Rigid Body Rotation

Chap. 11B - Rigid Body RotationA PowerPoint Presentation byPaul E. Tippens, Professor of PhysicsSouthern Polytechnic State University© 2007

Objectives: After completing thismodule, you should be able to:• Define and calculate the moment of of inertia forsimple systems.• Define and apply the concepts of of Newton’ssecond law, rotational kinetic energy, rotationalwork, rotational power, , and rotationalmomentum to to the solution of of physical problems.• Apply principles of of conservation of of energy andmomentum to to problems involving rotation ofofrigid bodies.

Inertia of RotationConsider Newton’s s second law for the inertia ofrotation to be patterned after the law for translation.F = 20 Na = 4 m/s 2F = 20 NR = 0.5 m = 2 rad/s 2Linear Inertia, m24 Nm =4 m/s 2 = 5kgRotational Inertia, I (20 N)(0.5 m)I = = = 2.5 kg m 4 m/s 22Force does for translation what torque does for rotation:

Rotational Kinetic EnergyConsider tiny mass m:v = RK = ½mv 2K = ½m(R) 2K = ½(mR 2 ) 2axism 1mm 4m 2m 3Sum to find K total:K = ½(mR 2 ) 2(½ 2 same for all m )Object rotating at constant Rotational Inertia Defined:I = mR 2

Common Rotational InertiasLLI1 3mL2I1 12mL2R R RI = mR 2 I = ½mR 2 2I2 5mRHoop Disk or cylinder Solid sphere

Example 2: A circular hoop and a diskeach have a mass of 3 kg and a radiusof 30 cm. . Compare their rotationalinertias.ImR2 2(3 kg)(0.2 m)RI = 0.120 kg m 2I = mR 2HoopRImR(3 kg)(0.2 m)1 2 122 2I = ½mR 2I = 0.0600 kg m 2Disk

Important AnalogiesFor many problems involving rotation, there is ananalogy to be drawn from linear motion.xfmA resultant force Fproduces negativeacceleration a fora mass m.IR4 kg 50 rad/s = 40 N mA resultant torque produces angularacceleration of diskwith rotational inertia I.FmaI

Newton’s s 2nd Law for RotationHow many revolutionsrequired to stop?= IFR4 kg 50 rad/sR = 0.20 mF = 40 NFR = (½mR 2 )22 f - 2 o2F2(40N) 0 mR (4 kg)(0.2 m)= 100 rad/s 20(50 rad/s)2 222 2(100 rad/s ) = 12.5 rad = 1.99 rev

Example 3: What is the linear accel-eration of the falling 2-kgmass?Apply Newton’s 2nd law to rotating disk:I TR = (½MR 2 )T = ½MRaT = ½MR( ) ;Rbuta = R; =andT = ½MaApply Newton’s 2nd law to falling mass:mg - T = ma mg - ½Ma T = ma(2 kg)(9.8 m/s 2 ) - ½(6 kg) a = (2 kg) a19.6 N - (3 kg) a = (2 kg) a a = 3.92 m/s 2aRR = 50 cmM6 kga = ?R = 50 cm6 kg+a2 kgTTmg2 kg

Work and Power for RotationWork = Fs = FRWork = FRsFWorkPower = =tt = ts = RFPower = Power = Torque x average angular velocity

Example 4: The rotating disk hasa radius of 40 cm and a mass ofs6 kg. . Find the work and power ifthe 2-kgmass is lifted 20 m in 4 s. sF2 kg 6 kgWork = = FR F=Ws 20 m = = = 50 rads = 20 mR 0.4 mF = mg = (2 kg)(9.8 m/s 2 ); F = 19.6 NWork = (19.6 N)(0.4 m)(50 rad)Work = 392 JWorkPower = =t392 J4sPower = 98 W

The Work-Energy TheoremRecall for linear motion that the work done is equalto the change in linear kinetic energy:2½2f0Fx ½mv mvUsing angular analogies, we find the rotational workis equal to the change in rotational kinetic energy:2½I2f0 ½I

Applying the Work-Energy Theorem:What work is neededto stop wheel rotating:FR 60 rad/sR = 0.30 mWork = r4 kgF = 40 NFirst find I for wheel: I = mR 2 = (4 kg)(0.3 m) 2 = 0.36 kg m 20 ½I ½IWork = -½I 2 2f0Work = -½(0.36 kg m 2 )(60 rad/s) 2Work = -648 J

Combined Rotation and Translationv cmv cmv cmNow consider a ball rolling withoutslipping. The angular velocity about the point P is same as fordisk, so that we write:First consider a disk slidingwithout friction. The velocity ofany part is equal to velocity v cmof the center of mass.RPv Or v RRv

Two Kinds of Kinetic EnergyKinetic Energyof Translation:K = ½mv2Kinetic Energyof Rotation: K = ½I 2RPvTotal Kinetic Energy of a Rolling Object:T1 2 1 22 2K mv I

Angular/Linear ConversionsIn many applications, you must solve an equationwith both angular and linear parameters. It isnecessary to remember the bridges:Displacement:Velocity:Acceleration:s Rs RvRv Rv RaR

Translation or Rotation?If you are to solve for a linear parameter, youmust convert all angular terms to linear terms: sRv 2 a I (?) mRR RIf you are to solve for an angular parameter, youmust convert all linear terms to angular terms:s R v Rv R

