Chapter 8 Rotational Equilibrium and Dynamics

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Chapter 8 Rotational Equilibrium and Dynamics
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8-1 Torque
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Torque (t)

• A quantity that measures the ability of a force
  to rotate an object around some axis.
• t ? Greek letter tau

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Net Torque produces rotation.

• Just like a net force causes motion/acceleration.

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Torque depends upon a force and the length of the
lever arm.
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• How easily an object rotates depends not only on
  how much force is applied, but also where the
  force is applied.

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Lever Arm

• The perpendicular distance from the axis of
  rotation to a line drawn along the direction of
  the force.

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Torque also depends upon the angle between a
force and the lever arm.
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l
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t Fd(sinq)

• Metric Units of torque are Newton-meters (N-m)

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Torque, like displacement and force, is a vector
quantity.

• Since torque is a rotational motion it has two
  directions

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1. Counterclockwise (CCW), which is positive.
2. Clockwise (CW), which is negative.

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Ex 1 Sean has a flat and is changing his tire,
to remove the lug nuts he applies a force of 25 N
on the tire iron at an angle of 87o. What is the
torque produced if the tire iron is 0.6 m long?
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• G d 0.6 m, F 25 N, q 87o
• U t ?
• E t Fd(sinq)
• S t (25 N)(0.6 m)(sin 87o)
• S t 14.98 N-m

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• Ex 2 What angle produces a torque of 400 N-m, if
  the force applied is 505 N at a distance from the
  axis of rotation of 0.82 m?

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• G t 400 N-m, F 505 N, d 0.82 m
• U q ?

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• G t 400 N-m, F 505 N, d 0.82 m
• U q ?
• E

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• G t 400 N-m, F 505 N, d 0.82 m
• U q ?
• E

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8-2 Rotation and Inertia
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Point mass

• Where all the mass is assumed to be located in
  one point.

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Center of Mass

• The point at which all the mass of a body can be
  considered to be concentrated when analyzing
  translational motion.

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Rotational and translational motion can be
combined.

• We use the the center of mass, as a reference, to
  analyze its translation motion.

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Center of Gravity

• The position at which the gravitational force
  acts on an extended object as if it were a point
  mass.

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Toppling

• If the Center of Gravity (CG) is above the
  support area, then the object will remain
  upright.
• If the CG extends outside the support area, the
  object will topple.

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Unstable Equilibrium

• Is when any movement/ displacement of a balanced
  object lowers the CG.

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Stable Equilibrium

• Is when any motion/displacement of a balanced
  object raises its CG.

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Neutral Equilibrium

• Is when any motion/displacement of a balanced
  object neither raises nor lowers its CG.

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Moment of Inertia (MOI)

• The tendency of a body to rotate freely about a
  fixed axis to resist a change in rotational
  motion.

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MOI is the rotational analog of mass.

• Very similar to mass, but MOI is not an intrinsic
  property of an object.

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It depends upon the objects mass and the
distribution of mass around the axis of rotation.
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• The farther the mass is, on average from the axis
  of rotation, the greater the the objects MOI and
  the more difficult it is to rotate the object.

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Calculating the MOI

• Pg 285 Table 8-1 has equations/formulas for a
  few common shapes.
• M mass in kilograms
• R radius in meters
• l length in meters

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Units for MOI are kg-m2
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• For an object to be in complete equilibrium,
  requires zero net force and zero net torque.

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If the net force is zero the object is in
translational equilibrium. (1st Condition of
Equilibrium)
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If the net torque is zero, than it is in
rotational equilibrium. (2nd Condition for
Equilibrium)
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8-3 Rotational Dynamics
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Newtons 2nd Law for Rotation

• F ma
• We know in a rotational system, torque is a
  function of force, also

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• 1. MOI replaces the mass.
• 2. Acceleration is replaced with angular
  acceleration.

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Net torque MOI x Angular acel.
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Remember that if the net torque is zero, that the
object can still be rotating, just at a constant
velocity.
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• Ex 4 Simon decides to ride the Gravitron, it
  has a radius of 3 m, if his mass is 79.4 kg. What
  is the net torque produced when his angular
  acceleration is 7 rads/sec2?

