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1
Using the Clicker

• If you have a clicker now, and did not do this
  last time, please enter your ID in your clicker.
• First, turn on your clicker by sliding the power
  switch, on the left, up. Next, store your student
  number in the clicker. You only have to do this
  once.
• Press the button to enter the setup menu.
• Press the up arrow button to get to ID
• Press the big green arrow key
• Press the T button, then the up arrow to get a U
• Enter the rest of your BU ID.
• Press the big green arrow key.

2
Torque

• Forces can produce torques. The magnitude of a
  torque depends on the force, the direction of the
  
• force, and where the force is applied.
• The magnitude of the torque is
  .
• is measured from the axis of rotation to the
  line of the force, and is the angle between
  and .
• To find the direction of a torque from a force,
  pin the object at the axis of rotation and push
  on it with the force. We can say that the torque
  from that force is whichever direction the object
  spins (counterclockwise, in the picture above).
• Torque is zero when and are along the same
  line.
• Torque is maximum when and are
  perpendicular.

3
Equilibrium

• For an object to remain in equilibrium, two
  conditions must be met.
• The object must have no net force
• and no net torque

4
Worksheet, part 1

• A uniform rod with a length L and a mass m is
  attached to a wall by a hinge at the left end. A
  string will hold the rod in a horizontal
  position the string can be tied to one of three
  points, lettered A-C, on the rod. The other end
  of the string can be tied to one of three hooks,
  numbered 1-3, above the rod. This system could be
  a simple model of a broken arm you want to
  immobilize with a sling.

5
Sling, part 1

• How would you attach a string so the rod is held
  in a horizontal position but the hinge exerts no
  force at all on the rod?
• A ? 1.
• A ? 2.
• A ? 3.
• B ? 1 or B ? 3.
• B ? 2.
• C ? 1.
• C ? 2.
• C ? 3.
• It cant be done.

6
Sling, part 2

• How would you attach a string so the rod is held
  in a horizontal position while the force exerted
  on the rod by the hinge has no horizontal
  component, but has a non-zero vertical component
  directed straight up?
• A ? 1.
• A ? 2.
• A ? 3.
• B ? 1 or B ? 3.
• B ? 2.
• C ? 1.
• C ? 2.
• C ? 3.
• It cant be done.

7
Sling, part 3

• How would you attach a string so the rod is held
  in a horizontal position while the force exerted
  on the rod by the hinge has no vertical
  component, but has a non-zero horizontal
  component?
• A ? 1.
• A ? 2.
• A ? 3.
• B ? 1 or B ? 3.
• B ? 2.
• C ? 1.
• C ? 2.
• C ? 3.
• It cant be done.

8
A balanced beam

• A uniform beam sits on two identical scales.
  Scale A is farther from the center than scale B,
  but the beam remains in equilibrium.
• Which scale shows a higher reading?
• How far to the left could scale B be moved
  without the beam tipping over?

9
A balanced beam

• A uniform beam sits on two identical scales.
  Scale A is farther from the center than scale B,
  but the beam remains in equilibrium.
• Which scale shows a higher reading?
• Scale B it is closer to the center of gravity.
• How far to the left could scale B be moved
  without the beam tipping over?

10
A balanced beam

• A uniform beam sits on two identical scales.
  Scale A is farther from the center than scale B,
  but the beam remains in equilibrium.
• Which scale shows a higher reading?
• Scale B it is closer to the center of gravity.
• How far to the left could scale B be moved
  without the beam tipping over?
• To the center of the beam, but no farther. The
  beams center of gravity must be between the
  supports to be stable.

11
A balanced beam

• Could you place a weight on the beam without the
  reading on scale A changing?
• Could you place a weight on the beam and cause
  the reading on scale A to decrease?

12
A balanced beam

• Could you place a weight on the beam without the
  reading on scale A changing?
• Yes place it directly over scale B.
• Could you place a weight on the beam and cause
  the reading on scale A to decrease?

13
A balanced beam

• Could you place a weight on the beam without the
  reading on scale A changing?
• Yes place it directly over scale B.
• Could you place a weight on the beam and cause
  the reading on scale A to decrease?
• Yes put it on the beam to the right of scale B.

14
A balanced beam

• What happens to the scale readings when scale B
  is moved to the right?
• 1. A is unchanged, B goes up.
• 2. A is unchanged, B goes down.
• 3. Both readings decrease.
• 4. Both readings increase.
• 5. A goes up, B goes down.
• 6. A goes down, B goes up.
• 7. None of the above.

