Chapter 10 Outline:  Angular

10.1: Angular Position, Velocity, and Acceleration:  radian, average angular speed, instantaneous angular speed, average angular acceleration, instantaneous angular acceleration 10.2: Rotational Kinematics:  Rotational Motion with Constant Angular Acceleration 10.3: Angular and Linear Quantities 10.4: Rotational Kinetic Energy:  moment of inertia, rotational kinetic energy

10.5: Calculation of Moments of Inertia:  moment of inertia of a rigid object, parallel-axis theorem 10.6: Torque:  moment arm 10.7: Relationship Between Torque and Angular Acceleration 10.8: Work, Power, and Energy in Rotational Motion Rolling Motion of a Rigid Object 10.9: Rolling Motion of a Rigid Object

10.1: Angular Position, Velocity, and Acceleration

sin (½dθ) = ½dx / r sin (½dθ) = ½ds / r ½dθ = ½ds / r dθ = ds / r s = rθ

As dθ approaches zero ½ds approaches ½dx which results in If θ is in radians, sin(θ) = θ if less than 10° (10°π / 180°)

vave = Δx /  Δt ωave = Δθ / Δt v = dx / dt ω = dθ / dt

a = Δv / Δt α = Δω / Δt a = dv / dt α = dω / dt

Example If the Merry-Go-Round accelerates for 5 seconds at a rate of 0.1 rad/s^2, what is the final angular velocity of the Unicorn. The Unicorn is at the outside edge 6 meters from the center? At this time what is the angular velocity of the Satyr on the inside edge which is 3 meters from the center?

α = Δω / Δt 0.1 = Δω / 5 Δω = ½ rad / sec (initial is zero…so the change is equal to the final) Same answer for both…angular velocity doesn’t depend on radius.

10.2: Rotational Kinematics

d = ½at^2  + v0t  + d0 θ = ½αt^2  + ω0t  + θ0

Example If the angular position is described by the following equation, θ = 6t^2 - 5t - 20, what is the angular acceleration at 4 seconds?

ω = dθ / dt α = dω / dt       α = d^2θ / dt^2 α = d^2(6t^2 - 5t – 20) / dt^2 α = 12 radians/s^2

Example If a frictionless pulley has an inertia of 1 kgm^2 (which can be modeled as a solid disk), where m1 = 8 kg, m2 = 12 kg, and R = 40 cm; what is the acceleration of the system?

τ                      =      I      α FNet r               =      I      α [m2(g-a) – m1(g+a)] r =      I  (a / r) a = (m2 - m1)g / (I/r^2 + m2 + m1) a = 1.53 m/s^2

Key Point This is a frictionless pulley, but NOT massless, so the tension on each side is NOT equal.

10.3: Angular and Linear Quantities

v =      dx     / dt s = r θ           dx = r dθ v = r  (dθ)   / dt ω = dθ / dt dθ = ω dt v = r (ω dt) / dt v = ω r

at = dv   / dt at = r dω/dt at = r    α at = α r at à tangential

ac =   v^2   / r ac = (ωr)^2/ r ac = ω^2r ac à centripetal

a^2 = at^2  +     ac^2 a^2 = (αr)^2 + (ω²r)^2 a = r √(α^2 + ω^4)

Example If the Merry-Go-Round accelerates for 5 seconds at a rate of 0.1 rad/s^2, what is the final (tangential) velocity of the Unicorn. The Unicorn is at the outside edge 6 meters from the center? At this time, what is the (tangential) velocity of the Satyr on the inside edge which is 3 meters from the center?

α = Δω / Δt 0.1 = Δω / 5 Δω = ½ rad / sec

v = ω r v = ½ rad/sec (6 meters) v = 3.0 m/s v = ω r v = ½ rad/sec (3 meters) v = 1.5 m/s

Demo:  Rotational Inertia:  ME-Q-RI

10.4: Rotational Kinetic Energy

Ki = ½ mivi²

All objects can be broken up into smaller components, thus rotational kinetic energy is the sum of the kinetic energies of all these smaller components, or

KR = ½Σ mivi² KR = ½Σ miri²ωi² KR = ½I ω²

where v = ωr so At this point we define a new term, Inertia. I = ½Σ miri²

K =   Kt    +    Kr           t à translational r à rotational K = ½mv² + ½Iω²

Demo:  Rotational Inertia Rods:  ME-Q-RR

10.5: Calculation of Moments of Inertia

Parallel Axis Theorem The moment of inertia about any axis parallel to and a distance D away from this axis is               I = ICM + MD^2

Figure (a) shows a disk that can rotate about an axis at a radial distance h from the center of the disk. Figure (b) gives the rotational inertia I of the disk about the axis as a function of that distance h, from the center out to the edge of the disk. The scale on the I axis is set by IA = 0.150 kg·m^2 and IB = 0.450 kg·m^2. What is the mass of the disk?

Calculate I at h = 0.1 m (or 0.2 m) (IB - IA = .3)  Ih=0.1 = 0.15 + 2.5div (0.3kgm^2 / 10div)    Ih=0.1 = 0.225 kgm^2

Then apply  Parallel Axis Theorem with the following given

Then write two Parallel Axis Theorem Equations.

