Chapter 8: Torque and Angular Momentum

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Chapter 8 Torque and Angular Momentum

• Concept. Questions 2, 4.
• Problems 5,13,18, 27, 39, 43, 55, 69, 73.
• Rotational Inertia Kinetic Energy
• Torque Angular Acceleration
• Torque Angular Momentum
• (Vector Nature of)

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Rotational Dynamics Newtons 2nd Law for
Rotation
Clockwise (CW) Counter-clockwise (CCW)
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Rotational Inertia Energy
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Central Axis
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Axis on End
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Calculated Rot. Inertias

• rotational inertias of solid objects can be
  calculated
• The calculated values are listed in your textbook
  on p.263
• /

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Ex Rotational Inertia A 0.3kg meter stick is
held horizontally from one end. Its rotational
inertia about one end is
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Torque lever-arm x force

• meter-newton
• ft-lb

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torque

• lever-arm is the shortest distance from axis to
  line of the force
• Torque (giam7-11)

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Ex Zero and Non-Zero Torque

• Zero Torque
• Large Torque

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Ex Torque due to Gravity A 0.3kg meter stick is
held horizontally from one end. The torque due to
gravity about the end is
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Ex Torque due to Gravity A 0.3kg meter stick is
rotated 85 deg from horizontal. The torque due to
gravity about the end is
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Ex Torque due to Gravity A 0.3kg meter stick is
rotated 5 deg from horizontal. The torque due to
gravity about the end is
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Newtons 2nd Law (Rotation)
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Ex Angular Acceleration A 0.3kg meter stick is
held horizontally from one end. Its angular
acceleration when released is
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Ex A merry-go-round has a rotational inertia of
100kgm2 and a radius of 1.0 meter. A force of
250 N is applied tangentially at its edge. The
angular acceleration is
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Equilibrium Problems

• Equilibrium is state when Net force 0 Net
  torque 0
• You can choose the axis anywhere, so we choose it
  where an unknown force acts.
• 1st Step Storque-ccw Storque-cw
• 2nd Step Sforce-up Sforce-down
• /

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Ex Torque due to Gravity A meter stick is in
horizontal equilibrium with mass m at 4cm,
65.5cm, 80cm, and fulcrum at 50cm. Check for
plausibility.
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Ex The drawing shows a person whose weight is
584N. Calculate the net force with which the
floor pushes on each end of his body.
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Rotational Kinetic Energy

• Rotational K ½(I)w2.
• Example Constant Power Source has 100kg, 20cm
  radius, solid disk rotating at 7000 rad/s.
• I ½MR2 ½(100kg)(0.2m)2 2kgm2.
• Rot K ½ (2kgm2)(7000/s)2 49 MJ

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Rotational Work-Energy Theorem

• (Work)rot tDq.
• Example torque of 50 mN is applied for one
  revolution.
• rotational work (50mN)(2prad) 314 J
• (Rotational Work)net DKrot.
• /

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Angular Momentum (L)

• analog of translational momentum
• L Iw kgm2/s
• Example Disk R 1m, M 1kg, w 10/s
• I ½MR2 ½(1)(1)2 0.5 kgm2
• L Iw (0.5kgm2)(10/s) 5kgm2/s

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Conservation of Angular Momentum

• For an isolated system
• (Iw)before (Iw)after
• Example Stationary disk M,R is dropped on
  rotating disk M, R, wi.
• (Iw)before (Iw)after
• (½MR2)(wi) (½MR2 ½MR2)(wf)
• (wf) ½ (wi)

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8 Summary

• Mass ? Rotational Inertia
• Force ? Torque
• Rotational KE
• Angular Acceleration
• Work and Energy
• Angular Momentum

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Concept Review

• Torque rotational action
• Rotational Inertia resistance to change in
  rotational motion.
• Torque force x lever-arm

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Mass-Distribution
Larger radius
Larger Speed
Larger Effort
?
?
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Torque (t) mN
F
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Rotational Inertia ( I )
kg(m)2
Example
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Problem 33

• Pivot at left joint, Fj ?, but torque 0.
• ccw (Fm)sin15(18) mg(26) cw
• ccw (Fm)sin15(18) (3)g(26) cw
• (Fm) (3)g(26)/sin15(18) 160N
• Note any point of arm can be considered the
  pivot (since arm is at rest)

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If ball rolls w/o slipping at 4.0m/s, how large
is the height h in the drawing?
rolling w/o slipping
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• Left force mg 30g, Right 25g
• mg 30g 25g m 55kg
• ccw mg(xcg) cw 30g(1.6)
• (55)g(xcg) 30g(1.6)
• (55)(xcg) 30(1.6)
• Xcg (30/55)(1.6)