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• G R 3 m, M 79.4 kg, a 7 rads/sec2
• U tnet ?
• E tnet Ia
• We are going to assume he is a point mass.

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MOI, I MR2

• Stnet MR2a
• S tnet (79.4)(3)2(7)
• S tnet 5002.2 N-m

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• Ex 5 In, 1995, a fully functional pencil with a
  mass of 24 kg and a length of 2.74 m was made.
  Suppose this pencil is suspended at its midpoint
  and a force of 1.8 N is applied perpendicular to
  its end, causing it to rotate. What is the
  angular acceleration of the pencil?

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• G M 24kg, F 1.8 N, l 2.74 m
• U a ?
• E t Ia ? a t / I
• We need to find the values for Torque and MOI.

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• Since the midpoint is the pivot point the lever
  arm is ½ l. Also, the force acts perpendicular,
  so the angle is 90 degrees.
• t Fd(sinq)
• t ½Fl (sinq)

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• t ½(1.8)(2.74)sin90
• t 2.466 N-m

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• We use the formula for MOI on pg 285 for a thin
  rod rotating around its midpoint.
• I 1/12 Ml 2
• I 1/12 (24)(2.74)2
• I 15.02 kg-m2

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• S a 2.466 /15.02
• S a 0.164 rads/s2

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Angular Momentum (L)

• The product of a rotating objects moment of
  inertia and the angular speed about the same axis.

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L Iw

• L angular momentum
• I Moment of Inertia
• w angular speed

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Angular Momentum Units

• Kg-m2/s

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Law of Conservation of Momentum

• When the net external torque acting on an object
  or objects is zero, the angular momentum does not
  change.
• Li Lf

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• Ex 6 A figure skater jumps into the air to
  perform a twisting maneuver. When she first jumps
  her moment of inertia is 86 kg-m2. While shes in
  the air she brings her arms in and decreases her
  momentum to 77 kg-m2. If her initial angular
  speed was 2 rads/sec what is her final angular
  speed, if momentum is conserved?

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• G wi 2 rads/sec, Ii 86 kg-m2 , If 77 kg-m2
• U wf ?
• Since momentum is conserved
• E Li Lf
• Iiwi Ifwf

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• wf Iiwi / If
• S wf (86)(2) / (77)
• S wf 2.23 rads/sec

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• Ex 7 A 65 kg student is spinning on a
  merry-go-round that has a mass of 525 kg and a
  radius of 2 m. She walks from the edge of the
  merry-go-round toward the center. If the angular
  speed is initially 0.2 rads/s, what is the
  angular speed when the student reaches a point
  0.50 m from the center?

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• G M 525 kg, Ri 0.2m, m 65 kg, Rf 0.5 m,
  wi 0.2 rads/s
• U wf ?

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• E Momentum is conserved
• Li Lf
• Lm,i Ls,i Lm,f Ls,f

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• For MOI, treat the merry-go-round as a solid disc
  and the student as a point mass.

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Rotational Kinetic Energy (KERot)

• Energy of an object due to its rotational motion.

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Calculating Rotational Kinetic Energy

• KErot ½ Iw2
• KErot rotational KE
• I moment of inertia
• w angular speed

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• EX 8 What is the MOI for a 0.75 kg top spinning
  at an angular speed of 1.75 rads/sec, if the
  Rotational KE is 12 Joules?

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• G M 0.75 kg, KErot 12 J, w 1.75 rads/sec
• U I ?
• E KErot ½ Iw2
• I 2 KErot /w2

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• S I 2(12)/(1.75) 2
• S I 7.84 kg-m2

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Remember Mechanical Energy may be conserved.

• ME is the sum of all types of KE and PE.
• ME KEtrans KErot PE
• ME ½ mv2 ½ Iw2 mgh

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If ME is conserved

• MEinitial MEfinal

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If ME is conserved

• MEinitial MEfinal

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