Rotational Motion

• Rotation (rigid body) versus translation (point particle) • Rotation concepts and variables • Rotational kinematic quantities ƒ Angular position and displacement ƒ Angular velocity ƒ Angular acceleration • Rotation kinematics formulas for constant angular acceleration

1

“Radian” “radian” : more convenient unit for angle than degree Definition: 360o 180o • 2π radians = 360 degree 1 radian = = = 57.3o 2π π s arc length ≡ s = r θ (in radians) θ ≡ rad r θ (in rad) s =×2π rr =θ 2π

r s θ’ θ

Example: r = 10 cm, θ = 100 radians Æ s = 1000 cm = 10 m.

1 Rigid body

Rigid body: A “rigid” object, for which the position of each point relative to all other points in the body does not change.

Example: Solid: Rigid body Liquid: Not rigid body

Rigid body can still have translational and rotational motion.

Angular position of rotating rigid body

• By convention, θ is measured CCW from the x-axis • It keeps increasing past 2π, can be y negative, etc. • Each point of the body moves around θ the axis in a circle with some specific Reference x line rotates radius with body

rotation axis “o” rigid body fixed to body parallel to z-axis

2 Angular displacement of rotating rigid body

y

θ Reference x line rota tes with body Angular displacement: rotation axis “o” rigid body • Net change in the angular coordinate fixed to body parallel to z-axis

Δθ ≡ θfinal − θinital (an angle in rad.)

Arc length: Δs • Measures distance covered by a point as it moves Reference line rotating with body y through Δθ (constant r) Δs = r Δθ

Δs ≡ rΔθ (a distance along a circular arc) θf r r

θo x

Rigid body rotation: angular & tangential velocity

Angular velocity ω: For any point, r is the perpendicular • Rate of change of the angular displacement distance to the rotation axis Δθ Δθ dθ ωave ≡ ωinst ≡ Lim ≡ Δt Δt → 0 Δt dt vT • Units: radians/sec. Positive in Counter-Clock-Wise sense

• Frequency f = # of complete revolutions/unit time r • f = 1/T T = period (time for 1 complete revolution θ = ωΔτ x ω = 2πf = 2π/T f = ω/2π

Tangential velocity vT: • Rate at which a point sweeps out arc length along circular path ΔΔs θ Æ Δ=Δs r θ Æ = r vrT = ω ΔΔtt

3 iClicker Quiz

1.1. The period of a rotating wheel is 12.57 seconds. The radius of the wheel is 3 meters. It’s angular speed is closest to:

A. 79 rpm B. 0.5 rad/s C. 2.0 rad/s D. .08 rev/s E. 6.28 rev/s

1.2. A point on the rim of the same wheel has a tangential speed closest to:

A. 12.57 rev/s B. 0.8 rev/s C. 0.24 m/s Δs ≡ rΔθ D. 1.5 m/s E. 6.28 m/s vT = ωr ω = 2πf = 2π/T

Rigid body rotation: angular acceleration

Angular acceleration α: Δω Δωωd • Rate of change of the angular velocity ααave ≡≡ inst Lim = ΔtdtΔ→t 0 Δt • Units: rad/s2 • CCW considered positive

• for CONSTANT α: ωf = ω0 + αΔt

4 1D and Angular Kinematics Equations (Same mathematical forms)

1D motion with Angular motion with constant acceleration a constant angular acceleration α x(t), v(t), a(t) variables θ(t), ω(t), α(t)

dx dv dθ dω v = a = Definitions ω = α = dt dt dt dt

vf (t) = v0 + at Kinematic ωf (t) = ω0 + αt Equations x (t) = x + v t + 1 at 2 θ (t) = θ + ω t + 1 αt2 f 0 0 2 f 0 0 2 v2 (t) = v2 + 2a[x − x ] 2 2 f 0 f 0 ωf (t) = ω0 + 2α[θf − θ0 ]

Rotational variables are vectors, having direction

The angular displacement, speed, and acceleration ( θ, ω, α ) are vectors with direction.

