d ) F = 2 Mω2 (a 2 + b 2 a ) (a 2 b 2 ) 3/2 (e ) F = 8 Mω2 (a 2 b 2 (ab) 2 ) (a 2 + b 2 ) 3/2 F A a k o C bearing a 2. What is the kinetic energy of the system? Solution below. (a ) T = 6 Mω a2 b 2 a 2 + b 2 (b ) T = 2 Mω2 a 2 b 2 a 2 + b 2 (c ) T = a2 b 2 2 Mω2 a 2 + b 2 (d ) T = a 2 b 2 4 Mω2 bearing a 2 + b 2 (e ) T = ω 2 a 2 b! 2 2 M a 2 + b 2 b The reference systems and the free-body diagram are shown in the figure j i A k o B F C F 2 a b! e

8 9 Kinetics of 3D rigid bodies - rotating frames 9-8 Note that the [i, j, k] is centered at the center of mass and is attached to the rectangle, and therefore rotates with respect to the fixed frame of reference with angular velocity ω. By applying the linear momentum principle, we can establish a relation between the two reaction forces at the bearing: Ṗ = m v C = F Since the center of mass is fixed, then v C =, and therefore: (9.24) F + F 2 = F = F 2 = F. (9.25) We need now another equation to determine the value of F. For this purpose, we can apply the principle of angular momentum w.r.t. point C. Being C the center of mass, we can write: H C = I C ω (9.26) The principal inertia tensor I C of a thin rectangle is I C = a 2 2 M b 2 (9.27) (a 2 + b 2 ) Before being able to differentiate (9.26) with respect to time, the angular velocity ω needs also to be written in terms of moving coordinates, as ω = ω cos αi ω sin αj (9.28) Plugging 9.27 and 9.28 into 9.26: H C = I xx ω cos αi I yy ω sin αj (9.29) Since the angular momentum is expressed w.r.t a moving/rotating coordinate frame, we need to accommodate for the variation of i, j, k with the former of differentiation with respect to moving frames: Ḣ C = H C + ω H C (9.3) I xx ω cos α ω cos α I xx ω cos α Ḣ C = I yy ω sin α + ω sin α I yy ω sin α I xx ω cos α (9.3) = I yy ω sin α ω 2 I yy sin α cos α + ω 2 I xx sin α cos α ω= = 2 M(a2 b 2 )ω 2 sin α cos αk In order to apply the angular momentum principle, the moment of external forces w.r.t. point C needs to be calculated. this is readily done: a2 +b2 a 2 +b M C = r CA F + r CB F 2 = F + F 2 (9.32) = a 2 + b 2 (in principal axis) F e 3 = a2 + b 2 F k

9 9 Kinetics of 3D rigid bodies - rotating frames 9-9 Finally, we can apply the angular momentum principle: 2 M(a2 b 2 )ω 2 sin α cos α = a 2 + b 2 F. (9.33) We can express sin α and cos α in terms of a and b: sin α = a a2 + b, cos α = b 2 a2 + b, (9.34) 2 so that the reaction force can be expressed as: F = 2 Mω2 (a 2 b 2 ab ) (9.35) (a 2 + b 2 ) 3/2 Note that: F rotates around the e axis and stays in the i j plane if a=b (square) F = Therefore the correct answer is answer (a). IV. Kinetic Energy The kinetic energy can be written according to its definition: T = 2 ωt I C ω + 2 MvT Cv C = 2 (I xxω 2 x + I yy ω 2 y) + = 2 ( 2 Ma2 ω 2 cos 2 α + 2 Mb2 ω 2 sin 2 α) = 2 Mω2 a 2 b 2 a 2 + b 2 (9.36)

10 9 Kinetics of 3D rigid bodies - rotating frames The gear shown in the figure has mass m and is mounted at an angle of φ on a shaft of negligible mass. A set of principal axes for the gear is indicated in the figure with [x, y, z]. Its principal moment of inertia are I z = 2I x, I x = I y. Assume that the shaft is rotating with a constant angular velocity ω. The frame [X, Y, Z] represents an inertial coordinate system. Determine the components of the reaction forces that the thrust bearing A and the journal bearing B exert on the shaft at the instant shown. (Hint: sin α cos α = sin(2α)) (a ) A X = A Y = mgl GB +.5I x sin(2φ)ω 2 L GA + L GB B X = B Y = mg mgl GB +.5I x sin(2φ)ω 2 L GA + L GB (b ) (c ) A X = B X = A Y = mgl GB +.5I x sin(2φ)ω (L GA + L GB ) 2 B Y = mg + mgl GB +.5I x sin(2φ)ω (L GA + L GB ) 2 A X = A Y = mgl GB + 2I x sin(φ)ω 2 L GA + L GB B X = B Y = mg mgl GB + 2I x sin(φ)ω 2 L GA + L GB (d ) A X = A Y = mgl GB +.5I x sin(2φ)ω 2 L GA L GB B X = B Y = mgl GB.5I x sin(2φ)ω 2 L GA + L GB (e ) A X = A Y = mg mgl2 GB +.5I x sin(2φ)ω 2 L GA L GB B X = B Y = mgl2 GB +.5I x sin(2φ)ω 2 L GA L GB A thrust bearing is able to withstand also axial forces

