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Intro Physics I (Phy.211) 2010 Fall

Unit 3 ; Topic Seven - Torque & Rotational Dynamics


Plan for Unit 3 :
-Topic 7:
      Torque by Force & Angular Kinematics : Ch.12 ; Ch.14 § 1,2
      . . . (HW due Tue.Oct.26)
      Torque & Angular Momentum change : Ch.13
      . . . (HW due Thr.Oct.28)
    Quiz Mon.Nov.01

-Topic 8:
      Oscillations : Ch.15 § 1,2,3
      . . . (HW due Tue.Nov.02)
      coupled Osc's & Pendulums : Ch.15 § 5 , 4
      . . . (2-pt HW due Wed.Nov.04)
    Quiz Fri.Nov.05

-Topic 9:
      Wave Propagation : Ch.16 § 1,2 ; Ch.17 § 1,2,3
      . . . (HW due Tue.Nov.09)
      Wave Interference : Ch.16 § 3,4 ; Ch.17 § 4,5
      . . . (HW due Thr.Nov.11)
  - plan for Exam 3 Mon.Nov.15 (1st hour)


Home-Work PRACTICE QUESTIONS for Topic 7 ...( NOT for grading ...):
      (numbers refer to Ohanian Physics for Engineers,3rd ed.)

Angular Kinematics : ch.12 prob. 1,3,5,7,9,11,15,13 ; 17,22,25 ; 27,29 ; 33,34,35,37,39,43,51,55,56,59,65,71
Torque Statics : ch.14 prob. 9,11,15,18,20,22,23,24,25,49,51,57,59,61,63,64 ; 33,37,39,45,46,47
L = r×p : (ch.9 prob. 38,39,41,43,44,67,69,72,73) ; Ch.13 prob. 47,49,51(L²/2mr²?) , 55,57,59,60 ; 61,62,63,64,66,67,69,73,74,75,76
Rotation Dynamics : ch.13 prob. 3,7,8,9,13,17,21,27,29,31,35,38,39,45 ; 79,81,102


Homework 12 (from Mon.Oct.25 - Wed.Oct.27 ... classes 36 - 37) - due Fri.Oct.29
  1. A 60[kg] gymnast (1.5[m]-long rod) doing a back-flip can change from ω=0 to ω=6[rad/s] during the Δt=0.3[s] jump.
          (a) calculate the angular acceleration, the angular Inertia around rod center, and Torque by the mat to cause this.
          (b) If the jump gives 2.5[m/s] upward velocity, thru what angle do they rotate before landing (c.o.m. to same h)?
        If the gymnast "tucks" during the flip, their length can effectively become ½ as long as when standing
          (c) what would their new Inertia be when tucked, and what angular speed would they have then?
          (d) to perform one (1) rotation, what portion of their ½ s should they be in "tuck" position?
  2. Suppose all 100 Million cars in the US (1500[kg] each) sped 25[m/s] Eastward at latitude 40° ... Earth's I = 1/3 MR²
          (a) calculate these cars' angular momentum, and compare it to Earth's angular momentum (around its spin axis).
          (b) Will the rest of Earth spin faster or slower because of this? How much faster or slower?
          (c) If all these cars drove 25[m/s] Southward, Earth's spin axis would tilt away (Northward) ... by what angle?
          (d) Angular Inertia Tables show Solid sphere I= 0.4 MR² , but Earth has I = .33 MR² ... comment! isn't Earth solid?
  3. A small merry-go-round (R =2[m], I =40[kg·m²]) sits at the origin on good bearings; a 40[kg] 6th grader runs 8[m/s] (x) along y=2[m] , from x=−2[m] to x=0.
          (a) use the right-hand rule to explicitly calculate the kid's angular momentum when he is at x=−2[m],y=2[m] .
          (b) use the right-hand rule to explicitly calculate the kid's angular momentum when he is at x=0,y=2[m] .
          (c) If the kid jumps onto the merry-go-round (and grabs a hold!), what is the angular velocity of kid-and-disk afterward?
          (d) calculate the change in KE for the system ... what Force did that Work? (frictionless bearings)
  4. It was suggested (70's Arab oil embargo) that 500[kg] commuter cars could be "powered" by 80[kg] disk flywheel, ½[m] diameter, spinning 1000 [rev/sec] .
          (a) calculate the angular Inertia of this flywheel, and its angular momentum, and its angular KE .
          (b) suppose the disk was vertical, with L pointed forward (along the car's velocity) : sketch a top view of the disk, with velocity of its top, and its L .
          . . . the driver wants to turn left 90° (=π/2 rad) ... suppose each front tire produces 1000[m×N] Torque (=1000[N] leftward Force at 1[m] from c.o.m.) for 1 second
          (c) what is the magnitude of the angular momentum change that they cause? What direction is the new L ? (angle!)
          (d) suppose the disk was horizontal, with L pointed upward ... and the level road became 7° downhill
          . . . draw a side view of this; if the car's c.o.m. was 1.5[m] from the rear wheels, how long before the L changes direction by 7°? Which way is the car tilting?