ICM  = Ih=0  = 0.15 kgm^2

ICM  + md^2 0.15 + md^2

h = 0.1 ½ mr^2 + m h^2 ½ mr^2 + m 0.1^2 0.15    +  0.225 ½ mr^2 + m 0.1^2 = 0.375

h = 0.2 ½ mr^2 + m h^2 ½ mr^2 + m 0.2^2 0.15    +  0.45 ½ mr^2 + m 0.2^2 = 0.6

Two Unknowns…two equations. To solve…subtract the two equations ½ mr^2 + m 0.2^2 = 0.6 ½ mr^2 + m 0.1^2 = 0.375 0    + m (0.03) = 0.225 solve for m m = 7.5 kg

Another example this time using the basic definition à r^2dm Example What's the moment of inertia through the central axis of a symmetrical homogenous cylinder?

I = ∫ r^2dm I = ∫ r^2 (2πρL r dr) I = 2πρ L ∫0 to R r^3dr I =   ½πρL R^4 I = ½ MR^2

I = (as Δmi approaches 0) I = Σ ri^2Δmi I = ∫r^2dm dm = ρ dV         ρ = m / V & by substituting ρ = M/V where V = πR^2L back into the equation we get

What role do center of mass of an object and inertia play? Below are three different pendulums

The one to the right is obvious, CM = L/2 (the middle of the meter stick), thus, h is zero to L/2 PEmax or Ug = mg(L/2)     and     KR = ½Iω² How do we calculate Inertia (r^2dm)? Linear density is λ = m / L (but L is the radius, so λ = m / r)

I = r^2  dm                 dm = λ dr I = r^2 λ dr I = λ ∫r^2dr I = λ r^3/3         from 0 to L I = λ L^3/3         λ = m / L I = (m/L) L^3/3 I = m L^2/3

Ug = mg(L/2)      KR = ½Iω^2 mg(L/2)            = ½Iω^2 mg(L/2)            = ½(m L^2/3)ω^2 ω = √(3g/L)

And if we want the linear speed at the bottom or the pendulum v = ωr       (r = L) v = ωL v = L √(3g/L) v = √(3gL^2/L) v = √(3gL)

Once again CM is very simple (it's the center of the ball since the rod is massless), and the Inertia doesn't even have to be calculated PEmax or Ug = mg(L)     and     KR = ½Iω^2 mgL = ½Iω^2 mgL = ½(m L^2)ω^2 ω = (2g/L)^½

This time we simply add together the Inertia of each portion of the rod-ball pendulum.

I = Irod + Iball I = m L^2/3 + m L^2 I = 4m L^2/3

Ug = mg(L)     and     KR = ½Iω^2 mgL      = ½       I      ω^2     mgL      = ½(4m L^2/3)ω^2 ω         = (3g/2L)^½

As you can see the center of mass, thus the inertia, for all of three of these changes.

Table 10.2

Demo:  Rolling Objects:  ME-Q-RO

10.6: Torque

Force times the lever arm (distance) is Torque, t t = Fd Where d = sinφ r        d is the perpendicular distance from the        pivot point to the force vector t = Fd where d = sinφ r t = r F sinφ Our Lab class…shows Torque is t = F^d, which is a little easier for beginning students to understand. t    =       F^               d t = sin60°(100N)    (0.1m) t = 8.66 Nm Proper METHOD t    =   F             d^ t = 100 N    sin60°(0.1 m) t = 8.66 Nm Doesn’t look too different…right?

10.7: Relationship Between Torque and Angular Acceleration

F = m a t = I a

What is the angular acceleration of a 100 gram meter stick (a hole is drilled at the 0 cm and the stick pivots about the 0 cm position) if a force of 0.2 Newtons is applied to the 100 cm end of the stick?

inertia of a solid rod is 1/3 mr^2           t       =        I        a             F    d  =   m  r^2/3    a         0.2N*1m = 0.1kg(1)^2/3 a a = 6 rad/s^2

10.8: Work, Power, and Energy in Rotational Motion Rolling Motion of a Rigid Object

dW = F      ds dW = F sinφ r dθ dW =     t      dθ           So the rate at which work is being done (just like the clock in our room where second hand requires work to be done all the time to keep it moving)

dW      = t   dθ       dW/dt = t dθ/dt             P = dW / dt Power  = t   dω                        dω = dθ/dt

Σ t    = I      α Σ t    = I dω / dt Σ t    = I dω/dθ (dθ/dt)  (apply chain rule) Σtdθ = I    dω    (   ω  )   dW = Σ t dθ   dW   = I ∫ ω dω Σ W   =    ΔKR Σ W   = ½I ωf² - ½I ωi²

10.9: Rolling Motion of a Rigid Object

vCM = ds / dt vCM =  R dθ / dt vCM =  R  ω

aCM = dvCM / dt aCM = R dω / dt aCM = R  α

Parallel Axis Theorem The moment of inertia about any axis parallel to and a distance D away from this axis is I = ICM + MD²  (see above 10.5)

Applying the parallel axis theorem

the Total Kinetic Energy (K = ½Ipω²) can now be expressed as     K = ½ICMω² + ½MR²ω²     K = ½ICMω² + ½M vCM²

Demo:  Double Cone and Plane:  ME-J-DC Actually belongs in Ch 9…but the center of mass is actually lowering…but it looks like the cone is going up…so could also be included in conservation of energy…so a good review for Total Kinetic Energy.