# 7 Kinematics and kinetics of planar rigid bodies II

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1 7 Kinematics and kinetics of planar rigid bodies II 7.1 In-class A rigid circular cylinder of radius a and length h has a hole of radius 0.5a cut out. The density of the cylinder is ρ. Assume that the cylinder rolls without slipping on the floor. Compute the kinetic energy and the potential energy of the cylinder using the generalized coordinate θ shown. Solution 0. Notation We use the following notation for vectors: v = vi I. Choose the reference system Set the reference system as shown below. II. Moment of inertia The mass of the cylinder is M = ρh [ πa 2 π(0.5a) 2] = 3 4 πa2 ρh. (7.1) The moment of inertia of a cylinder of radius R without the hole is I O = π 2 R4 ρh. (7.2) 7-1

2 7 Kinematics and kinetics of planar rigid bodies II 7-2 Considering a new reference system as shown below, the center of mass of the cylinder is X C = 0 because of symmetry, (7.3) Y C = πa2 (a) π( a 2 )2 (a + a 2 ) πa 2 π( a 2 )2 = 5 6 a. (7.4) The center of mass of the cylinder without the hole and the center of mass of the hole are denoted by O, O respectively. Applying the parallel axis theorem and the additive property of the moment of inertia, we get ( [ π π ( a ) 4 ( ]) a 2 I C = ρh 2 a4 + πa 2 r OC 2 + π ro 2 2 2) C 2 = πa4 ρh. (7.5) III. Potential and kinetic energy The potential energy of the cylinder is V = Mgy C = 3 4 πa2 ρgh(a r OC cos θ) = 3 4 πa2 ρgha The kinetic energy of the cylinder is ( 1 cos θ ). (7.6) 6 T = 1 2 Mv2 C I Cω 2. (7.7) The angular velocity is ω(t) = θ(t)k. (7.8) Since point A is the instantaneous center of rotation (v A = 0) v C (t) = θ(t) r AC. (7.9) Note that r AC depends on time. Considering the geometry (7.9) becomes v C (t) = θ(t) r OC 2 + r OA 2 2 r OC r OA cos θ(t), (7.10)

3 7 Kinematics and kinetics of planar rigid bodies II 7-3 v C (t) = θ(t) (a 6 ) 2 + a2 2 a 37 a cos θ(t) = a θ(t) 6 36 Substituting all the known variables into (7.7) yields T (t) = 1 2 cos θ(t) 3. (7.11) ( ) [ ] πa2 ρh a θ(t) 37 cos θ(t) + 1 ( ) πa4 ρh θ(t) 2, (7.12) ( ) T (t) = πρha cos θ(t) θ(t). (7.13) 64 8

4 7 Kinematics and kinetics of planar rigid bodies II In-class A rod of mass m, length 2a and centroidal moment of inertia I C = 1 3 ma2 is dropped onto the edge of a table as shown. The rod is horizontal, has zero angular velocity and has downward velocity v 0 at the moment just before touching the table. (a) Determine, in terms of v 0, the angular velocity of the rod just after impact, assuming that energy is conserved in the collision. (b) Under the same assumptions, determine the velocity of the end of the rod that touched the table just after the impact. Does your result seem reasonable? Explain. Solution 0. Notation We use the following notation for vectors: v = vk Note that A denotes the contact point on the table and A denotes the contact point on the rod. I. Choose the reference system Set the reference system as shown below. II. Draw the free-body diagram III. Draw the configurations

