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Transcript
Torque, Equilibrium, and
Stability
Dual Credit Physics
Montwood High School
R. Casao
Torque
• A force is needed to produce a change in rotational
motion.
• The rate of change depends on the magnitude of
the force and on the perpendicular distance of its
line of action from the axis of rotation.
• The line of action of a force is an imaginary line
extending through the force vector arrow (the line
along which the force acts.
• The figure shows that r = r∙sin , where r is the straight
line between the axis of rotation and the point at which
the force and  is the angle between the line of r and the
force F.
• The perpendicular distance r is called the moment arm
or lever arm or torque arm.
• Previously, I taught you to resolve the force into a
component parallel to the axis of rotation and a
component perpendicular to the axis of rotation.
– The parallel force component passes through
the pivot and does not produce a rotation.
– The perpendicular force component
produces the rotation of the object around
the pivot point.
• Torque equation:  = r∙F = r∙F·sin  = r∙F
angle  is the angle between r and F
• Unit: m·N; this is the same as the unit of work
and energy, N·m = J. The torque unit is generally
written in reverse order m·N to avoid confusion.
Torque is not work and the unit is not a Joule.
• Rotational acceleration is not always produced
when a force acts on a stationary rigid body.
– When the force acts through the axis of rotation so that
 = 0º, the  = 0.
– When the force acts perpendicular to r,  = 90º and the
 is a maximum.
• The angular acceleration depends on where a
perpendicular force is applied and on the length of
the lever arm.
– The torque produced by the vector
T1 at radius R1 is greater than the
torque produced by vector T2 at
radius R2.
• Torque in rotational motion is similar to force in
translational motion.
– An unbalanced or net force changes translational
motion; an unbalanced or net torque changes rotational
motion.
• Torque is a vector; its direction is always perpendicular
to the plane of the force and the moment arm vectors.
– Torque is a vector cross product (the multiplcation of
two vectors to produce a resultant third vector):
 
  rxF

– The direction is given by a right-hand rule.
– The figure shows a particle at point A in the xy plane
with a single force F acting on the particle
• The particle’s position relative to the origin O is given
by the position vector r.
• To find the direction of the torque, slide the force vector
without changing its direction until its tail is at the origin
O so that it is tail to tail with the position vector.
• Use the right hand rule to rotate the position vector r into
the force vector F. Point fingers of right hand in
direction of r and rotate fingers toward F. The thumb
points in the direction of the torque.
Equilibrium
• Equilibrium refers to a condition in which the net
forces acting on a system are balanced.
– Unbalanced forces produce translational accelerations;
balanced forces produce translational equilibrium.
– Unbalanced torques produce rotational accelerations;
balanced torques produce rotational equilibrium.
• Translational equilibrium occurs when the sum of
the forces acting on a body is 0 N; the body
remains at rest (static equilibrium) or in motion
with a constant velocity (dynamic equilibrium).
– Condition for translational equilibrium: F = 0 N.
• Forces with lines of action through the
same point are called concurrent forces (a
and b in the figure to the right).
• When the vector sum of concurrent forces
is 0 N, the body is in translational
equilibrium.
• In part c of the figure, the F = 0 N, but the
opposing forces will cause the object to
rotate about the pivot; the object will not be
in static equilibrium.
• A pair of equal and opposite forces that do
not have the same line of action is called a
couple.
• The condition F = 0 N is a necessary but
not sufficient condition for equilibrium.
• Condition for rotational equilibrium:  = 0 m·N
• If the sum of the torques acting on an object is 0,
the object is in rotational equilibrium and either
remains at rest (no rotation) or rotates with a
constant angular velocity  ( = 0 rad/s2).
• A body is in mechanical equilibrium when the
conditions for both translational and rotational
equilibrium are satisfied:
F = 0 N and  = 0 m·N
• A rigid body in mechanical equilibrium may be
either at rest or moving with a constant linear
and/or angular velocity.
• Static equilibrium is the condition that exists
when a rigid body remains at rest; v = 0 m/s and
 = 0 rad/s.
• Translational static equilibrium example 8.4
• Torque is a vector and has a direction.
– We can designate torque directions as positive and
negative depending on the rotational acceleration
they produce.
– Rotational directions are taken as clockwise or
counterclockwise around the axis of rotation.
– A torque that produces a counterclockwise rotation is
positive; a torque that produces clockwise rotation is
negative.
• Rotational static equilibrium Example 8.5
• Static equilibrium Example 8.6
Stability and Center of Gravity
• The equilibrium of a particle or a rigid body can
be stable or unstable in a gravitational field.
• For rigid bodies, the equilibrium is analyzed in
terms of the center of gravity of the body.
• The center of gravity is the point at which all of
the weight of an object can be considered to be
acting.
• When the acceleration of gravity is constant, the
center of gravity and the center of mass coincide.
• If an object is in stable equilibrium, any small
displacement results in a restoring force or torque which
tends to return the object to its original equilibrium
position.
• The ball in the bowl is in stable equilibrium. Any movement of the ball within the bowl returns the ball to the
bottom of the bowl.
• The pointed object is also in stable equilibrium.
• Any slight displacement of the boject raises its center of
gravity and a restoring gravitational force tends to return
the object to the position of minimal potential energy
(with the base flat on the surface – the equilibrium
position).
• The gravitational force actually produces a restoring
torque that is due to a component of the weight that tends
to rotate the object about a pivot point back to its original
position. This will only work as long as the center of
gravity remains to
the left of the pivot
point.
• If an object is in unstable equilibrium, any small displacement
results in a torque that tends to rotate the object farther away from
its equilibrium position.
• The ball on top of the bowl is in unstable equilibrium. Any
movement of the ball from the top of the bowl causes the ball to
leave its equilibrium position, decreasing its potential energy.
• The pointed object is also in unstable equilibrium. Any
displacement of the center of gravity from directly over the pivot
will result in a decrease in the object’s potential energy.
• An object is in stable equilibrium as long as its
center of gravity after a small displacement still
lies above and inside the object’s original base of
support. There will always be a restoring
gravitational torque.
– The line of action of the weight intersects the original
base of support.
• When the center of gravity or center of mass falls
outside the base of support, the object falls over
because of a gravitational torque that rotates it
away from the equilibrium position.
– The line of action of the weight does not intersect the
original base of support.
• Rigid bodies with wide bases and low centers of
gravity are most stable and least likely to tip over.
Ex. High speed racing cars
• Rigid bodes with a narrow
base and a higher center of
gravity are less stable and more likely to tip over.
Ex. SUV’s