LE 01-10b

Physics 231Topic 8: Rotational MotionAlex BrownOctober 21-26 2015

MSU Physics 231 Fall 2015#

MSU Physics 231 Fall 2015#

MSU Physics 231 Fall 2015#Key Concepts: Rotational MotionRotational KinematicsEquations of Motion for RotationTangential MotionKinetic Energy and Rotational InertiaMoment of InertiaRotational dynamicsRolling bodiesVector quantitiesMechanical EquilibriumAngular Momentum

Covers chapter 8 in Rex & WolfsonMSU Physics 231 Fall 2015#Review: Radians & RadiusrCircumference= 2rs s = r in radians!360o=2 rad = 6.28 rad (rad) = 2/360o (deg)MSU Physics 231 Fall 2015#Review: Angular speed

Average angularvelocityInstantaneous Angular velocityAngular velocity : rad/s MSU Physics 231 Fall 2015#Review: Angular Linear Velocities

linear (tangential) velocity

MSU Physics 231 Fall 2015#Angular accelerationDefinition: The change in angular velocity per time unit

Average angularaccelerationInstantaneous angularaccelerationUnit: rad/s2MSU Physics 231 Fall 2015#Angular Linear Accelerations

velocitychange in velocityChange in velocity per time unitlinear or tangential acceleration

MSU Physics 231 Fall 2015#Equations of motionLinear motionAngular motionx(t) = xo + vot + at2

v(t) = vo + at

(t) = o + ot + t2

(t) = o + t

Angular motion = Rotational Motion!MSU Physics 231 Fall 2015#example=0A person is rotating a wheel. Thehandle is initially at rest at = 0o. For 5s the wheel gets a constant angularacceleration of 1 rad/s2. After thatThe angular velocity is constant.Through what angle will the wheelhave rotated after 10s. tFirst 5s: (5)= 0 +0t + t2

= 0 + (0)(5) + 1(5)2 = 12.5

(5) = 0 + t = 0 + (1)(5) = 5 rad/s

Next 5s: (5) = 12.5 + (5)(5) + 0 = 37.5

37.5 rad = 2150o = 6.0 rev

MSU Physics 231 Fall 2015#A Rolling Coin d0 = 18 rad/s = -1.8 rad/sHow long does the coin roll before stopping?What is the average angular velocity?c) How far (D) does the coin roll before coming to rest?Da) (t) = 0 + t 0 = 18 - 1.8t t=10 sb) = (0 + (10))/2 = (18 + 0)/2 = 18/2 = 9 rad/sd = 0.02 m = area under vs t curve = (10)(18)/2 = 90 rad

c) D = s = r = (90) (0.02/2) = 0.9 mMSU Physics 231 Fall 2015#TorqueIt is much easier to swing thedoor if the force F is appliedas far away as possible (d) fromthe rotation axis (O).Torque: The capability of a force to rotate an object aboutan axis.Torque = F d (Nm)Torque is positive if the motion is counterclockwise (angle increases)Torque is negative if the motion is clockwise (angle decreases)Top viewFdOMSU Physics 231 Fall 2015#Multiple forces causing torque.0.6 m0.3 mF1 = 100 NF2 = 50 NTwo persons try to gothrough a rotating doorat the same time, one onthe l.h.s. of the rotator andone the r.h.s. of the rotator.If the forces are applied asshown in the drawing, whatwill happen?Top view1 = -F1 d1 = -(100)0.3 = -30 Nm2 = F2 d2 = (50)0.6 = 30 NmNothing will happen! The 2torques are balanced.0 NmMSU Physics 231 Fall 2015#Center of gravity (center of mass)Fpulldpull(side view)Fgravitydgravity? = Fpulldpull + FgravitydgravityWe can assume thatfor the calculationof torque due to gravity,all mass is concentratedin one point:The center of gravity:the average position ofthe massdcg=(m1d1+m2d2++mndn) (m1+m2++mn) 1 2 3nMSU Physics 231 Fall 2015#Center of mass; more general

The center of gravity

center of gravity = center if massNote 2: freely rotating systems (I.e. not held fixed at some point)must rotate around their center of gravity.x1,y1x2,y2x3,y3m1gm2gm3gOxcg,ycgMSU Physics 231 Fall 2015# examples: (more in the book)

Oxygen molecule relative mass O is 16 H is 1

Where is the center of gravity?

