Torque and Angular Momentum

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Torque and Angular Momentum
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Rotational Mechanics -- Torque

• Turning force to cause rotation
• Torque perpendicular force x lever arm
• Lever arm is length of object from force to axis
• Perpendicular force is that component
  perpendicular to the arm

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Torque

• Quantity measuring how effectively a force causes
  rotation
• Formula- t rF sin f where f is the angle
  between F and r directions
• Maximum force at f 90 degrees, no torque at
• f 0 degrees (r and F same direction)
• Counterclockwise motion is positive A vector
  quantity
• SI unit newton-meter British system
  foot-pound

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Examples 9.3/9.4
The instructions for replacing the head gasket on
an automobile engine say that the bolts should be
torqued down to 90 N m. If you use a wrench that
is 45cm long, how much force must you apply in a
direction perpendicular to the wrench handle to
accomplish this? (Too much torque will pinch the
gasket so that it will not seal properly.) Since
t rF sin f , then F t / r sin f f 90o
sin f 1.000 F (90 Nm) / 0.45 m 200 N
The crank arm of a bicycle pedal is 16.5cm long.
If a 52.0 kg woman put all her weight on one
pedal how much torque is developed (a) when the
crank is horizontal and (b) when the pedal is 15o
from the top? Fmg a. t rF sin f f
90o t (0.165)(52.0)(9.81)(1) 84.2Nm b. f
180-15 165o t (.165)(52.0)(9.81)(sin165)
21.8 Nm
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Static/Dynamic Equilibrium

• For equilibrium, vector sum of the forces on a
  body must 0
• Object in translational equilibrium may still
  rotate
• For body to be in rotational equilibrium, sum of
  torques must be zero
• Static equilibrium object is stationary, does
  not rotate about any axis at all
• Dynamic equilibrium object moving with neither
• translational nor rotational acceleration

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Torque

• Balanced torques happen on a seesaw when two
  children of different weights
• sit at different positions
• Balance used in science works by balanced torques

7
Center of Gravity/Mass

• Point of Rotation of Projectiles
• Ball or bat follows one point within during its
  flight
• Center of gravity is this point

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Center of Gravity/Mass

• Located at objects average position of weight
• May not be at geometric center in an irregularly
  shaped object
• May not be at geometric center if object made of
  two or more materials of different density (Clown
  toy with lead bottom)

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Center of Mass Obeys Inertia

• Spinning wrench follows straight path
• Exploding fireworks - debris follows
  parabola

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Center of Mass

• Only slightly different - measures point of
  average mass position
• In zero gravity, center of mass is the point
  considered
• Center of mass of solar system is slightly
  outside of the sun, causing it to wobble
• Astronomers search for wobbling stars to find
  planets

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Locating Center of Mass

• Of a uniform object is at its mid point
• Irregular object center of mass can be found by 3
  plumb-bob readings
• Center of mass can be outside the object or where
  no material exists (basketball, chair)

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Torque and Center of Mass

• Toppling occurs when center of mass is not over
  or under its support.
• This is due to torque created
• A football kicked below its center of mass will
  tumble because the force
• applied a torque to it

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Human Center of Mass

• Within the body near the navel
• Higher in men than women
• Higher in children than adults
• To avoid toppling - stand so that your center of
  mass is above area enclosed
• by your feet

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Example 9.5

• A 5 0-kg mass(m2)and
• an unknown mass(m3)
• hang from a 1.0 m rod of
• 2.0-kg mass(m1), as shown.
• The rod is supported on a
• knife edge fulcrum at a
• distance 35cm from one end. How large is the
  mass m3 if the rod and masses are to
• balance on the knife-edge?
• Torques that tend to rotate things
  counterclockwise are positive those that tend to
  rotate things clockwise are negative. m1 2.0
  kg, m2 5.0 kg, m3?
• Therefore the sum of the torques about point A
  is
• - r2m2g rF Fful- r1m1g - r3m3g 0 -r2m2g rF
  (m1g m2g m3g) - r1m1g - r3m3g.
• -r2m2 rF (m1 m2 m3) - r1m1 - r3m3 0
• -(0.05 m)(5.0 kg) (0.35 m)(5.0kg 2.0kg m3)
  - (0.50 m)(2.0 kg) - (0.85 kg)(m3 ) 0
• -0.25 m-kg 2.45 m-kg (0.35 m)(m3) - l.0 m-kg -
  (0.85 m)(m3) 0
• (0.50m)(m3) - 1.20 m-kg 0, m3 2.4 kg.