The directions are given by the right-hand rule:

Fingers of right hand curl along the angular direction (See Fig.)

Then, the direction of thumb is thdihe direct ion o fhf the angu lar quantity.

5 Example: A grindstone is rotating with constant angular acceleration about a fixed axis in space. Initial conditions 2 at t = 0: α = 0.35 rad/s ω0 = - 4.6 rad/s

When is Δθ = 0 again in addition to t=0? Positive directions: right hand rule

Example: Wheel rotating and accelerating

At t = 0, a wheel rotating about a fixed axis at a constant angular acceleration has an angular velocity of 2.0 rad/s. Two seconds later it has turned through 5.0 complete revolutions. Find the angular acceleration of this wheel?

ωf (t) = ω0 + αt θ (t) = θ + ω t + 1 αt2 f 0 0 2 2 2 ωf (t) = ω0 + 2α[θf − θ0 ]

6 Rigid body rotation: radial and tangential acceleration

Centripetal (radial) acceleration ac or ar • Radial acceleration component, points toward rotation axis v2 T 2 Fma= arr == ωω (use v = r ) rr a r T T vT r ac ω,α

Tangential acceleration aT: x • Tangential acceleration component • Proportional to angular acceleration α and also to radius r • Units: length / time 2

aT = rα Fmatangential = T

Rotation variables: angular vs. linear

s = rΔθ

vrT = ω

arT = α

v 2 a = T =rω 2 r r

7 A ladybug sits at the outer edge of a merry-go-round, and a gentleman bug sits halfway between her and the axis of rotation. The merry-go-round makes a complete revolution once each second. The gentleman bug’s angular velocity is

A. half the ladybug’s. B. the same as the ladybug’s. C. twice the ladybug’s. D. impossible to determine

A ladybug sits at the outer edge of a merry-go-round, and a gentleman bug sits halfway between her and the axis of rotation. The merry-go-round makes a complete revolution once each second. The gentleman bug’s velocity is

A. half the ladybug’s. B. the same as the ladybug’s. C. twice the ladybug’s. D. impossible to determine

8 Rotational Dynamics

We want something like “F=ma” for rotational motion…..

• Moment of inertia – rotational analog of mass • Torque – rotational analog of force

17

Something like mass for rotational motion: Moment of Inertia, I

G L

Kinetic energy of ladybug and gentlemanbug

11 1 1 Kmvmvmr=+22 =()ωω 2 + mr () 2 22LL GG 2 L L 2 G G

1122 22 1 2 2 2 1 2 =+mrLLω mr GGωωω =+=( mr LL mr GG ) I 22 2 2 22 I =+mrL LGG mr 1 Generally, Imrmrmr=+++222... Kinetic energy: K = Iω 2 11 2 2 33 2

9 Example: Find moment of inertia for a crossed dumbbell

•Four identical balls as shown: m m = 2 kg •Connected by massless rods: 2 d d length d = 1 m. mmdd A BC d Rotational inertia I depends on axis chosen m

A) Choose rotation axis perpendicular to figure through point “A”

B) N ow ch oose axi s perpendi cul ar t o fi gure th rough poi nt “B”

C) Let rotation axis pass through points “B” and “C”

Calculation of Moment of inertia for continuous mass distributions requires “Integration, a kind of calculus”. We will just use the result.

10 Moments of Inertia of Various Rigid Objects

Now we want to define “torque, τ”, so that “τ = I α”.

F Newton’s Law along tangential direction θ F T m FTT= ma= m r α r Multiplying “r”, so that we have “I” on right side rp axis 2 rFT == m r α Iα

So, let’s define torque as τ ≡ rFT Then we got τ = Iα

Since FFT = sinθ and rrp = sinθ

τ ==rFTp rFsinθ = r F (could be positive or negative)

11 F τ ==rFTp rFsinθ = r F θ F T m If r = 0, torque is zero. r

rp axis If theta = 0 or 180 degree, the torque is zero.