MSU Physics 231 Fall 2015#QuizConsider a hoop of mass 1 kg.The center of gravity is located at:The edge of the hoopThe center of the hoopDepends on how the hoop is positioned relative to the earths surface.MSU Physics 231 Fall 2015#Object in equilibrium?CGFpFpddTop viewNewtons 2nd law: F=ma Fp+(-Fp)=ma=0No acceleration of the center of massup or downBut the block starts to rotate! = Fpd + Fpd = 2Fpd (both clockwise)

There is rotation!MSU Physics 231 Fall 2015#Translational equilibrium: F = ma = 0 The center of gravity does not moveRotational equilibrium: =0 The object does not rotate

Mechanical equilibrium: F = ma = 0 & =0 No movement! Torque: = Fd

Center ofGravity:MSU Physics 231 Fall 2015#question20N20N20NIs this freely rotating object in mechanical equilibrium?a) Yes b) No1m1mCOGFreely rotating: can only rotate around its COG:Rotational eq.: = -20x1 + 20x0 + 20x1 =0Translational eq.: F= 20 20 +20 =+20

No translational equilibrium: no mechanical equilibrium!MSU Physics 231 Fall 2015#quiz0-212 m20N40N40NA wooden bar is initially balanced. Suddenly, 3 forces areapplied, as shown in the figure. Assuming that the bar canonly rotate, what will happen (what is the sum of torques)?the bar will remain in balancethe bar will rotate counterclockwisethe bar will rotate clockwiseTorque: =FdMSU Physics 231 Fall 2015#Somethings to keep in mindIf an object is in equilibrium, it does not matter whereone chooses the axis of rotation: the net torque must always be zero.

If it is given that an object is at equilibrium, you can choose the rotation axis at a convenient location to solvethe problem.

MSU Physics 231 Fall 2015#Weight of board: wWhat is the tension in each of thewires (in terms of w) given that theboard is in equilibrium?Translational equilibriumF = ma = 0T1+T2-w = 0 so T1=w-T2Rotational equilibrium = 0T10 (0.5)w + (0.75)T2=0T2=0.5/(0.75 w) = 2/3wT1=1/3wT2=2/3wNote: I chose the rotation point at T1, but could have chosen anywhere else. By choosing T1 the torque due to T1Does not contribute to the equation for rotational equilibrium wT1T200.75m0.25mMSU Physics 231 Fall 2015#rFt = matTorque and angular accelerationmFtNewton 2nd law: F=maFtr = mrat

Ftr = mr2 Used at = r

= mr2 = I (used = Ftr)The angular acceleration goes linear with the torque.

I = mr2 = moment of inertia for one point mass MSU Physics 231 Fall 2015# = IMoment of inertia I:I = (miri2) ri is the distance to the axis of rotation

I = mr2 = when all mass is at the same distance rcompare with: F=maThe moment of inertiain rotations is similar tothe mass in Newtons 2nd law.

m1m2m3r1r2r3More General for rotation around a common axisMSU Physics 231 Fall 2015#Extended objects - hoopI=(miri2) =(m1+m2++mn)R2 =MR2MFrmiMSU Physics 231 Fall 2015#A homogeneous stickRotation pointmmmmmmmmmmF=(m1r12+m2r22++mnrn2) =(miri2) =I

Moment of inertia I:

I=(miri2)

MSU Physics 231 Fall 2015#Two inhomogeneous sticks mmmmmmmm5m5mF5mmmmmmm5mmmF18m18 m=(miri2) 118mr2=(miri2) 310mr2rEasy to rotate!Difficult to rotateMSU Physics 231 Fall 2015#

Demo: fighting sticksMSU Physics 231 Fall 2015#A simple exampleA and B have the same total mass. If the sametorque is applied, which one accelerates faster? FFrrAnswer: A=IMoment of inertia I:I=(miri2)ABMSU Physics 231 Fall 2015#The rotation axis matters!I = (miri2)=(0.2) 0.52 + (0.3) 0.52 + (0.2) 0.52 + (0.3) 0.52= 0.5 kg m2I = (miri2)= (0.2) 0 + (0.3) 0.52 + (0.2) 0 + (0.3) 0.52= 0.3 kg m2

0.5 m0.2 kg0.3 kgMSU Physics 231 Fall 2015#=I (compare to F=ma)

Moment of inertia I: I=(miri2)