5 7 Kinematics and kinetics of planar rigid bodies II 7-5 IV. Conservation of energy Assume that the collision occurs at t = t 1. A vertical impulse ( P y ) acts on the end of the rod (point A ) at t = t 1. As a result, velocity of the center of mass would be v 1 and angular velocity ω 1 just after the impact. Energy is conserved in the collision 1 T (t 1 ) + V (t 1 ) = T (t + 1 ) + V (t + 1 ). (7.14) Since gravity does not have enough time to act V (t 1 ) = V (t + 1 ). (7.15) Substituting (7.15) into (7.14) yields T (t 1 ) = T (t + 1 ), (7.16) 1 2 m(v 0) 2 = 1 2 m(v 1) I C(ω 1 ) 2, (7.17) (v 0 ) 2 = (v 1 ) a2 (ω 1 ) 2. (7.18) V. Angular momentum principle In order to get a second relation, we apply angular momentum about point A on the table 2 Ḣ A + v A P = M ext A. (7.19) Since v A = 0 and gravity has no time to act, angular momentum is conserved about point A. H A (t 1 ) = H A (t + 1 ). (7.20) VI. Angular momentum transfer formula Applying angular momentum transfer formula to point A and C H A = H C + P r CA = H C + mv C r CA. (7.21) Using (7.20) and (7.21), we get H C (t 1 ) + mv C (t 1 ) r CA (t 1 ) = H C (t + 1 ) + mv C (t + 1 ) r CA (t + 1 ), (7.22) 0 mav 0 k = 1 3 ma2 ω 1 k mav 1 k. (7.23) From (7.23), we get v 0 = v aω 1. (7.24) 1 Note that this statement is not valid in general, only if the collision is totally elastic. Elastic collision is defined as a collision in which kinetic energy is conserved. In several problems, this is a fair approximation. Elastic collisions occur only if there is no net conversion of kinetic energy into other forms. 2 Please remember: we can choose any arbitrary point as reference to apply angular momentum principle. The point is not necessarily on the body (lecture notes page 52).

6 7 Kinematics and kinetics of planar rigid bodies II 7-6 Using (7.24) and (7.18), we get v 1 = v 0 2. The angular velocity of the rod just after the impact ω 1 = 3v 0 2a. (7.25) (7.26) (b) VII. Velocity transfer formula In order to determine the velocity of point A just after the impact, we use velocity transfer formula with respect to point A and C. v A (t + 1 ) = v 1 + ω 1 r CA. (7.27) v A (t + 1 ) = v 1 j + ( ω 1 k) ( ai). (7.28) v A (t + 1 ) = v ( 0 2 j + ( 3v 0 k) ( ai) = v 0 2a 2 + 3v ) 0 j = v 0 j. (7.29) 2 We know that just before the impact the velocity of point A was v 0 j. Therefore our result seems reasonable. Since the magnitude of the velocity of A on the rod is conserved and changes just the direction in the collision. This shows an elastic collision which is expected when energy is conserved.

7 7 Kinematics and kinetics of planar rigid bodies II In-class A rigid, uniform flat disk of mass m and radius R is moving in the plane towards a wall with central velocity v 0 while rotating with angular velocity ω 1, as shown. Assuming that the collision in the normal direction is elastic and no slip occurs at the wall, find the velocity of the (center of the) disk after it collides with the wall. Solution 0. Notation We use the following notation for vectors: ω = ωk Note that B denotes the contact point on the wall and B denotes the contact point on the disk. I. Choose the reference system Set the reference system as shown below. II. Draw the free-body diagram III. Draw the configurations

8 7 Kinematics and kinetics of planar rigid bodies II 7-8 IV. Velocity transfer formula Assume the disk collides with the wall at point B, where an impulse ( P x, P y ) acts on it at t = t 1. Collision in the normal direction (y) is elastic so the magnitude of the velocity in the normal direction is conserved (v C ) y (t + 1 ) = (v C ) y (t 1 ), (7.30) (v C ) y = ( v 1 ) y = v 0 cos θ. (7.31) Since no slip occurs at the wall (v B ) x (t + 1 ) = 0. (7.32) Using the velocity transfer formula, we get (v B ) x (t + 1 ) = (v C ) x (t + 1 ) + (ω 1 r CB ) x = 0, (7.33) (v C ) x (t + 1 ) = (v 1 ) x = Rω 1. (7.34) V. Angular momentum principle Applying angular momentum principle about point B is Ḣ B + v B P = M ext B. (7.35) Since there is no force which produces external torque on B and v B momentum is conserved about B = 0,the angular Ḣ B = 0 H B (t 1 ) = H B (t + 1 ). (7.36) Using the angular momentum transfer formula (7.36) becomes Ḣ B = 0 H C (t 1 ) + P (t 1 ) r CB (t 1 ) = H C (t + 1 ) + P (t + 1 ) r CB (t 1 ). (7.37) lim r CB = lim r CB = R. t 1 t 1 t + 1 t mr2 ωk + mv 0 R sin θk = 1 2 mr2 ω 1 k + mr(v 1 ) x k. (7.38) Using (7.38) and (7.34) the x-component of the velocity of the disk after the impact is (v 1 ) x = 2 3 v 0 sin θ + 1 Rω. (7.39) 3 Using (7.31) and (7.39) the velocity of the disk after the collision is (2 v 1 = 3 v 0 sin θ + 1 ) 2 3 Rω + (v 0 cos θ) 2. (7.40)