11 9 Kinetics of 3D rigid bodies - rotating frames 9- Solution I. Free-Body Diagram and Coordinate system Consider the reference system shown in the figure below. The origin coincides with the center of mass of the gear G, which is a fixed point. The body reference system xyz is a the principal reference frame, fixed with the gear in and rotating with it so. With the XY Z we denote the inertial reference system. Note that the bearing at point A is able also to sustain an axial force A Z. II. Kinematics The angular velocity ω of the gear is constant in magnitude and is always directed along the axis X of the shaft AB. The corresponding components in the xyz axes are ω x =, ω y = ω sin φ, ω z = ω cos φ. (9.37) These components remain constant for any general orientation of the xyz body frame, and therefore ω x = ω y = ω z =. Also note that since the body reference frame is rotating with the body, we know that ω must have a constant magnitude and direction (+Z) thus ω = (i.e. ω x = ω y = ω z = ). Furthermore, since G is a fixed point, r G =. III. Equations of Motion Being G the center of mass of the body, the angular momentum principle about point G writes: H G = I G ω (9.38) where I x ω = ω sin φ and I G = I y. (9.39) ω cos φ I z Substituting (9.39) into (9.38) yields I x H G = I G ω = I y ω sin φ = I z ω cos φ I y ω sin φ. (9.4) I z ω cos φ

12 9 Kinetics of 3D rigid bodies - rotating frames 9-2 By realizing that the body fixed reference frame is rotating with the same angular velocity of the body leads to Ḣ G = H G + ω H G = I y ω sin φ + ω sin φ I y ω sin φ, (9.4) I z ω cos φ ω cos φ I z ω cos φ which expresses the time derivative of the vector H G as ( I z + I y ) sin φ cos φω 2 Ḣ G = I y ω sin φ. (9.42) I z ω cos φ Since ω = and I z = 2I x, I y = I x the previous equation is equivalent to.5i x sin(2φ)ω 2 Ḣ G =. (9.43) The moment M ext G M ext G = about G with respect to the moving frame is given by A Y L GA + B Y L GB A X L GA cos φ B X L GB cos φ. (9.44) A X L GA sin φ B X L GB sin φ Note that the external moment is written w.r.t. the moving frame in the configuration shown in the figure. It is now possible to rewrite the angular momentum principle component-wise A Y L GA + B Y L GB =.5I x sin(2φ)ω 2 (9.45) A X L GA cos φ B X L GB cos φ = (9.46) A X L GA sin φ B X L GB sin φ =. (9.47) So far, we have 4 unknowns and 2 equations, since (9.46) and (9.47) leads to the same result. By applying the linear momentum principle in the X and Y directions we are able to obtain the required extra equations F x = m(a G ) x A X + B X = (9.48) F y = m(a G ) y A Y + B Y mg =. (9.49) Solving the linear system of equations gives A X = B X =, A Y = mgl GB +.5I x sin(2φ)ω 2 L GA + L GB (9.5) B Y = mg mgl GB +.5I x sin(2φ)ω 2 L GA + L GB. (9.5) The correct answer is therefore (a).

13 9 Kinetics of 3D rigid bodies - rotating frames A cube of side length 2a and mass M is moving with an initial velocity v along a frictionless table. When the cube reaches the end of the table, it pivots on a short lip and begins to rotate. What is the minimum speed magnitude v such that the cube falls off the table? Assume that the collision is completely inelastic. 8 (a ) v > 3 g( 3 ) 6 (b ) v > 5 ag( 4 + ) 3 (c ) v > 3 ag( 2 + ) (d ) v > 3 ag( 3 ) 6 (e ) v > 3 ag( 2 ) Solution The reference system is shown below, together with the free-body diagram at the instant of impact: By applying angular momentum principle about point B, we obtain Ḣ B + v B P = M ext B. (9.52) Three external forces, namely N, F, mg, act on the cube. However, only N and mg produce torque about point B. Since forces N and mg are not impulsive, the limit of the integral of the angular momentum principle between t and t + (the time instant right before and right after the impact) leads to M ext B = r BC (N + mg) lim t t + t+ t M ext B dt =. (9.53)

14 9 Kinetics of 3D rigid bodies - rotating frames 9-4 Since v B =, the angular momentum with respect to B is conserved. We can thus write H B (t ) = H B (t + ), Ma v = I B ω,. from which the angular velocity of the cube right after the collision is readily obtained (9.54) (9.55) ω = Ma v. (9.56) I B For the block to tip over the lip, its center of mass must be able to be aligned with a vertical line originating from the lip itself. Its elevation has to be a( 2 ), as shown in the figure below: For the block to attain such a configuration, the energy of the rotational motion (just after the impact) has to be large enough to raise the center of mass with a( 2 ). 2 I B ω 2 > Mga( 2 ) (9.57) Substituting (9.56) into (9.57) 2 I M 2 a 2 v 2 B > Mga( 2 ). (9.58) IB 2 2I B g( 2 ) v >. (9.59) Ma The centroid moment of inertia of a cube which has mass m and edge k is I = 6 mk2. (9.6) Substituting mass M and edge 2a, we get I C = 2 3 Ma2. (9.6) By using parallel axis theorem, we obtain the moment of inertia with respect to point B I B = 2 3 Ma2 + M(a 2) 2 = 8Ma2. (9.62) 3 Substituting I B into (9.59) yields 2 8Ma2 g( 2 ) 3 6 v > v > Ma 3 ag( 2 ). (9.63) The correct answer is answer (e). Note that the energy of the rotational motion transforms to potential energy, therefore ω continuously decreases till the center of mass reaches its maximum height.