: angular acceleration

I depends on the choice of rotation axis!!MSU Physics 231 Fall 2015#Some common casesRRLLhoop/thin cylindricalshell: I=MR2disk/solid cylinder I=1/2MR2Long thin rod withrotation axis throughcenter: I=1/12ML2solid sphere I=2/5MR2thin spherical shell I=2/3MR2Long thin rod withrotation axis throughend: I=1/3ML2MSU Physics 231 Fall 2015#Lmass: mFg=mgCompare the angular acceleration for 2 bars of differentmass, but same length. = I = mL2/3 also = Fd = mg(cos)L/2 So mg(cos)L/2 = mL2/3 and = 3g(cos)/(2L)independent of mass!Compare the angular acceleration for 2 bars of same mass,but different length = 3g(cos)/(2L) so if L goes up, goes down!

Ibar=mL2/3Fp = Fgcos = mg cos perpendicular componentFalling BarsMSU Physics 231 Fall 2015#Rotational kinetic energyConsider a object rotatingwith constant velocity. Each pointmoves with velocity vi. The totalkinetic energy is:

KEr = I2Conservation of energy for rotating object must include:rotational and translational kinetic energy and potential energy

[PE+KEt+KEr]initial= [PE+KEt+KEr]finalmirivMSU Physics 231 Fall 2015#Convenient way of writing KE for rotationobject Ik ACylindrical shellMr21 BSolid cylinder(1/2) mr20.5 CThin spherical shell(2/3) mr2(2/3) DSolid sphere(2/5) mr2(2/5) I = kmr2MSU Physics 231 Fall 2015#

rolling motion (no slipping) this means that =v/rMSU Physics 231 Fall 2015#

KE for rolling motion (no slipping) r = radius KE = KE (translational) + KE (rotational) = [mv2 + I2] = [mv2 + (kmr2)v2/r2] = [mv2+kmv2] = (1+k)mv2

used: I=kmr2 and =v/r KE = (1+k)mv2MSU Physics 231 Fall 2015#Example.1mConsider a ball (sphere) and a block going down the same 1m-high slope.The ball rolls and both objects do not feel friction. If bothhave mass 1kg, what are their velocities at the bottom (I.e.which one arrives first?). The radius of the ball is 0.4 m.Block: [mv2+mgy]initial = [mv2 + mgy]final

so v = 4.43 m/s 0 + mgh = mv2 + 0 v=2ghMSU Physics 231 Fall 2015#Example.1mConsider a ball and a block going down the same 1m-high slope.The ball rolls and both objects do not feel friction. If bothhave mass 1kg, what are their velocities at the bottom (I.e.which one arrives first?). The radius of the ball is 0.4 m.Ball: [m(1+k)v2+mgy]initial= [m(1+k)v2 + mgy]final

so v = 3.74 m/s slower than the block since part of the energy goes into rotational KE 0 + mgh = m(1+k)v2 + 0 v=[2gh/(1+k)] with k = 2/5 (No mass dependence)MSU Physics 231 Fall 2015#Which one goes fastest (h=1m)objectkvfastestBlock sliding04.43Cylindrical shell13.134Solid cylinder1/23.612Thin spherical shell2/33.423Solid sphere2/53.741The larger the moment of inertia, the slower it rolls!

The available Ekin is spread over rotational and translational (linear) Ekin

Or: besides moving the object, you spend energy rotating itMSU Physics 231 Fall 2015#Angular momentum

Conservation of angular momentum: If the net torque equals zero, the angular momentum L does not change

Ii i = If fMSU Physics 231 Fall 2015#Conservation laws:In a closed system:

Conservation of energy E

Conservation of linear momentum p

Conservation of angular momentum LMSU Physics 231 Fall 2015#Neutron star

Sun: radius: Rs = (7) 105 km Period of rotation: 25 daysSupernova explosionNeutron star: radius: Rn = 10 km What is frequency of rotation ?Isphere=(2/5)MR2 s = sun n = neutron star(assume no mass is lost so Ms=Mn)

s = 2/(25 days x 24 x 3600) = (2.9) 10-6 rad/s

Conservation of angular momentum:Iss = Inn n = (Rs/Rn)2 s = 1.4E+04 rad/s

fn = (n /2 ) = 2200 Hz !