9 7 Kinematics and kinetics of planar rigid bodies II Homework A cube with sides of length 2a and a mass M is moving with an initial speed v 0 along a frictionless table. When the cube reaches the end of the table it is caught abruptly by a short lip and begins to rotate. What is the minimum speed v 0 such that the cube falls off the table? (The collision is not elastic.) Solution 0. Notation Note that B denotes the contact point on the lip. I. Choose the reference system Set the reference system as shown below. II. Draw the free-body diagram at t = t 1 III. Angular momentum principle Applying angular momentum principle about point B Ḣ B + v B P = M ext B. (7.41) Three external forces (N, F, mg) act on the cube while only N and mg produce torque about point B. Since forces N and mg are not impulsive, we get M ext B = r BC (N + mg) t+ t M ext B dt = 0. (7.42) Since v B = 0 angular momentum with respect to B is conserved so H B (t 1 ) = H B (t + 1 ), Ma v 0 = I B ω. (7.43) (7.44)

10 7 Kinematics and kinetics of planar rigid bodies II 7-10 From which the angular velocity of the cube just after the collision ω = Ma v 0 I B. (7.45) For the block to tip over the lip, its center of mass must end up a distance a( 2 1) above its original position. The energy of the rotational motion 1 (just after the impact) has to be large enough to raise the center of mass with a( 2 1). 1 2 I B ω 2 > Mga( 2 1) (7.46) Substituting (7.45) into (7.46) 1 2 I M 2 a 2 v0 2 B v 0 > I 2 B > Mga( 2 1). (7.47) 2I B g( 2 1). (7.48) Ma The centroid moment of inertia of a cube which has mass m and edge k is I = 1 6 mk2. (7.49) Substituting mass M and edge 2a, we get I C = 2 3 Ma2. (7.50) Using parallel axis theorem I B = 2 3 Ma2 + M(a 2) 2 = 8Ma2 3 Substituting I B into (7.48) yields 2 8Ma2 g( 2 1) 3 v 0 > v 0 > Ma. (7.51) 16 3 ag( 2 1). (7.52) 1 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.

11 7 Kinematics and kinetics of planar rigid bodies II Homework A pendulum consists of a rod of length L with a frictionless pivot at one end. The pendulum is suspended from a flywheel of radius R which rotates with fixed angular velocity ω, as shown below. (a) Determine the angular velocity of the rod in terms of ω and the generalized coordinate θ indicated in the sketch. (b) Calculate the velocity of the mid point C of the rod Solution 0. Notation We use the following notation for vectors: v = ve 1 I. Choose the reference system Set the reference system as shown below. II. Draw the reference and displaced configuration III. Angular velocity of the rod To find the angular velocity of the rod, compare orientation of AB to A B ω rod = [ θ + ϕ]k = [ θ + ω]k. (7.53)

12 7 Kinematics and kinetics of planar rigid bodies II 7-12 IV. Draw the displaced configuration V. Velocity transfer formula The velocity transfer formula v C = v A + ω rod k r AC, (7.54) where v A is v A = ω rod r OA = ωre ψ. (7.55) Now the velocity of point C is v C = ωre ψ + (ω + θ) L 2 e θ. (7.56) Expressing e ψ and e θ in terms of i and j e ψ = sin ψi + cos ψj, (7.57) [ ( π )] [ ( π )] e θ = cos θ 2 ψ i sin θ 2 ψ j, (7.58) e θ = sin [θ + ψ]i + cos [θ + ψ]j. (7.59) Substituting (7.59) and (7.57) into (7.56) yields v C = ( ωr sin ψ (ω + θ) L 2 sin [θ + ψ])i+(ωr cos ψ +(ω + θ) L cos [θ + ψ])j. (7.60) 2

