Transcript Class 6 - Department of Physics | Oregon State University

Here are some of the direct analogies between (linear) translational
and rotational motion:
Quantity or Principle
Linear
Rotation
Displacement
x

Velocity
v

Acceleration
a

Inertia (resistance to
acceleration)
mass (m)
moment of
inertia (I)
Momentum
P = mv
L = I
Momentum rate of change
dP/dt = Fnet
Stated as Newton’s 2nd Law:
F = ma
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Oregon State University PH 212, Class 6
dL/dt = net
 = I
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What is a torque?
It is a “twisting” force that causes angular acceleration in a rigid
body—usually caused by a linear force acting at some point on
the body, causing it to rotate about some axis. If the body is not
constrained (“pinned”), the rotation axis will be the body’s center
of mass (c.m.).
Torque is a vector (r x F), positive when its twisting effect is
counter-clockwise; negative if clockwise.
An easy way to compute this is  = (F·sin)d where d is the
distance from the axis of rotation to the point where F is acting
on the body, and  is the angle between F and d.
Torque has units of force·distance (N·m).
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Oregon State University PH 212, Class 6
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You are using a wrench to tighten a lugnut on a car tire. Which of these four
arrangements shown is most effective?
A.
B.
C.
D.
E.
1
2
3
4
All the same
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Oregon State University PH 212, Class 6
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A straight, rigid, mass-less rod of length 1.00 m
connects the point masses shown. Where must
you locate a fulcrum (support point) so that the
object balances (does not tip) with gravity?
10.0 kg
30.0 kg
1. x = 0.333 m
2. x = 0.500 m
3. x = 0.667 m
4. x = 0.750 m
5. None of the above.
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Oregon State University PH 212, Class 6
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Notice that in locating the fulcrum at the center of mass,
we effectively balanced the torques acting on the rod.
That is, we prevented the stick from accelerating
rotationally—i.e. in the angular sense.
Indeed, Newton’s First Law must include rotational
motion, too: A rigid body is in total mechanical
equilibrium only when it is not accelerating along any
translational axis; and when it is not accelerating around
any rotational axis.
In other words, it’s in total mechanical equilibrium when:
F = 0
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and
Oregon State University PH 212, Class 6
 = 0
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And notice how summing both torques and forces works together.
Consider, for example, item 3d in the Prep 3 set…
Follow-up: What force (magnitude and direction) is the fulcrum
exerting on the board so that the board remains at rest?
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Oregon State University PH 212, Class 6
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And of course, the two parts of Newton’s 2nd Law addresses all
cases in general—when a rigid body is not necessarily in total
mechanical equilibrium:
F = ma
and
 = I
As always, notice how summing both torques and forces works
together. Consider, for example, item 4a in the Prep 3 set…
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Oregon State University PH 212, Class 6
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