Download Definition of Torque Center of Gravity Finding the Center of Gravity

Definition of Torque
Chapter 8: Rotational Equilibrium and Dynamics
Torque, , is the tendency of a force to rotate an object about some axis
Force vs. Torque
• Forces cause accelerations
• Torques cause angular
accelerations
Torque = (Magnitude of Force) * (Lever arm)
=Fd
Lever arm is the perpendicular distance from the
axis of rotation to a line drawn along the direction of
the force.
Torque is a vector quantity.
Direction of torque: right-hand rule
= r F sin
points from the point of reference (rotational
axis) to the location of the force. Obviously, the
magnitude of a torque depends on where we assume
the axis of rotation to be. F sin is the tangential
component of the force.
Unit of torque:
Multiple torques, net torque,
and an example
• When two or more torques are acting on an object, the torques
are added as vectors. The sum is the net torque.
Find the net torque (magnitude and
direction) produced by the forces F1
and F2 about the rotational axis shown
in the drawing. The forces are acting
on a thin rigid rod, and the axis is
perpendicular to the page.
N•m
Center of Gravity
The net torque produced by the force of gravity:
• The object is divided up into a
large number of very small
particles of weight (mig)
• Each particle will have a set of
coordinates indicating its
location (xi, yi)
• The torque produced by each
particle is migxi
• The total (net) torque is (migxi)
Now locate a point (xcg, ycg), so that the torque
produced by the gravity of the whole mass on
that point is equal to the total torque:
(mig) xcg = (migxi)
Finding the Center of Gravity
• The center of gravity of a homogenous,
symmetric body must lie on the axis of
symmetry.
• Often, the center of gravity of such an object
is the geometric center of the object.
• If a single force can be used to support an
object in equilibrium, it must exactly
counteract the effect of gravitational force.
The force must lie on a line through the
center of gravity of that object.
Two Conditions for Equilibrium
A rigid body is in equilibrium if it has zero translational acceleration and
zero angular acceleration. In equilibrium, the sum of the externally
applied forces is zero, and the sum of the externally applied torques is
zero:
and
Note that if the net force is zero, the net torque becomes independent of
the choice of rotational axis. In this case, we can use any rotational axis to
sum up the net torque. The nature of the problem will often suggest a
convenient location for the axis.
A zero net torque does not mean the absence of rotational motion. An
object that rotates at uniform angular velocity can be under the influence
of a zero net torque. This is analogous to the translational situation where
a zero net force does not mean the object is not in motion.
CG could be in a “hollow”!
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Example of Equilibrium
Solving Equilibrium Problems
• Draw a diagram of the system
– Include coordinates and choose a rotation axis
• Isolate the object being analyzed and draw a free body diagram
showing all the external forces acting on the object
– For systems containing more than one object, draw a separate
free body diagram for each object
• Apply the Second Condition of Equilibrium
– This will yield a single equation, often with one unknown
which can be solved immediately
• Apply the First Condition of Equilibrium
– This will give you two more equations
• Solve the resulting simultaneous equations for all of the unknowns
More Example
If the 70-kg fireman goes any higher than as shown, the ladder, which is
8.0-m long and has an evenly distributed mass of 20 kg, will slip. The
wall is smooth. What is the coefficient of static friction between the
ladder and the floor?
A woman of mass m=55.0 kg sits
on the left end of the seesaw, a
plank of length L=4.00 m and mass
m=12.0 kg, pivoted in the middle.
A man of mass M=75.0 kg sits on
the right side of the pivot to balance
the seesaw. (a) Where should the
man sit? (b) Find the normal force
exerted by the pivot.
Consequence of Net Torques: Angular Accel.
For a point object rotating about an axis a distance r away from the object,
rFT = = mr2
FT = ma = mr
Now define Moment of Inertia of a point object going around an external:
I = mr2
then:
= I For a rigid body rotating about a fixed axis
Net external torque = ( moment of intertia) *
(angular acceleration)
Note that a centripetal force leads
to no angular acceleration.
Moment of Inertia
= I The relationship is analogous to F = ma
Moments of Inertia
For a rigid system consisting of many particles,
1 = (m1r12)
2 = (m2r22)
3 = (m3r32)
.