13 7 Kinematics and kinetics of planar rigid bodies II Homework A rigid cylinder of radius R is moving to the right such that its center C has velocity v. There is no slipping between the cylinder and the bar BD, but there is slipping between the cylinder and the ground. In the position shown (a) Determine the angular velocity of the bar BD. (b) Determine the velocity of the cylinder at the point where it contacts the ground. Solution 0. Notation We use the following notation for vectors: v = vi I. Choose the reference system Set the reference system as shown below. II. Set up the variables III. Angular velocity of the rod The velocity of point C v C = v C i = v C cos θi + v C sin θj. (7.61) Since there is no slip between the cylinder and the bar v A,cyl = v A,bar. (7.62)

14 7 Kinematics and kinetics of planar rigid bodies II 7-14 The velocity of point A on the bar is v A,bar = ω bar r BA = r BA θj. (7.63) The velocity of point A on the cylinder can be determined from v C transfer formula by using velocity v A,cyl = v C,cyl + ω cyl r CA = v C cos θi + v C sin θj Rω cyl I. (7.64) Substituting (7.63) and (7.64) into (7.62), we get r BA θj = v C cos θi + v C sin θj Rω cyl I, (7.65) 0 = (v C cos θ Rω cyl )I + (v C sin θ r BA θ)j. (7.66) Equating separately the coefficients of the I-components and the coefficients of the J- components, yields ω cyl = v C cos θ R θ = v C sin θ r BA From the geometry, we know that., (7.67) (7.68) r CA r BA = tan θ 2 r BA = R cot θ 2. (7.69) Using all the already expressed variables, the angular velocity of the bar is ω bar = θk = v C sin θ k = v ( ) C θ sin θ tan k = 2v ( ) C θ r BA R 2 R sin2 k. (7.70) 2 (b) The velocity of the cylinder at the point where it contacts the ground can be determined from v C by using velocity transfer formula [ v E,cyl = v C,cyl + ω cyl r CE = v C + v ] C cos θ R R i = v C (1 + cos θ)i. (7.71)

15 7 Kinematics and kinetics of planar rigid bodies II Homework A uniform rod of mass M and length 2b is pivoted at a point O, a distance s above the center of mass (CM). The rod is struck with a rapid impulsive force perpendicular to the rod at a point A, a distance a below the center of mass.the magnitude of the impulse is P = F t. Find the value of a such that there is no horizontal (N) reaction at the pivot, during the impact. (The moment of inertia of a uniform rod with mass M and length L about an axis through its center perpendicular to its longer side is I CM = ML 2 /12.) Solution I. Choose the reference system Set the reference system as shown below. II. Draw the free-body diagram III. Linear momentum principle Applying linear momentum principle in the x-direction P x = F ext x Mẍ = F N. (7.72) IV. Angular momentum principle Applying angular momentum principle Ḣ O + v O P = M ext O. (7.73) Since v O = 0, we get Ḣ O = M ext O I O θ = (s + a)f, (7.74) where the moment of inertia with respect to O can be determined by parallel axis theorem, such as [ ] [ ] (2b) I O = I CM + Ms 2 2 b 2 I O = M 12 + s2 = M 3 + s2. (7.75)

16 7 Kinematics and kinetics of planar rigid bodies II 7-16 Now we have two equations and three unknowns ẍ, θ, N. To solve these for N, we need to add a third equation. Considering the geometry and using small-angle approximation 1 we can write sin θ θ = x s ẍ = s θ. (7.76) Using (7.72), (7.74), (7.75) and (7.76) yields N = F Mẍ = F b 2 3 sa b s2. (7.77) Therefore the horizontal normal force will be zero if we set a = b2 3s. (7.78) 1 The small-angle approximation is a useful simplification of the basic trigonometric functions which is approximately true in the limit where the angle approaches zero. They are truncations of the Taylor series for the basic trigonometric functions to a first-order approximation.