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Example 9.6

• Calculate the mass m3 by computing torques about
  the fulcrum.
• Choose the axis of rotation to be about the
  fulcrum, then only three torques to consider the
  counterclockwise torque due to m2 and the
  clockwise torques due to m1 and m3. The torque
  due to the fulcrum force becomes zero because the
  moment arm is now zero. The sum of the torques
  becomes
• r2m2g r1m1g r3m3g 0.
• Upon rearranging, we find m3 r2m2 - r1m1
• r3
• r2 30 cm, r1 15 cm, and r3 50 cm. When
  numerical values are inserted into the equation
  for m3,
• m3 (30)(5) -(15)(2) 2.4 kg
• (50)

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Example 9.7

• A sign weighing 400 N is suspended at the
• end of a 350-N uniform rod that is hinged at
• the wall. (a) What is the tension in the support
• cable that makes an angle of 35o with the rod?
  
• The sum of the torques becomes (T sin q
  )L - w(½ L) W L 0.
• Remove the common factor of L and rearrange this
  equation to get the tension T as
• T Substitute the numerical values, we
  get
• T ½(350 N) 400 N 1000 N
• sin 350
• (b) If the cable is moved so that the angle it
  makes with the rod is 550, the analysis in part
  (a) doesn't change. Substitute the new value for
  the angle,
• T ½(350 N) 400 N 700 N.
• sin 550

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Example 9.8

• The average mass of a womans arm is
  approximately 5 of her whole body mass. When
  the arm is lifted as shown, the distance between
  the center of mass of the arm and the center of
  rotation of the shoulder is approximately 0.15 m.
  In a simplified model, consider the deltoid
  muscle to pull horizontally with a lever arm of
  about 3 cm. With what force, in terms of the
  woman's body weight, must the deltoid muscle pull
  in order to hold the arm in equilibrium in the
  position shown?
• The arm is in a condition of static equilibrium
  for both translation and rotation. Calculate
  torques about the center of rotation of the
  shoulder. The downward force on the arm is a
  percentage of the woman's total weight, w mg.
  Using the symbols shown,
• S ti FD d - lFA 0 where FA 0.05 w. Upon
  rearranging,
• FD l FA (0.15 m)(0.05w) 0.25w.
• d 0.03 m

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Homework

• Assign 11,2,3,4,5
• Assign 211,14, 17,
  18, 23

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Torque and Angular Acceleration

• t F r ma r , and since a a r then
• t (mr2)a where the quantity in parentheses
• represents the Rotational Inertia, I
• Rotational inertia is not like regular inertia,
  since the shape of an object can affect it

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Rotational Inertia

• A rotating object tends to continue rotation at
  same angular speed and direction
• Rotational Inertia depends not just on mass but
  also how the mass is distributed

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Rotational Inertia

• Rotational Inertia can be varied by varying
  distribution of mass
• Shorten a pendulum inertia changes
• Bend legs when running inertia changes
• General Formula I m r2
• Rolling occurs faster with smaller rotational
  inertia - solid cylinder accelerates more than
  hollow cylinder

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Rotational Inertia and Gymnastics

• Human body has 3 axes
• Head to toe spin axis used by ice skaters
• Transverse axis - forward/backward axis used in
  tumbling, mostly gymnastics
• Medial axis - side rotation, used in doing
  cartwheels
• Gymnasts learn to take advantage of which axis to
  use and to which to apply torque

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Rotational Kinetic Energy

• Rotational kinetic energy is KErot ½ Iw2.
• The total kinetic energy of a rolling body
  separates into the translational kinetic energy
  of the center of mass and rotational kinetic
  energy about the center of mass KEtot KEtrans
  KErot ½ mv2 ½ Iw2.

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Conservation of Energy

• Both translational and rotational energies must
  be conserved--the sum together makes the Total
  Energy

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Angular Momentum

• Measure of rotational inertia, just like regular
  momentum measures inertia
• Angular momentum rotational inertia x rotational
  velocity or mvr (linear momentum x radius)
• An object or system of objects will maintain its
  angular momentum unless acted upon by an
  unbalanced external torque
• A moving bicycle is thus easier to balance than a
  still bicycle

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Angular Momentum

• Can be separated into two parts one term that
  depends on the properties of the body and the
  other term that is the angular velocity

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Conservation of Angular Momentum

• If no unbalanced torque acts on a rotating
  system, the angular momentum of that system is
  constant
• Spinning skater spins slowly with arms out,
  faster with arms in, faster yet with arms up
• Reducing r makes w greater to have same angular
  momentum

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Conservation of angular momentum

• When the net applied torque on an object is zero,
  its angular momentum is conserved. If an object
  rotates with a large angular momentum, its axis
  of rotation remains relatively stationary in
  space unless a large torque is applied transverse
  to the initial rotation axis.

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Homework

• 29, 37, 41, 42, 47, 53