MSU Physics 231 Fall 2015#The spinning lecturerA lecturer (60 kg) is rotating on a platform with i=2 rad/s (1 rev/s). He is holding two 1 kg masses 0.8 m away from hisbody. He then puts the masses close to his body (R=0.0 m). Estimate how fast he will rotate (marm=2.5 kg).0.4m0.8mIi = 0.5 MlecR2 (body) +2(MwRw2) (two weights) +2(0.33Marm0.82) (two arms) = 1.2 + 1.3 + 1.0 = 3.5 kg m2

If = 0.5 MlecR2 = 1.2 kg m2

MSU Physics 231 Fall 2015#The spinning lecturerA lecturer (60 kg) is rotating on a platform with i=2 rad/s (1 rev/s). He is holding two 1 kg masses 0.8 m away from hisbody. He then puts the masses close to his body (R=0.0 m). Estimate how fast he will rotate (marm=2.5 kg).0.4m0.8m Conservation of angular mom.

Ii I = If f f = (Ii /If) i

f = (3.5/1.2) 2 = 18.3 rad/s (approx 3 rev/s)

For details see next slide KEi = KEf ? A=yes B=noMSU Physics 231 Fall 2015#Details of spinning lecturer

Iinitial=Ibody+2Imass+2IarmsIbody= 0.5MlecRlec2 (solid cylindrical shell) M=60, R=0.2Imass=MmassRmass2 (point-like mass at radius R) M=1 R=0.8Iarms=1/3MarmRarm2 (arm is like a long rod with rotation axis through one of its ends) M=2.5 kg R=0.8Iinitial=0.5MlecRlec2+2(MmassRmass2)+2(0.33MarmRarm2) =1.2+1.3+1.0= 3.5 kgm2Ifinal =0.5MlecR2=1.2 kgm2 (assumed that masses and arms do not contribute to moment of inertia anymore) Conservation of angular mom. Iii=Iff3.5*2=1.2*f i= was given to be 2 rad/s f=18.3 rad/s (approx 3 rev/s)

KEi = (1/2)Ii (i )2 = 69 KEf = 201MSU Physics 231 Fall 2015#rFt = matOrbital motionmFtI = mr2 = moment of inertia one point mass L = I

For a point-like mass orbiting a central axis:I=mr2 and =v/r

So L = (mr2 ) (v/r) and thus L=mrv MSU Physics 231 Fall 2015#

MSU Physics 231 Fall 2015#

MSU Physics 231 Fall 2015#

MSU Physics 231 Fall 2015#

MSU Physics 231 Fall 2015#

MSU Physics 231 Fall 2015#question

ABdAdBTwo people are on a see saw. Person A is lighter than person B. If person A is sitting at a distance dA from the center of the seesaw, is it possible to find a distance dB so that the seesaw is in balance?A) YesB) NoIf MA=0.5MB and dA=3 m, then to reach balance, dB=A) 1 m B) 1.5 m C) 2 m D) 2.5 m E) 3 mMSU Physics 231 Fall 2015#Translational equilibrium (Hor.)Fx = ma = 0n-Tx = n - Tcos37o = 0 so n = Tcos37oTranslational equilibruim (vert.)Fy = ma = 0sn w w + Ty=0sn - 2w + Tsin37o = 0 (use n=Tcos37o)sTcos370 - 2w + Tsin370 = 00.4T - 2w + 0.6T=0T = 2w Rotational equilibrium: = 0xw + 2w - 4Tsin370 = 0 (4sin37o) = 2.4 T = 2ww (x + 2 - 4.8)=0x = 2.8 ms=0.5 coef of friction between the wall andthe 4.0 meter bar (weight w). What is the minimum x where you can hang a weight w for which the bar does not slide?wsnnwTTyTx(x=0,y=0)37oxMSU Physics 231 Fall 2015#The direction of the rotation axisThe conservation of angular momentum not only holdsfor the magnitude of the angular momentum, but alsofor its direction. The rotation in the horizontal plane isreduced to zero: There must have been a largenet torque to accomplish this! (this is why youcan ride a bike safely; a wheel wants to keep turningin the same direction.)LLMSU Physics 231 Fall 2015#Rotating a bike wheel!LLA person on a platform that can freely rotate is holding a spinning wheel and then turns the wheel around. What will happen?Initial: angular momentum: Li = IwheelwheelClosed system, so L must be conserved.Final: Lf = - Iwheel wheel + Ipersonperson = Li = Iwheelwheelperson= 2Iwheel wheel/Iperson MSU Physics 231 Fall 2015#