Shape Rotational Inertia, I I = M1r + M2r + ... (Point Masses) I = ∫ R Dm

Chs. 10 - Rotation & Angular Motion

October 26, 2015

Shape Rotational Inertia, I

I one aspect of shape is the center of mass I How diﬃcult is it to spin an object about (c.o.m.) any axis?

I another is how mass is distributed about the I Formulas: (distance r from axis) c.o.m. 2 2 I = m1r1 + m2r2 + ... I an important parameter describing shape is called, (point masses) 1. Rotational Inertia (in our book) Z 2. Moment of Inertia (in most other books) I = r2 dm (solid object) 3.2 nd Moment of Mass (in engineering, math books) Solution: 8 kg(2 m)2 + 5 kg(8 m2) + 6 kg(2 m)2 = 96 kg · m2

P 2 Rotational Inertia: I = miri Two you should memorize:

Let’s calculate one of these: I mass M concentrated at radius R

I = MR2

I Disk of mass M, radius R, about center 1 I = MR2 solid disk 2

Parallel Axis Theorem Active Learning Exercise Problem: Rotational Inertia of Point We can calculate the rotational inertia about an Masses axis parallel to the c.o.m. axis Four point masses are located at the corners of a square with sides 2.0 m, 2 I = Icom + m × shift as shown. The 1.0 kg mass is at the origin. What is the rotational inertia of the four-mass system about an

I Where M is the total mass of the body axis passing through the mass at the origin and pointing out of the screen I And shift is the distance between the axes (page)? I REMEMBER: The axes must be parallel! What is so useful about I? Rotation

I radians!

It connects up linear quantities to angular I right-hand rule quantities, e.g., kinetic energy angular speed v ω = rad r s I v = rω 2 2 I ac = rω 1 2 1 2v 1 2 mv → mr → Iω ang. acceleration r2 2 2 2 a rad α = t r s2

Kinematics: Linear → Angular Analogy Dynamics: Linear → Angular Analogy

Linear Translation Rigid Rotation Linear Translation Rigid Rotation

1 2 1 2 ∆x = vo t + a t ∆θ = ω t + α t F~ = m~a τ~ = I α~ 2 o 2 1 ∆v = a t 1 2 2 ∆ω = α t Ktran ≡ m v Krot ≡ I ω 2 2 2 2 v = v + 2 a · ∆x 2 2 x o ω = ωo + 2 α · ∆θ Z 2 Z θ2 ~ d~r dθ W ≡ F · d~r W ≡ τ · dθ ~v = ω = x1 θ dt 1 dt ~p ≡ m ~v ~ d~v d2~r dω d2θ L ≡ I ω~ ~a = = α = = (momentum) (ang. momentum) dt dt2 dt dt2 Solution: -24 rad/s2 .

Rigid Body Acceleration Rigid Body Acceleration

I Ignoring bodies that break apart: vr = 0 at = αr p |a| = a2 + a2 a = ω2r t c c p = r α2 + ω4

I There you have it: The α and ω of physics!

Periodic Motion: three indicators Active Learning Exercise Time Period T the number of seconds that elapse per complete cycle (or trip) Problem: Angular Kinematics around a circle. The angular position of a wheel rotating about a ﬁxed axis is given by Frequency f number of cycles per second,s −1 or Hz. θ = 3.0 rad − (2.0 rad/s3) t3 . Automotive units rpm: revolutions-per-minute What is the wheel’s angular acceleration at t = 2.0 s ? Angular Speed ω radians-per-second requires inserting “2π” here or there: 2π ω = or ω = 2πf T Solution: Solution: 20 m/s2 , 160 turns. 20 rad/s

Active Learning Exercise Active Learning Exercise Problem: Helicopter Problem: Helicopter

ω blade of a helicopter A 4.0 m blade whirs up turns through 40 to 100 rad/s in 20s radians in 4.0 s. from rest. Calculate its Assuming constant

angular acceleration, I tangential accel. what is its angular velocity at the end of I number of turns Starting from rest, the 4.0 s ?

Rotational Dynamics Forces & Torque

Allowed Calculating Torque I produces a torque Forces Given ~r going from pivot to F~ , τ τ~ = ~r × F~ F u friction: ~f = |r||F | sin r I if no slip, v = ωr ∠ ~ = r⊥F u , I force applied at r Gravity: Fg torque causes angular acceleration; its formula Normal: N~ ~ ~ is Spring: Fsp I ﬁngers curl ~r into F τ~ = ~r × F~ Tension: T~ I thumb = τ~ kinetic energy increases as an object spins faster; I depends on choice of pivot! rotational kinetic energy is added to mv2/2 Solution: 32 mN

Active Learning Exercise Review Problem: Torque We now have a lot of ways to solve mechanical problems: A single force of magnitude F = 88 N is applied 1. If problem says “no slip”, what eqn. is as shown to a point on the rim of a wheel with suggested? radius R = 0.72 m. The magnitude of the resulting torque about the center of the wheel is v = ωr closest to 2. If torque is known what can you calculate?

Acceleration! τ = Iα a = αr 3. What torque occurs for rope unwinding a spool?

See Yo-Yo problem! τ = RFT

Kinetic Energy of Motion Rotational Motion A few ways to solve the problem: How much energy in generic, spinning body? 2 kinematics solve eqns. as in Ch.2 for x, v, a but parts: instead for θ, ω, α, remembering where units of radians should be used and 1. A linear part, and know how radius is used to convert back to linear motion quantities 2. a spinny part dynamics use forces to calculate a torque, then 1 1 obtain α from τ = Iα K = m v2 + I ω2 2 com 2 energy use conservation of energy; potential | {z } | {z } energies are calculated the same as linear part spinny part before, but kinetic energy may now be 1 2 both linear 2mv , in addition to 1 2 rotational 2Iω Active Learning Exercise Rotation∗ Problem: Wiley 10.18 Name Deﬁnition Eqns. / Examples Angle orientation of particle relative to 0◦ θ (SN), write down the Angular Displacement change in angle over time interval ∆θ = θf − θi Angular Velocity change in angle over time ω = ∆θ = θf −θi eqns. for determining ∆t tf −ti Instant. Ang. Veloc. angular velocity at one time ω = dθ the speed of the block dt Instant. Ang. Speed angular speed at one time dθ after it has descended dt Avg. Ang. Acceleration change in ω over time α = ∆ω = ωf −ωi 1 m. Use an energy ∆t tf −ti Instant. Ang. Accel. change in ω at one time α = dω principle, namely, dt Kinematics 1-d constant ang. accel. ω = ωo + α · t consider all forms of ω + ωo equations of motion θ = θ + t o 2 1 2 1. Kinetic Energy θ = θo + ωot + 2 αt 2 2 2. Potential Energy ω = ωo + 2 ∆θ · α In Symbolic Notation ∗compare: Chapter 2 Summary!