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60, z-axis

• Each mass has r2 1.52 2.52.
• I sum mr2 (2314)(1.52 2.52)

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65

• First with no frictional torque, then with
  frictional torque as specified in problem.
• M 0.2kg, R 0.15m, m1 0.4, m2 0.8

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83

• Pulley M, R. what torque causes it to reach ang.
  Speed. 25/s in 3rev?
• Alpha use v-squared analog eqn.
• Torque Ia (½MR2)(a)

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89, uniform sphere part

• Rolling at v 5m/s, M 2kg, R 0.1m
• K-total ½mv2 ½Iw2.
• ½(2)(5x5) ½(2/5)(2)(0.1x0.1)(5/0.1)2.
• 25 10 35J
• Roll w/o slipping, no heat created, mech energy
  is conserved, goes all to Mgh.
• 35 Mgh h 35/Mg 35(19.6) 1.79m

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111

• Ice skater, approximate isolated system
• Therefore
• (Iw)before (Iw)after
• (100)(wi) (92.5)(wf)
• (wf) (100/92.5)(wi)
• K-rot increases by this factor squared times new
  rot. Inertia x ½.

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Example Thin rod formulas.
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Angular Momentum

• Symbol L Unit kgm2/s
• L mvr m(rw)r mr2w Iw.
• v is perpendicular to axis
• r is perpendicular distance from axis to line
  containing v.

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Angular Momentum

• Symbol L Unit kgm2/s
• L mvr m(rw)r mr2w Iw.
• v is perpendicular to axis
• r is perpendicular distance from axis to line
  containing v.

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13) Consider a bus designed to obtain its motive
power from a large rotating flywheel (1400. kg of
diameter 1.5 m) that is periodically brought up
to its maximum speed of 3600. rpm by an electric
motor at the terminal. If the bus requires an
average power of 12. kilowatts, how long will it
operate between recharges? Answer 39.
minutes Diff 2 Var 1 Page Ref Sec. 8.4
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6) A 82.0 kg painter stands on a long horizontal
board 1.55 m from one end. The 15.5 kg board is
5.50 m long. The board is supported at each
end. (a) What is the total force provided by both
supports? (b) With what force does the support,
closest to the painter, push upward?
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28) A 4.0 kg mass is hung from a string which is
wrapped around a cylindrical pulley (a
cylindrical shell). If the mass accelerates
downward at 4.90 m/s2, what is the mass of the
pulley? A) 10.0 kg B) 4.0 kg C) 8.0 kg D) 2.0
kg E) 6.0 kg
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19) A solid disk with diameter 2.00 meters and
mass 4.0 kg freely rotates about a vertical axis
at 36. rpm. A 0.50 kg hunk of bubblegum is
dropped onto the disk and sticks to the disk at a
distance d 80. cm from the axis of
rotation. (a) What was the moment of inertia
before the gum fell? (b) What was the moment of
inertia after the gum stuck? (c) What is the
angular velocity after the gum fell onto the disk?
(a) 2.0 kg-m2 (b) 2.3 kg-m2 (c) 31. rpm
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1. A pair of forces with equal magnitudes and
opposite directions is acts as shown. Calculate
the torque on the wrench.
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3. The drawing shows the top view of two doors.
The doors are uniform and identical. The mass of
each door is M and width as shown below is L. How
do their rotational accelerations compare?
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A Ring, a Solid-Disk, and a Solid-Sphere are
released from rest from the top of an incline.
Each has the same mass and radius. Which will
reach the bottom first?
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5. The device shown below is spinning with
rotational rate wi when the movable rods are out.
Each moveable rod has length L and mass M. The
central rod is length 2L and mass 2M.
Calculate the factor by which the angular
velocity is increased by pulling up the arms as
shown.
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Rotational Review
(angles in radians)
4 kinematic equations
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Angular Momentum Calculation
L Iw
Example Solid Disk M 2kg R 25cm Spins
about its center-of-mass at 35 rev/s
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4. A one-meter-stick has a mass of 480grams. a)
Calculate its rotational inertia about an axis
perpendicular to the stick and through one of its
ends. b) Calculate its rotational inertia about
an axis perpendicular to the stick and through
its center-of-mass. c) Calculate its angular
momentum if spinning on axis (b) at a rate of
57rad/s.
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Conservation of Angular Momentum

• Example 50 grams of putty shot at 3m/s at end of
  200 gram thin 80cm long rod free to rotate about
  its center.
• Li mvr (0.050kg)(3m/s)(0.4m)
• Lf Iw (1/12)(0.200kg)(0.8m)2
  (0.050kg)(0.4m)2(w)
• final rotational speed of rodputty

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