Download When the applied force is not perpendicular to the crowbar, for

When the applied force is not perpendicular to the
crowbar, for example, the lever arm is found by drawing
the perpendicular line from the fulcrum to the line of
action of the force.
We call torques
that produce
counterclockwise
rotation positive,
and torques that
produce
clockwise rotation
negative.
What is the net torque acting on the merrygo-round?
a)
b)
c)
d)
e)
+36 N·m
-36 N·m
+96 N·m
-60 N·m
+126 N·m
96 N·m (counterclockwise)
- 60 N·m (clockwise)
= +36 N·m (counterclockwise)
How far do we have to place the 3-N
weight from the fulcrum to balance the
system?
a)
b)
c)
d)
2 cm
27 cm
33 cm
53 cm
F=3N
 = +1 N·m
l=/F
= (+1 N·m) / (3 N)
= 0.33 m = 33 cm
The center of gravity
is the point about which the weight of the object itself
exerts no torque.
We can locate the center of gravity by finding the point
where it balances on a fulcrum.
What’s the center of gravity of a disk?
A. any point on the edge of the disk.
B. Center of the disk
C. Any point half way between the center and the edge.
The center of gravity
For a more complex object, we locate the center of gravity by
suspending the
object from two
different points,
drawing a line straight
down from the point of
suspension in each
case, and locating the
point of intersection of
the two lines.
1J-21 Center of Gravity of an Irregular Lamina
How can we find the Center of Gravity of Irregular shapes?
Why does the mass
on the string hang
straight down ?
Where is the Center
of Gravity of the cut
board ?
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If the center of gravity
lies below the pivot
point, the object will
automatically regain
its balance when
disturbed.
The center of gravity
returns to the position
directly below the
pivot point, where the
weight of the object
produces no torque.
1J-23 Corks & Forks
Can the Center of Gravity lie at a point not on the object?
How difficult is it to
balance this system on
a sharp point ?
Where is the C
of G ?
THE CENTER OF GRAVITY IS NOT ON THE OBJECT. IT ACTUALLY LIES ALONG THE
VERTICAL BELOW THE SHARP POINT.
WHEN THE FORK IS MOVED THE CM RISES AND THIS MEANS THE SYSTEM IS IN STABLE
EQUILIBRIUM .
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1J - 24 double cone
What happens to the
center of mass ?
A). Going down the hill
B). Going uphill
C). Stay at rest
D). Depend on the object shape
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1J-16 Walk the Plank
What happens when a mass is placed at the end of a massive plank?
Sum of Torque about Pivot
XMg–xmg=0
m=MX/x
Can you safely
walk to the
end of the
plank ?
One can solve for either M or m, if
the other quantity is known
EVEN WITH A MASS AT THE END OF THE PLANK, THE SYSTEM CAN STILL BE IN
EQUILIBRIUM
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1J-28 Wine Bottle Holder
Balance a Bottle and a Wooden Holder by Eliminating Net Torque
M
How does this
system Balance?
m1
x1
x2
m2
Sum of Torque about Pivot
m1x1g - m2x2g = 0
THE CENTER OF GRAVITY OF THE BOTTLE PLUS THE WOOD MUST LIE DIRECTLY OVER AND
WITHIN THE BOUNDARY OF THE SUPPORT (PIVOT). FOR BALANCE THERE CAN BE NO
NET TORQUE ON SYSTEM.
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How far can
the child walk
without tipping
the plank?
• For a uniform plank, its center of gravity is at its geometric center.
• The pivot point will be the edge of the supporting platform.
• The plank will not tip as long as the counterclockwise torque from the
weight of the plank is larger than the clockwise torque from the
weight of the child.
• The plank will verge on tipping when the magnitude of the torque of
the child equals that of the plank.
An 80-N plank is placed on a dock as shown. The
plank is uniform in density so the center of gravity of
the plank is located at the center of the plank. A
150-N boy standing on the plank walks out slowly
from the edge of the dock. What is the torque
exerted by the weight of the plank about the pivot
point at the edge of the dock?
a)
b)
c)
d)
e)
f)
+80 N·m
-80 N·m
+160 N·m
-160 N·m
+240 N·m
-240 N·m
Rotational Inertia and Newton’s
Second Law
• In linear motion, net force and mass determine the
acceleration of an object.
• For rotational motion, torque determines the rotational
acceleration.
• The rotational counterpart to mass is rotational inertia or
moment of inertia.
– Just as mass represents the resistance to a change in linear
motion, rotational inertia is the resistance of an object to change
in its rotational motion.
– Rotational inertia is related to the mass of the object.
– It also depends on how the mass is distributed about the axis of
rotation.
Simplest example:
a mass at the end of a light rod
• A force is applied to the mass
in a direction perpendicular to
the rod.
• The rod and mass will begin to
rotate about the fixed axis at
the other end of the rod.
• The farther the mass is from
the axis, the faster it moves for
a given rotational velocity.
Simplest example:
a mass at the end of a
light rod
• To produce the same rotational
acceleration, a mass at the end
of the rod must receive a larger
linear acceleration than one
nearer the axis.
• F = ma
– It is harder to get the system
rotating when the mass is at the
end of the rod than when it is
nearer to the axis.
– I case the distance are equal, it’s
harder to move a heavier mass.
Rotational Inertia and Newton’s
Second Law
• Newton’s second law for linear motion:
Fnet = ma
• Newton’s second law for rotational motion:
∙R=m∙
∆
∆
∙
= ∙

∙R=m∙
∙
net = I
– The rotational acceleration produced is equal to the torque
divided by the rotational inertia.