Example (a): Find velocity v of a disk ifgiven its total kinetic energy E.Total energy: E = ½mv 2 + ½I E mv I; I mR ; 1 2 1 2 1 22 2 2 v E mv mR E mv mvR21 2 11 21 2 1 22 2 2 2 ;2 4E 23mv4Eor v4 3mvR

Example (b) Find angular velocity of a diskgiven its total kinetic energy E.Total energy: E = ½mv 2 + ½I E mv I; I mR ; vR1 2 1 2 1 22 2 2E m( R) mR ; E mR mR 1 2 1 1 2 2 1 2 2 1 2 22 2 2 2 4E2 23mR4Eor 4 3mR2

Strategy for Problems• Draw and label a sketch of the problem.• List givens and state what is to be found.• Write formulas for finding the moments ofinertia for each body that is in rotation.• Recall concepts involved (power, energy,work, conservation, etc.) and write anequation involving the unknown quantity.• Solve for the unknown quantity.

Example 5: A circular hoop and a circulardisk, each of the same mass and radius,roll at a linear speed v. . Compare thekinetic energies.Two kinds of energy:K T = ½mv 2 K r = ½I 2Total energy: E = ½mv 2 + ½I = vRDisk:222½ ½ ½ v E mv mR 2 R E = ¾mv 2Hoop:222½ ½ v E mv mR 2 R E = mv 2vv

Conservation of EnergyThe total energy is still conserved forsystems in rotation and translation.However, rotation must now be considered.Begin: (U + K t + K R ) o = End: (U + K t + K R ) fHeight?Rotation?velocity?mgh o½½mv2o=mgh f½f½mv2fHeight?Rotation?velocity?

Example 6: Find the velocity of the 2-kgmassjust before it strikes the floor.mgh o½½mv2o=mgh f½f½mv2fR = 50 cm6 kg2 kgh = 10 mmgh mv I1 2 1 20 2 221 2 1 1 2 v mgh0 2mv 2( 2MR) 2 R (2)(9.8)(10) (2) v (6) vI121 2 1 22 4MR22.5v 2 = 196 m 2 /s 2v = 8.85 m/s

Example 7: A hoop and a disk roll from thetop of an incline. What are their speeds atthe bottom if the initial height is 20 m? mmgh o = ½mv 2 + ½I 2 Hoop: I = mR 222 2 v mgh0 ½mv ½( mR ) 2 R mgh o = ½mv 2 + ½mv 2 ; mgh o = mv 220 m Hoop: v = 14 m/s2v gh 0(9.8 m/s )(20 m)Disk: I = ½mR 2 ; mgh o = ½mv 2 + ½I 24 3 022 2 v mgh0 ½mv ½(½ mR ) 2 R vghv = 16.2 m/s

Angular Momentum DefinedConsider a particle mmoving with velocity vin a circle of radius r.Define angular momentum L:L = mvrSubstituting v= r, gives:L = m(r) r = mr 2 For extended rotating body:L = (mr 2 ) v = raxism 1mm 4m 2m 3Object rotating at constant Since I = mr 2 , we have:L = IAngular Momentum

Example 8: Find the angularL = 2 mmomentum of a thin 4-kgrod oflength 2 m if it rotates about itsmidpoint at a speed of 300 rpm. m = 4 kg112112For rod: I = mL 2 = (4 kg)(2 m) 2I = 1.33 kg m 2 rev 2 rad 1 min 300 31.4 rad/s min 1 rev 60 s L = I(1.33 kg m 2 )(31.4 rad/s) 2L = 1315 kg m 2 /s

Impulse and MomentumRecall for linear motion the linear impulse is equal tothe change in linear momentum:F t mv mvUsing angular analogies, we find angular impulse tobe equal to the change in angular momentum:f0 t I If0

Example 9: A sharp force of 200 N is applied tothe edge of a wheel free to rotate. The force actsfor 0.002 s. What is the final angular velocity?I = mR 2 = (2 kg)(0.4 m) 2I = 0.32 kg m 2Applied torque FRt = 0.002 sRF2kg 0 rad/sR = 0.40 mF = 200 NImpulse = change in angular momentumt = I f o0FR t = I ffFRt(200 N)(0.4 m)(0.002 s) 2I0.32 mf = 0.5 rad/s

Conservation of MomentumIn the absence of external torque the rotationalmomentum of a system is conserved (constant).0I f f o = tI f f oI o = 2 kg m 2 ; = 600 rpm I f = 6 kg m 2 ; = ?fI00If2(2 kg m )(600 rpm)6 kg m2 f = 200 rpm

Summary – Rotational AnalogiesQuantity Linear RotationalDisplacement Displacement x Radians Inertia Mass (kg) I (kgm 2 )Force Newtons N Torque N·mVelocity v “ m/s ” Rad/sAcceleration a “ m/s 2 ” Rad/s 2Momentum mv (kg m/s) I (kgm 2 rad/s)

Analogous FormulasLinear MotionF = maRotational Motion = IK = ½mv 2 K = ½I 2Work = FxPower = FvFx = ½mv f2-½mv o2Work = Power = I = ½I f2 -½I o2

Summary of Formulas: I = mR 2K12I2Work I I o o f f½I2½I2f0 Power tHeight?Rotation?velocity?mgh o½½mv2o=mgh f½f½mv2fHeight?Rotation?velocity?

CONCLUSION: Chapter 11BRigid Body Rotation