Homework 11 (from Fri.Oct.22 - Mon.Oct.25 ... classes 34 - 35) - due Tue.Oct.26

practice writing these angular symbols on the homework problems, to reinforce the linear-angular analogies.

  1. A 1× audio CD spins 210 revolutions per minute when playing the outermost track, at radius 58[mm].
          (a) write its angular speed ω in [rad/s] ... what is the linear speed of that track as it passes the reading laser?
          (b) find the angular speed, and spin frequency, that makes a 23[mm]-radius track have the same linear speed.
       If you don't skip any songs, it takes about 70 minutes to play from first track (inside) to last (outside)
          (c) what's the average angular acceleration of the CD during its play? {watch ± signs!}
          (d) estimate the average angular speed of the CD during its play ... through what angle [rad!] did it spin?
  2. A motor turns a pulley with radius 22[mm] at 1725 r.p.m. ; it is connected by a belt to a pump's 90[mm] radius pulley. Tension in the belt's bottom section is 10[N] , but Tension is 8[N] in belt's top part (see Ch.14.prob.47).
          (a) what is the frequency (or angular speed - specify by symbol!) of the pump axle ([r.p.m.] is okay)
          (b) what is the torque caused by the drive belt on the motor pulley?
          (c) what torque is caused by the drive belt on the pump pulley?
          (d) calculate the Work per time (i.e, Rotational Power) transfered from motor to drive belt
          (e) calculate the Power transferred from drive belt to the pump pulley ... comment!
  3. A 55-gallon drum weighing 300 lbs is held on a 30° ramp by a cable pullling parallel the ramp.
          (a) which of these torque axis locations can NOT show that friction ≠ 0 :
          {drum center, ramp contact, drum bottom (below center), cable contact}
          (b) Sum Torques around the ramp contact to determine the Tension [lbs].
          (c) Sum Torques around the drum center to determine the friction Force [lbs].
          (d) If the Tension is increased by 2 lbs, does the friction Force also increase?
  4. exercise pose: from the shoulder, the humerus bone extends mostly forward but downward 30° to the elbow;
       biceps attach 55[mm] along the fore-arm from the elbow , and pull approximately parallel to humerus.
       Consider the forearm at 20° above horizontal , holding a 20[kg] "weight" at distance 350[mm] from the elbow.
          (a) What Force do the biceps apply to the radius bone? {vector components!}
          (b) What Force does the humerus bone apply to the radius bone (at elbow)?
          (c) If the biceps Force is increased by 1% , what angular acceleration of the arm-and-weight would ensue?
          . . . treat the fore-arm as a 3[kg] rod that is 350[mm] long, pivoted at the elbow.
          (d) at that α , how long would it take to change that 20° above horizontal to 77° above horizontal?
             

Topic 1 Reminders :

(to topic 1 summary)

USE UNITS all thru your answer ... don't just append them onto the final number !

Put symbols on a sketch, as you read . . . use meaningful phrases : "v_match" , "catch" , "ground" "turn-around"
. . . show process spans as brackets or vector arrows , with words like   "average" , "losing" , "gaining" , "t=0→2" ...

Write statements as symbols first ... each statement has a SUBJECT ... manipulate symbols before plugging numbers.
. . . keep track of adjectives such as   "initial" , "average" , "final" , "stopped" . . . "change"

Topic 2 Reminders :

(to topic 2 summary)

subscripts for "total" or "system" , "part A" ...

keep track of the SOURCE for the external Forces applied to the (passive) object

Recognize whether you are predicting based on theory . . . reasoning from causes to their effects ... (what should it do?)
      or whether you are deducing from observation . . . (what does this imply about it?)

