Chapter 5 Kinetics of Rigid Bodies

Chapter 5 Kinetics of Rigid Bodies

Chapter 5 Kinetics of Rigid Bodies Question 5–1 A homogeneous circular cylinder of mass m and radius r is at rest atop a thin sheet of paper as shown in Fig. P5-1. The paper lies flat on a horizontal surface. Suddenly, the paper is pulled with a very large velocity to the right and is removed from under the cylinder. Assuming that the removal of the paper takes place in a time t, that the coefficient of dynamic friction between all surfaces is µ, and that gravity acts downward, determine (a) the velocity of the center of mass of the cylinder and (b) the angular velocity of the cylinder the instant that the paper is removed. r g ω O x Friction (µ) Paper Figure P5-1 Solution to Question 5–1 First, let be the ground. Then, it is convenient to choose the following coordinate F system fixed in reference frame : F Origin at O Ex To The Right = Ez Into Page = Ey Ez Ex = × 182 Chapter 5. Kinetics of Rigid Bodies Next, let be the cylinder. Then, choose the following coordinate system fixed in D reference frame : D Origin at O er Fixed in = D ez Ez = eθ ez er = × Now, in order to solve this problem, we need to apply linear impulse and momen- tum to the center of mass of the cylinder and angular impulse and momentum about the center of mass of the cylinder. In order to apply these two principles, we use the free body diagram shown in Fig. 5-1 where N Reaction Force of Paper (Surface) on Disk = mg Force of Gravity = F Force of Friction f = O mg P Ff N Figure 5-1 Free Body Diagram of Disk Now from the geometry we have that N NEy = mg mgEy = (5.1) v F µ N rel f = − k k v k relk Now we need to determine vrel. Denoting the point of contact between the disk and the paper by P (see Fig. 5-1), we note that v FvP Fv (5.2) rel = − paper where v is the velocity of the paper in reference frame . Since the paper is F paper F pulled in the positive Ex-direction, we have that Fv v Ex (5.3) paper = paper 183 Next, we need to determine FvP . From kinematics of rigid bodies we have that FvP FvO F ωR (rP rO) (5.4) − = × − where denotes the reference frame of the cylinder and ω is the angular ve- R F R locity of reference frame in reference frame . From the geometry we have that R F F ωR ωEz (5.5) = and rP rO r Ey (5.6) − = Consequently, FvP FvO ωEz r Ey rωEx (5.7) − = × =− Furthermore, rO xEx (5.8) = which implies that F d FvO (rO) x˙Ex (5.9) = dt = We then have that FvP FvO rωEx (x˙ rω)Ex (5.10) = − = − Now, since the paper is “suddenly” pulled to the right, it implies that the paper is pulled such that its speed is extremely large. Therefore, v x˙ rω (5.11) paper ≫ − which implies that x˙ rω v 0 (5.12) − − paper ≪ Now we have that v vP v (x˙ rω)Ex v Ex (x˙ rω v )Ex (5.13) rel = − paper = − − paper = − − paper But from Eq. (5.12) we see that v x˙ rω v Ex (5.14) rel = −| − − paper| which implies that vrel Ex (5.15) v =− k relk The force of friction is then given as F µ N Ex (5.16) f = k k Now that we have expressions for all of the forces, we move on to the application of linear impulse and momentum and angular impulse and momentum 184 Chapter 5. Kinetics of Rigid Bodies (a) Velocity of Center of Mass of Cylinder at Instant When Paper is Removed We have that F mF¯a mFaO (5.17) = = Differentiating Eq. (5.9) in reference frame , we have F FaO x¨Ex (5.18) = Furthermore, using the result of the force resolution from above, we have F N mg F NEy mgEy µ N Ex (N mg)Ey µ N Ex (5.19) = + + f = + + k k = + + k k Equating F and mFaO, we obtain (N mg)Ey µ N Ex mx¨Ex (5.20) + + k k = which yields the following two scalar equations: N mg 0 + = (5.21) µ N mx¨ k k = Therefore, N mg (5.22) =− which implies that N mgEy (5.23) =− which further implies that N mg (5.24) k k= We then have that µmg mx¨ (5.25) = This last equation can be integrated from 0 to ∆t to give ∆t ∆t µmgdt mxdt¨ (5.26) Z0 = Z0 We then have that µmg∆t mx(˙ ∆t) mx(t˙ 0) mvO(∆t) mvO(t 0) (5.27) = − = = − = Noting that the disk is