For multiple forces

τ net =+++τττ123...

τ net = Iα

m1 m2

m1=100 kg adult, m2=10 kg baby. Distance to fulcrum point is 1 m and 11 m respectively. The seesaw st art s at h ori zont al positi on f rom rest. Which direction will it rotates? (a) Counter-Clockwise (b) Clockwise (c) No rotation (d) Not enough information Example: Find the net torque, moment of inertia, and initial angular acceleration. Choose axis of rotation through fulcrum point.

12 Example: second law for rotation

PP10606-49*: When she is launched from a springboard, a diver's angular speed about her center of mass changes from zero to 6.20 rad/s in 220 ms. Her rotational inertia about her center of mass is constant at 12.0 kg·m2. During the launch, what are the magnitudes of (a) her average angular acceleration and (b) the average external torque on her from the board? G G τnet = Itot α

5 N tangential force is applied at the edge of a uniform disk of radius 2 m and mass of 8 kg. Find angular acceleration. F=5 N

Axis of rotation

1 Formula: I = MR2 2

13 5 N tangential force is applied at 1 m from the center of a uniform disk of radius 2 m and mass of 8 kg. Find angular acceleration.

FF5=5N N

Axis of rotation

1 Formula: I = MR2 2

5 N force is applied at 1 m from the center of a uniform disk of radius 2 m and mass of 8 kg. Find torque. 45 degree F=5 N

Axis of rotation

14 Torque on extended object by gravitational force Æ Assume that the total gravitational force effectively acts at the center of mass.

Gravitational potential energy of extended object Æ M g H, where H is the height of the center of mass and M is the total mass.

iClicker Q

Axis of rotation

HiHorizonta l unif orm rod of length L & mass M Find the torque by gravitational force.

A. LMg B. (L/2)Mg C. 2LMg D. (3/2)LMg E. None of the above 1 Find the angular acceleration. I = ML2 end, rod 3

15 Example of energy conservation

Axis of rotation

Horizontal uniform rod of length L & mass M is released from rest.

Find its angular speed at the lowest point, assuming no friction between axis of rotation and the rod.

Example A thin uniform rod (length = 1.2 m, mass = 2.0 kg) is pivoted about a horizontal, frictionless pin through one end of the rod. (The moment of inertia of the rod about this axis is ML2/3.) The rod is released when it makes an angle of 37° with the horizontal. What is the angular acceleration of the rod at the instant it is released?

16 Example: Torque and Angular Acceleration of a Wheel

•Cord wrapped around disk, hanging weight • Cord does not slip or stretch •Let m = 1.2 kg, M = 2.5 kg, r =0.2 m • Find acceleration of mass m, r find angular acceleration α for disk, tension, and torque on the a disk 1 Formula: I = Mr2 2 mg

Rolling : motion with translation and rotation about center of mass ω KKKtotal=+ rot cm v 1 cm KI= ω 2 rot2 cm 1 KMv= 2 cm2 cm For rolling without slipping UMgravity= gh cm ΔsR=Δθ

vRcm = ω Emech=+KU tot

Δ=EWmech non− conservative

17 Example: Use energy conservation to find the speed of the bowling ball as it rolls w/o slipping to the bottom of the ramp Given: h=2m ƒ Formula: For a solid sphere I = 2 MR2 cm 5 Hint:

ƒ Rotation accelerates if there is friction between the sphere and the ramp ‰ Friction force produces the net torque and angular acceleration. ‰ There is no mechanical energy change because the contact point is always at rest relative to the surface, so no work is done against friction

iClicker Q: A solid sphere and a spherical shell of the same radius r and same mass M roll to the bottom of a ramp without slipping from the same height h.

True or false? : “The two have the same speed at the bottom.”

A) True B) False. Shell is faster. C) False. Solid sphere is faster. D) Not enough information.

I_(cm, spherical shell) = (2/3) MR^2 I_(cm, solid sphere)=(2/5) MR^2

18