.
.
The moment of inertia for a (rigid) system consisting
of many particles is found to be
Note: distance to the axis of rotation (a line)!
Distance parallel to the axis does not matter.
Obviously, the moment of inertia of an object
depends on the chosen axis of rotation.
Moment of inertia depends on the
location of the rotational axis
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Example
Rotational Kinetic Energy
A solid, frictionless cylindrical reel of mass
M=3.00 kg and radius R=0.400 m is used to
draw water from a well. A bucket of mass
m=2.00 kg is attached to a cord that is
wrapped around the cylinder. (a) Find the
tension T in the cord and the acceleration a of
the bucket. (b) If the bucket starts from rest
at the top of the well and falls for 3.00 s before
hitting the water, how far does it fall?
The rotational kinetic energy KER of a rigid
object rotating with an angular speed about a
fixed axis and having a moment of inertia I is
Total Mechanical Energy
The total kinetic energy of an object that moves with both a linear
velocity and a rotational velocity is the sum of the two contributions.
Here v is the linear velocity of the center of mass of the object and is the angular velocity about its center of mass.
of center of mass
The total mechanical energy of the object is then the sum of its
kinetic energy and potential energy. In the absence of work due to
non-conservative forces, the total mechanical energy is conserved.
of center of mass
Conservation, and Work-Energy
Theorem in a Rotating System
• Conservation of Mechanical Energy
(KEt+ KEr+PE)i = (KEt+ KEr+PE)f
– Remember, this is for conservative forces, no
dissipative forces such as friction can be present
– Potential energies of any other conservative forces
could be added
• In the case where there are dissipative forces such as friction, use
the generalized Work-Energy Theorem instead of Conservation of
Energy:
Wnc = KEt + KEr + PE
Energy concepts can be useful for simplifying the analysis
of rotational motion.
Example Problems
A thin-walled hollow cylinder (mass = mh, radius = rh ) and a
solid cylinder (mass = ms, radius = rs) start from rest at the top of
an incline. Both cylinders start at the same vertical height h0. All
heights are measured relative to an arbitrarily chosen zero level
that passes through the center of mass of a cylinder when it is at
the bottom of the incline. Ignoring energy losses due to retarding
forces, determine which cylinder has the greater translational
speed upon reaching the bottom.
Angular Momentum
The angular momentum L of a rigid body rotating about a fixed axis is the
product of the body’s moment of inertia I and its angular velocity with respect
to the axis:
L = I ,
and
For a point object, L=I =mr2 =rmv.
Changes in angular momentum are due to rotational impulse: t .
Internal forces of a system cannot lead to a net torque on the system, because of
the action and the reaction act in opposite directions, but ALONG THE SAME
LINE.
If the net external torque acting on the system is zero, the total
angular momentum of a system remains constant, i.e. it is
conserved.
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Conservation of Angular Momentum, Examples
• With hands and feet drawn
closer to the body, the
skater’s angular speed
increases
– L is conserved, I
decreases, increases
Example Problems of Angular Momentum
Movie
The puck has a mass of 0.120 kg. Its original distance from the center of
rotation is 40.0 cm, and the puck is moving with a speed of 80.0 cm/s. The
string is pulled downward 15.0 cm through the hole in the frictionless table.
Determine the work done on the puck. (Hint: Consider the change of kinetic
energy of the puck.)
Rotational Analog of Translational Concepts
Item
Rotational
Translational
Displacement
s
Velocity
v
Acceleration
a
Cause of Accel.
Torque Force F
Inertia
Moment of Inertia I
Mass m
Newton’s 2nd Law
= I
F = ma
Work
Fs
Kinetic Energy
I2 / 2
mv2 / 2
Momentum
Angular momentum
I
Linear Momentum
mv
Chapter 8 Summary
Torque; Conditions for equilibrium; Center of gravity;
Newton’s second law for rotational motion; Moment of inertia;
Rotational work; Rotational kinetic energy; Conservation of
total mechanical energy; Angular momentum; Conservation of
angular momentum.
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