Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. � � 1 Finding Moments of Inertia 2.003J/1.053J Dynamics and Control I, Spring 2007 Professor Thomas Peacock 3/14/2007 Lecture 11 2D Motion of Rigid Bodies: Finding Moments of Inertia, Rolling Cylinder with Hole Example Finding Moments of Inertia Figure 1: Rigid Body. Figure by MIT OCW. IC = mi|ρi|2 i = mi(x 2 i + yi 2) i IC is the Moment of Inertia about C.Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. � � � 2 Finding Moments of Inertia Example: Uniform Thin Rod of Length L and Mass M Figure 2: Uniform thin rod of length L and mass M. Figure by MIT OCW. IC = mi(xi 2 + yi 2)For very thin rod, yi is small enough to neglect. i ≈ mixi 2 i Rod has mass/length = ρ. Convert to integral. IC = x 2dm rod dm = ρdx � L/2 IC = x 2ρdx −L/2 � 3 �L/2 L3x= ρ = ρ 3 12 −L/2 We know that mass M = ρL. Therefore: ML2 IC = 12Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. � � � 3 Example: Rolling Cyinder with a Hole Example: Uniform Thin Disc of Radius R Figure 3: Uniform thin disc of radius R. Figure by MIT OCW. Let ρ = mass/area. Consider a sliver that is a distance r from the center on this disc of radius R. IC = mi(x 2 i + yi 2) i = (x 2 + y 2)dm disc � R = r 22πrρdr 0 R R4 = ρ2πr3dr = 2πρ 0 4 R2 = πR2ρ 2 Mass of Disc: M = πR2ρ. Thus, MR2 IC = 2 Example: Rolling Cyinder with a Hole Find the equation of motion for a cylinder with a hole rolling without slip on a horizontal surface. In the hole with center A, R2 = R/2.Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. � 4 Example: Rolling Cyinder with a Hole Figure 4: Rolling cylinder with hole shown at 2 distinct positions. Figure by MIT OCW. Kinematics 2 Constraints: 1. Rolling on surface 2. No slip condition Use 1 generalized coordinate θ to describe the motion Only need 1 equation. For this example, we will use the work-energy principle to obtain the equation. 1. Gravity is a potential force. 2. Normal force on object: At point of contact, velocity is zero so no work done. No work done by external forces therefore T + V = constant. T = 12 M|vC |2 + 12 IC |ω|2 . Need center of mass. Where is the center of mass? Below O, because of hole. Kinetics Center of Mass Calculation First find position of center of mass. We know the center of mass of disc without hole: Point O. Can think of the hole to be “negative mass.” Consider moments about OX at point O πR2 πR2 R ρπR2(0) = ρ(πR2 − )OC − ρ 4 4 2 Distance from O to O is zero. ρ(πR2 − πR2 )OC: Mass moment of cylinder with the hole. 4Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. � � � � 5 Example: Rolling Cyinder with a Hole ρ πR2 R : Mass moment of the hole. 4 2 3ππR R (OC) = ⇒ OC = 4 8 6 Calculation of 21 M|vC |2 We know what rC is. rC is from point B to point C. R xC = Rθ − sin θ 6 R yC = R − cos θ 6 Differentiate: x˙C = Rθ˙ − Rθ˙ cos θ 6 y˙C = Rθ˙ sin θ 6 1 1 � R2 � 2 2Mv2 = πR2ρ − π ρ ( ˙xC + ˙yC )2 C 2 4 1 3 R2 R2θ˙2 R2θ˙2 = πR2ρ R2θ˙2 + θ˙2 cos 2 θ − cos θ + sin2 θ 2 4 36 3 36 11 3 1 1 Mv2 = πR4θ˙2ρ 1 − cos θ + 2 c 2 4 3 36 Calculation of 12 IC |ω|2 2nd term of kinetic energy is 21 IC |ω|2 . What is IC ? We want to find ICcwh . cwh: cylinder with hole mc: missing cylinder cyl: cylinder First find Icwh around O. Then shift to C with the Parallel-Axis Theorem. O cyl = Icwh + Imc IOO OCite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. � � � � � � 6 Example: Rolling Cyinder with a Hole Icwh = Icyl − Imc O OO Icyl =1 ρπR2R2 O 2 Imc = Imc + Mmc(OA)2 ← Parallel Axis Theorem O A 1 � R2 � R2 � R2 � R2 = ρπ + ρπ 2 4 4 4 4 3 = ρπR4 32 Icwh 1 3 13 = ρπR4 − ρπR4 = ρπR4 O 2 32 32 Now: Icwh = Icwh + Mcwh(OC)2 ← Parallel Axis Theorem O C Icwh = Icwh − M cwh(OC)2 C O 13 � πR2 � R2 37 Icwh = ρπR4 − ρπR2 − ρ = ρπR4 C 32 4 36 96 So we have that: 1 137 IC |ω|2 = ρπR4θ˙2 2 2 96 T +V =3 ρπR4θ˙2 37 − 1 cos θ + 37 ρπR4θ˙2 +V = ρπR4θ˙2 37 − 1 cos θ +V 8 36 3 192 64 8 Calculation of Potential Energy (V) What is V ? 3 R V = mgh = πR2ρ R − cos θ 4 6 Finding Equation of Motion So we have T + V = Constant d (T + V ) = 0 dt Differentiate:Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. � � � 7 Example: Rolling Cyinder with a Hole 2ρπR4θ˙θ ¨ 37 − 1 cos θ +1 ρπR4θ˙2 sin θθ˙+1 πR3ρg sin θθ˙= 0 64 8 8 8 Equation of motion: motion is complicated. Alternative Approach: Using Angular Momentum (Sketch) Angular momentum about moving point B. B is not on cylinder. B is not on ground. B is contact point between ground and cylinder. d τ B = HB + vB × P dt τB : Torque due to gravity HB = HC + rC × P vB : Moving Point (Rθ˙) P = MvC 1. Still have to find velocity and location of center of mass. 2. Still have to find IC . 3. But, even more work because need to take