Get un-stuck (often) by wondering "why isn't it the same as it used to be?"

Topic 3 Reminders :

it's the Sum of external Forces that cause an object's mass to accelerate subscripts for "total" or "system" , "part A" ...

once you separate vectors into components, keep each component separate from the others!
ΣF = ma , for each component separately.

Topic 4 Reminders :

Name each Force by the source of that Force ... its cause ... a function of the environment.

spring Force : Fspring = − k s ... opposite the stretch vector .

Pressure Force : Fspring = P A   generally ... in a fluid , δP = ρ g δh .

Gravity : every mass in the universe contributes to the gravity field   g   at any place of interest.
... add "nearby" contributions as vectors ... gby M = G M / r²M-to-point (toward M) . . . then use   Fgravity, on m = mg ;       NOTICE : m ≠ M !


Topic 5 Reminders :

Energy is conserved quantity : Σ Ebefore   +   Σ Wby F's not in PE list   =   Σ Eafter .
  KE = ½ m v² = ½ p·v = ½ p²/m   , in any of these equivalent forms .

Each Force either does Work during motion , or has an associated PE function ... not both .
=> Wby F = Δs .
=> − dU / dr = Fr , the component along that direction (r)

  PEgravity,local = m g h , with h measured upward (z-direction) from a "zero-height reference" point
  PEelastic = ½ k s²   . . . stretch s must be measured from its relaxed length
  PEPressure = P·V   ; it is okay to measure from a non-zero reference pressure , but be consistent !
  PEGravitation = − m GM / r .
  There IS NO PE for Friction Force ... must be explicitly included as a "Work by Forces not in the PE list"

Topic 6 Reminders :

Some Energy forms are somewhat complicated to calculate, at a fundamental level, such as Chemical PE or Nuclear PE.
. . . they can be treated as experimentally-measured quantities which are "released" (transformed) during some conversion process (i.e, burning).
      more material holds more Energy ; it is the PE density (PE/Volume) or the specific PE (PE/mass) that is measured.

efficiency is the fraction (or %) of the input Energy that is transformed into the "desired form".
. . . because total Energy is consserved, the sum of efficiencies in all output forms must be 100%.

Power is the rate that Energy is transformed (or transfered) into some other type (or to some other object)
=> P = ΔE / Δt , reported in [Watt] = [J/s] .

Because KE goes as v² = vx² + vy² + vz² , motion along each component adds in an independent manner .
. . . in particular, radial and angular motion components can be treated separately for objects in orbit
=> L = r × p is conserved then , so that   KEangular = L²/2mr²   is a useful formula .


Topic 7 Summary :

Even if the Sum of all Forces (applied to an object) is zero, the object's motion might be changing rotationally.
. . . a thing's spinning motion tends to stay the same <=> analogous to linear motion
. . . angular location, or angular position, angular orientation   φ [radians] = (arc length)/r <=> x, y, or z .
      in the direction that a right-hand screw would move (if right fingers curl along arc's arrow, thumb points in direction of φ)
      . . . counter-clockwise arc has φ out of page . . . clockwise motion has Δφ into page .
. . . angular velocity ω [radian/sec] = Δφ / Δt = v/r is analogous to linear velocity
. . . angular acceleration α [radian/s²] = Δω / Δt = a/r is analogous to linear acceleration .
The analogy to any kinematic formula is obtained by directly replacing each linear quantity by its angular sibling :
      φi + ωavg Δt = φf   ;   Δφ = ωi Δt + ½ αavgt)² ;   ωi + αavg Δt = ωf   ;   ωi² + 2 αavg·Δφ = ωi²   ; etc.


Rotational Dynamics statements are almost a straight-forward analogy to the familiar linear (translational) statements .