15
The human spine

• Equilibrium ideas can be applied to the human
  body, including the spine. If you bend your upper
  body so it is horizontal, you put a lot of stress
  on the lumbrosacral disk, the disk separating the
  lowest vertebra from the tailbone (the sacrum).
  Picking something up is even worse.
• Simulation
• Treat the spine as a pivoted bar. There are
  essentially three forces acting on this bar
• The force of gravity, mg, acting on the upper
  body (this is about 65 of the body weight). The
  tension in the back muscles. This can be
  considered as one force T that acts at an angle
  of about 12 to the horizontal when the upper
  body is horizontal. The support force F from the
  tailbone, which also acts at a small angle
  measured from the horizontal.

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The human spine

• At equilibrium, take torques about the tailbone.
  T is applied about 10 farther from the tailbone
  than the force of gravity.
• The components of the support force F can be
  found by summing forces.
• For a person with a weight of 600 N, the upper
  body weighs almost 400 N, while T and F are
  around 1700 N.
• Picking up something with your arms that has a
  weight of 100 N increases both T and F by about
  600 N!
• The moral of the story bend your legs instead of
  your back. Picking up a 100 N bag of groceries by
  bending at the knee produces a force of about 500
  N on the bottom disk in the spine. The force is
  4-5 times larger if you bend your back!

17
Center of gravity

• The center-of-gravity of an object is the point
  that moves as though the weight of the object is
  concentrated there.
• The center-of-gravity is given by
• In a uniform gravitational field, the
  center-of-mass and the center-of-gravity are the
  same point. They're different if the
  gravitational field is non-uniform.
• The center-of-gravity can be found by hanging an
  object from a support. In stable equilibrium, the
  center-of-gravity is directly below the support.

18
Red and blue rods

• Two rods of the same shape are held at their
  centers and rotated back and forth. The red one
  is much easier to rotate than the blue one. What
  is the best possible explanation for this?
• 1. The red one has more mass.
• 2. The blue one has more mass.
• 3. The red one has its mass concentrated more
  toward the center the blue one has its mass
  concentrated more toward the ends.
• 4. The blue one has its mass concentrated more
  toward the center the red one has its mass
  concentrated more toward the ends.
• 5. Either 1 or 3 6. Either 1 or 4
• 7. Either 2 or 3 8. Either 2 or 4
• 9. Due to the nature of light, red objects are
  just inherently easier to spin than blue objects
  are.

19
Newtons First Law for Rotation

• An object at rest tends to remain at rest, and an
  object that is spinning tends to spin with a
  constant angular velocity, unless it is acted on
  by a nonzero net torque or there is a change in
  the way the object's mass is distributed.
• The net torque is the vector sum of all the
  torques acting on an object.
• The tendency of an object to maintain its state
  of motion is known as inertia. For straight-line
  motion mass is the measure of inertia, but mass
  by itself is not enough to define rotational
  inertia.

20
Rotational Inertia

• How hard it is to get something to spin, or to
  change an object's rate of spin, depends on the
  mass, and on how the mass is distributed relative
  to the axis of rotation. Rotational inertia, or
  moment of inertia, accounts for all these
  factors.
• The moment of inertia, I, is the rotational
  equivalent of mass.
• For an object like a ball on a string, where all
  the mass is the same distance away from the axis
  of rotation
• If the mass is distributed at different distances
  from the rotation axis, the moment of inertia can
  be hard to calculate. It's much easier to look up
  expressions for I from the table on page 291 in
  the book (page 10-15 in Essential Physics).

21
A table of rotationalinertias
22
The parallel axis theorem

• If you know the rotational inertia of an object
  of mass m when it rotates about an axis that
  passes through its center of mass, the objects
  rotational inertia when it rotates about a
  parallel axis a distance h away is

23
Worksheet, part 2

• When a system is made up of several objects, its
  total rotational inertia about a particular axis
  is the sum of the rotational inertias of the
  individual objects for rotation about that axis.
• What is the systems rotational
• inertia in the first case?
• Each block has a mass of m/3, and the rod, of
  negligible mass, has a length L.

24
Worksheet, part 2

• When a system is made up of several objects, its
  total rotational inertia about a particular axis
  is the sum of the rotational inertias of the
  individual objects for rotation about that axis.
• What is the systems rotational
• inertia in the first case?

25
Worksheet, part 2

• In the second case, do we expect the rotational
  inertia to be larger, smaller, or the same as the
  rotational inertia in the first case?
• What is the systems rotational
• inertia in the second case?

26
Worksheet, part 2

• In the second case, do we expect the rotational
  inertia to be larger, smaller, or the same as the
  rotational inertia in the first case? Larger
  the mass is farther from the axis.
• What is the systems rotational
• inertia in the second case?

27
Worksheet, part 2

• In the second case, do we expect the rotational
  inertia to be larger, smaller, or the same as the
  rotational inertia in the first case? Larger
  the mass is farther from the axis.
• What is the systems rotational
• inertia in the second case?

The parallel-axis theorem gives the same result.
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Whiteboard