initially stationary, we have that vO(t 0) 0 (5.28) = = which implies that µmg∆tmvO(∆t) (5.29) Solving this last equation for vO(∆t), we obtain vO(∆t) µg∆t (5.30) = Therefore, the velocity of the center of mass at the instant that the paper is com- pletely pulled out is FvO(∆t) µg∆tEx (5.31) = 185 (b) Angular Velocity of Cylinder at Instant When Paper is Removed Applying Euler’s law about the center of mass of the cylinder, we have that ¯ Fd ¯ Fd M FH MO FHO (5.32) = dt ⇐⇒ = dt Now since N and mg pass through point O, the moment about O is due to only Ff . Consequently, ¯ M (rP ¯r) F (5.33) = − × f Substituting the earlier expressions for rP ¯r and F , we obtain − f ¯ M r Ey µmgEx rµmgEz (5.34) = × =− Furthermore, ¯ ¯ FH IR F ωR (5.35) = · Now since er , eθ, ez is a principal-axis basis, we have that { } ¯ IR Irr er er Iθθeθ eθ Izzez ez (5.36) = ⊗ + ⊗ + ⊗ ¯ Using the angular velocity F ωR from Eq. (5.5), we obtain FH as 2 ¯ mr FH ωez (5.37) = 2 ¯ Differentiating FH in reference frame , we have F 2 Fd ¯ mr FH ω˙ ez (5.38) dt = 2 Then, applying Euler’s second law, we have mr 2 rµmg ω˙ (5.39) − = 2 Integrating this last equation from 0 to ∆t, we obtain ∆t ∆t mr 2 rµmgdt ωdt˙ 2 (5.40) Z0 − = Z0 We then obtain mr 2 mr 2 rµmg∆t ω(∆t) ω(t 0) (5.41) − = 2 − 2 = Noting that the disk is initially stationary, we have that ω(t 0) 0 (5.42) = = 186 Chapter 5. Kinetics of Rigid Bodies Therefore, mr 2 rµmg∆t ω(∆t) (5.43) − = 2 Solving for ω(∆t), we obtain 2µg ω(∆t) ∆t (5.44) =− r The angular velocity of the disk at the instant that the paper is pulled out is then given as 2µg F ωR(∆t) ∆tez (5.45) =− r 187 Question 5–2 A collar of mass m1 is attached to a rod of mass m2 and length l as shown in Fig. P5-2. The collar slides without friction along a horizontal track while the rod is free to rotate about the pivot point Q located at the collar. Knowing that the angle θ describes the orientation of the rod with the vertical, that x is the horizontal position of the cart, and that gravity acts downward, determine a system of two differential equations for the collar and the rod in terms of x and θ. x m1 Q l g θ m2 Figure P5-2 Solution to Question 5–2 Preliminaries For this problem it is convenient to apply the following balance laws: • Newton’s 2nd law to the collar • Euler’s 1st law to the center of mass of the rod • Euler’s 2nd law about the center of mass of the rod In order to use the aforementioned balance laws, we will need the following kine- matic quantities in an inertial reference frame: • The acceleration of the collar • The acceleration of the center of mass of the rod • The rate of change of angular momentum of the rod relative to the center of mass of the rod 188 Chapter 5. Kinetics of Rigid Bodies Kinematics First, let be a fixed reference frame. Then, choose the following coordinate sys- F tem fixed in : F Origin at Collar When x 0 = Ex To The Right = Ez Out of Page = Ey Ez Ex = × Next, let be a reference frame fixed to the rod. Then, choose the following coor- R dinate system fixed in : R Origin at Collar er Along Rod = Ez Out of Page = eθ Ez er = × The relationship between the bases Ex, Ey , Ez and er , eθ, ez is shown in Fig. 5- { } { } 2 Using Fig. 5-2, we have that Ey eθ θ Ez, ez Ex θ er Figure 5-2 Relationship Between Bases Ex, Ey , Ez and er , eθ, ez for Question { } { } 6.2. er sinθ Ex cosθ Ey = − eθ cosθ Ex sinθ Ey = + (5.46) Ex sinθ er cosθ eθ = + Ey cosθ er sinθ eθ = − + In terms of the basis Ex, Ey , Ez , the position of the collar is given as { } r xEx (5.47) = 189 Therefore, the velocity of the collar in reference frame is given as F Fv x˙Ex (5.48) = Furthermore, the acceleration of the collar in reference frame is given as F Fa x¨Ex (5.49) = Next, the position of the center of mass of the rod relative to the collar is given as l ¯r r er (5.50) − = 2 In addition, the angular velocity of in reference frame is given as R F ˙ F ωR θEz (5.51) = Differentiating ω in Eq.

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