17 7 Kinematics and kinetics of planar rigid bodies II Homework A rigid block of height H, length L, depth D and mass m rests on a rigid cylinder of mass M and radius R, as shown in the sketch. The cylinder rolls on the floor without slipping and the block rolls on the cylinder without slipping as well. Determine the kinetic and potential energy of the system. Solution 0. Notation We use the following notation for vectors: v = vi I. Choose the reference system Set the inertial reference system as shown below. II. Set up the variables Note that point O denotes the center of the cylinder and r OC = x OC i + y OC j. III. Angular velocity The angular velocity of the cylinder ω cyl = ϕk. (7.79) The angular velocity of the block ω block = θk. (7.80)

18 7 Kinematics and kinetics of planar rigid bodies II 7-18 IV. Velocity of the block and the cylinder Since there is no slip between the cylinder and the floor v O = R ϕi. (7.81) The velocity of the center of mass is v C = R ϕi + ẋ CO i + ẏ CO j = (R ϕ + ẋ CO )i + ẏ CO j. (7.82) Since there is no slip between the block and the cylinder v B,block = v B,cyl. (7.83) The velocity of point B on the cylinder can be determined using velocity transfer formula. v B,cyl = v O + ω cyl r OB = R ϕi + ( ϕk) (R sin θi + R cos θj) (7.84) v B,cyl = R ϕ(1 + cos θ)i R ϕ sin θj (7.85) The velocity of point B on the block can be determined using velocity transfer formula. v B,block = v C + ω block r CB (7.86) v B,block = (R ϕ + ẋ CO )i + ẏ CO j + ( θk) [(R sin θ x CO )i + (R cos θ y CO )j] (7.87) v B,block = (R ϕ + ẋ CO + R θ cos θ θy CO )i + (ẏ CO R θ sin θ + θx CO )j (7.88) V. Position of the center of mass Substituting (7.85) and (7.88) into (7.83) and equating separately the coefficients of the i-components and the coefficients of the j-components, yields R ϕ + ẋ CO + R θ cos θ θy CO = R ϕ(1 + cos θ), (7.89) ẏ CO R θ sin θ + θx CO = R ϕ sin θ. (7.90) Multiplying (7.89) by sin θ and (7.90) by cos θ and adding them together yields ẋ CO sin θ + ẏ CO cos θ θy CO sin θ + θx CO cos θ = 0. (7.91) Now we can realize that (7.91) can be written as d dt (x CO sin θ + y CO cos θ) = 0. (7.92) Integrating (7.92) with respect to time yields x CO sin θ + y CO cos θ = constant. (7.93) Substituting θ = 0, x CO = 0 and y CO = R + H 2 initial conditions into (7.93) yields x CO sin θ + y CO cos θ = R + H 2. (7.94)

19 7 Kinematics and kinetics of planar rigid bodies II 7-19 Multiplying (7.89) by cos θ and (7.90) by sin θ and subtracting them yields R( ϕ θ) = ẋ CO cos θ ẏ CO sin θ θy CO cos θ θx CO sin θ. (7.95) Now we can realize that (7.95) can be written as d dt (R(ϕ θ)) = d dt (x CO cos θ y CO sin θ), (7.96) Integrating (7.96) with respect to time yields R(ϕ θ) + C 1 = x CO cos θ y CO sin θ + C 2. (7.97) Substituting θ = 0, ϕ = 0, x CO = 0 and y CO = R + H 2 C 1 = C 2, so initial conditions into (7.97) yields R(ϕ θ) = x CO cos θ y CO sin θ. (7.98) In order to get x CO we multiply (7.94) by sin θ and (7.98) by cos θ and add them together. [ x CO = R + H ] sin θ + R(ϕ θ) cos θ. (7.99) 2 In order to get y CO we multiply (7.94) by cos θ and (7.98) by sin θ and subtract them. [ y CO = R + H ] cos θ R(ϕ θ) sin θ. (7.100) 2 VI. Energy of the system The potential energy of the system is V = mgy CO + mgy O = mgy CO, (7.101) ([ V = mg R + H ] ) cos θ R(ϕ θ) sin θ. 2 (7.102) The kinetic energy of the system is T = 1 2 Mv2 O I cylω 2 cyl Mv2 C I blockω 2 block, (7.103) T = 1 2 M(R ϕ) ( ) 1 2 MR2 ϕ M[(R ϕ+ẋ CO) 2 +ẏ 2 CO]+ 1 2 ( ) 1 12 m(h2 + L 2 ) θ 2. (7.104)