. . . the rotational analogy to Force is Torque   τ [m×N] = r × F . . . "r cross F" . . . (direction via right-hand-rule)
      Torque around a rotation axis depends on the distance from the axis (tail of r) to the point the Force is applied
      . . . and only the component of F that is  |  to r provides torque
      . . . or, use the entire Force at its lever-arm distance : the distance from the axis  |  to the Force's line of action
      generally, τ [m N] = r × F = r F sin(φ) ; here, φ is the angle from the r-vector direction to the F-vector direction
=> r × F = τ . . . right-hand rule : right rist along r , fingers flip thru angle φ till point along F ; Thumb points ( | ) along Torque
            is the pattern for any vector cross-product   a × b .
      . . . you should use the right-hand rule to verify that a × b = c   implies that   b × a = − c . . . <= ORDER MATTERS !

. . . if the total Torque is zero, the conserved rotational quantity is angular momentum L [m N s] = r × p . . . (cross product!)
      of course the anguLar momentum is around the same rotation axis that has Στ =0 . . .
=> L = r_|_ p = r p_|_ = r p sin(φ) ... rite rist along r, pingers point to p , thumb shows L .


The "story-board" statement of how Nature behaves is just as valid in rotational form as in linear form :
      "starting anguLar momentum is changed , by total Torque applied for a while , into the anguLar momentum afterward."
=> Σ Li   +   Σ τ Δt   =   Σ Lf
  .


If you try to isolate the more easily-measured kinematic quantities , the analogies are not generally as useful
. . . the analogy to mass is rotational Inertia or "angular Inertia" or "(second) Moment of Inertia" :
=> I [kg m2] = Σ m r² .
      ... so it depends not only on the amount of mass , but also depends on how far from the axis the mass is
      ... because its velocity is   v = − r × ω , more distant mass has more momentum

if the mass is spread out in some Volume, then the sum must become an integral of the mass density ρ
=> I = ∫ r² dm = ∫ r² ρ dV . . . over all the Volume that has non-zero mass density .
. . . the mass density is allowed to be different at different places : ρ(r) as a function of location .
. . . the differential Volume is   dV = dx dy dz   , in "rectangular" coordinates
      in "spherical polar coordinates" dV = r cos θ dφ r dθ dr . . . measure θ from spin axis z , measure φ around the spin axis
      if the mass density has rotational symmetry (same ρ at any φ) , then ∫ = 2π . . . dV = 2πr cos θ r dθ dr .
      if the mass is all in the x-y plane , θ = 0 so cos θ = 1 , and ∫ r dθ = t , the thickness of the sheet . . . dV = 2πr t dr .
      ... in that situation, ρ t = σ , called the surface density , is often given ... σ = m/Area [kg/m²]

. . . in any case, once the Inertia is known, the anguLar momentum is that Inertia I multiplied by the angular velocity vector ω :
=> L = I ω


If some mass's distance (from the axis) changes , it's essentially a "soft rotational collision" between the moving masses and the others.
. . . If you know the initial and final locations, you may compute the initial rotational Inertia and the different final Rotational Inertia
      Ii ωi + Σ τ = If ωf

If the object does NOT change shape, and the rotation axis does not change,
      then the moment of Inertia (around that axis) does NOT change ;
. . . then any change in angular momentum comes from change in angular velocity ω
. . . so the total external torque causes the object's rotational Inertia to have angular acceleration   α :
=>   ΣτA = I ΔωA / Δt = I αA ... where "A" is the axis , which the Inertia will rotate around ...
      ... a mass has tangential acceleration component   aφ = − r × α   , but radial acceleration component   ar = − rω² .


because the different parts of the object do not have the same speed, there is an additional Kinetic Energy term :
=> KErotation = ½ I ω² calso be written   KErot = ½ L ω   , or   KErot = L²/2I .
. . . this KE of mass around the rotation axis is added to the linear KE as if it was moving at the speed that the axis has .

this rotational KE is changed by rotational Work :
=> Wrot = τ ·Δθ . . . parallel components of Torque and angular displacement .

an implication of this is that an object's moment of inertia is smallest around its center-of-mass,
. . . more rotational inertia the farther the axis is from the c.o.m. ; if axis is distance d from c.o.m. :
=> Iaxis = Ic.o.m. + M d² .


You STILL get to choose which objects in the scene are included INSIDE your system
. . . use "bigger" systems if you intend to use angular momentum during a "rotational collision"
      so that internal Forces do not produce external torques to Sum ... but the Total Inertia inside stays the same.


(to topic 8 summary)



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