For rigid body planar kinetics, the same principle of work and energy
and conservation of energy that we learned from particle kinetics apply.
As we learned in particle kinetics, the principle of work and energy states that
the total external work done to the system equals to the change in the kinetic energy of the system.
When there are only conservative forces doing work, then we can derive the conservation of energy
which states that the total energy which includes kinetic energy and potential energy of the system stays constant.
The potential energy commonly includes the gravitational potential energy as well as the elastic potential energy.
These two equations also apply to a system of bodies as well.
Now we need to revisit some key concepts defined for rigid body kinetics.
If the rigid body is subjected to a force, and during motion the point of application follows a path,
then during this procedure the work done by this force is defined as the integration of
the dot product of the force vector and the differential position vector.
Keep in mind that this force
is a variable that could change in both magnitude and direction. And since we know that the dot product of two vectors is a scalar,
therefore we can write the scalar form, which is the integration
of the magnitude of the force times cosine theta d_s -- theta is the angle made by the line of action of the force,
and the tangent line along the path,
integrated from the initial point s_1 to the final point s_2.
And as you can probably tell this is the same as what we've learned in particle kinetics.
If the rigid body is subjected to a couple moment, and it rotates within the plane by angle theta, then the work done by this
couple moment is defined as the integration of the dot product of the moment vector and d_theta, the differential angular position vector.
But for planar motion don't forget the moment vector and the angular position vector are both perpendicular to the plane,
therefore they are either of the same direction or opposite direction. Therefore this vector equation can be simplified to this scalar equation:
the work done by this moment is positive when the two vectors are of the same direction, and
the work done by this moment is negative when they are of opposite direction.
A special case will be when the moment has a constant magnitude, then its work is simply M times the difference between
the initial angular position and the final angular position of this rigid body.
Now let's look at the kinetic energy of a rigid body.
For a rigid body undergoing planar motion, from what we've learned in rigid body kinematics that it'll have an angular velocity
and every point on this rigid body will have different linear velocity.
But we can always find the linear velocity of its gravitational center, point G.
Therefore,
the kinetic energy of this rigid body is defined as one half m v_G squared plus one half I_G omega squared.
m, again is the mass of this rigid body,
v_G is the linear velocity of its gravitational center, I_G is the mass moment of inertia of this rigid body with respect to an axis that
passes through point G and is perpendicular to the plane, and omega is the angular velocity.
As you can see this equation clearly captures contributions from both translation as well as rotation.
This is the general equation that applies to all three cases of planar motion.
But if we have translation,
that means that the angular velocity omega is zero, therefore
the kinetic energy of the rigid body simply equals to ½ m v_G squared. As you can see
this is the same as the kinetic energy for a particle.
For rotation about a fixed axis,
the general equation also applies, but we can also rewrite the equation into T equals to ½ I_O omega squared.
I_O is the mass moment of inertia with respect to point O, which is the
center of rotation.
I_O can be determined through parallel axis theorem
from I_G, the mass moment of inertia about the gravitational center.
And lastly for general plane motion, once again the general equation still applies, but
if we can locate the instantaneous center of zero velocity, then
the kinetic energy can be also determined through ½ I_IC omega squared.
Once again, I_IC is the mass moment of inertia about the instantaneous center of zero velocity.
For rigid body kinetics, the gravitational potential energy is defined the same way as in particle kinetics.
Here since the gravitational force always applies at the gravitational center, point G, therefore the displacement of this weight force
is always measured from the gravitational center of this rigid body.
And the elastic potential energy is also defined the same way as in particle kinetics. Once again, for the elastic potential energy
the displacement s of this spring is always measured with respect to its own unstretched neutral position.
And the elastic potential energy is always non- negative.
Let's look at this system. We have two identical rods and two identical disks connected together by pins,
as well as a spring that is initially unstretched when theta equals to 60°.
If the system is released from rest when theta equals to 60°, from our general knowledge we
know that under the weight of the two rods, the two rods are going to fall and
the two disks will be pushed to roll apart. And the disks roll without slipping,
and we need to determine the angular velocities of the rods and the disks when theta becomes 30°.
Since the disks roll without slipping, that indicates that the static frictional force
between the disks and the ground is doing no work during this process,
and since we also know that the normal force exerted by the ground does no work, therefore during motion for our system we only have
conservative forces doing work, which include the gravitational forces as well as the spring force.
Therefore we can apply the conservation of energy to this entire system.
Now let's look at each of these four terms individually.
First we have the total kinetic energy of the system at the initial state. But since the system was released from rest, therefore
the total kinetic energy at the initial state is zero.
Then let's look at the total potential energy
at the initial state.
If we draw a line passing through the centers of the two disks as our datum,
and since the two disks have centers of gravity along the datum, their gravitational potential energy is zero.
And for the gravitational forces of the two rods,
their gravitational center is located at 0.5 times sine 60° above the datum,
and also initially the spring is unstretched, therefore the elastic potential energy of the spring is zero as well.
Therefore from here we can determine
the total potential energy of the system at state one to be 340 joule.
And then we can sketch the final state of this system and determine the final potential energy of the system, which includes the gravitational
potential energy of only the two rods as well as the elastic potential energy,
since the spring now has been stretched.
Therefore that is 203 joule, so we are only left with the last term, which is the final kinetic energy of the system
that includes the kinetic energy of the two rods as well as the two disks.
Don't forget we only have one equation which is the conservation of energy, therefore we can only
solve for one unknown from that equation. So our goal is to rewrite this equation here
into an expression with only one unknown.
Because of the symmetry we only need to focus on the left half of this system which includes one rod and one disk, both of which
are undergoing general plane motion at the final state when theta equals to 30°.
For the rod it has the angular velocity omega_rod, and we can sketch the linear velocity of its gravitational center at point B.
Right now we don't know the direction of this velocity.
For the disk it also has an angular velocity omega_disk, but from the rigid body kinematics that we learned before
we know that the direction of the linear velocity of its gravitational center is horizontal pointing to the left.
As you can see we have two linear velocities here and two angular velocities,
all of which are needed to determine the final kinetic energy of the system.
Therefore we need to explore the relations between these parameters
Let's look at the disk first.
Again because the disks roll without slipping, from the rigid body kinematics that we learned before we know that the point of contact
is the instantaneous center of zero velocity for this disk.
And this is the radius from point C to its IC.
And we learned already that the linear velocity of any point on this rigid body can be determined by omega,
its angular velocity times the radius with respect to IC. Therefore,
v_C can be determined as omega_disk times 0.2 m.
But notice that point C also belongs to the rigid body ABC the rod,
therefore if we can find the instantaneous center of zero velocity for the rod ABC,
then we can relate the linear velocity v_C to the angular velocity omega_rod. How do we do that?
From the symmetry, from observation, we know that the linear velocity of point A is vertical down. Therefore
we draw the two perpendicular lines,
then we can determine the instantaneous center of zero velocity
for the rod.
This distance, the radius of point B with respect to the IC of the rod, from the geometry can be determined to be 0.5 m.
This distance is also 0.5 m.
Therefore
v_C equals to omega_rod time 0.5, but v_C was determined to be 0.2 times omega_disk,
therefore from here we can determine that omega_rod is 0.4 omega_disk.
Therefore,
v_B equals to omega_rod times its radius with respect to IC_rod, which is 0.5 omega_rod again, which is also 0.2 omega_disk.
So now v_B, v_C and omega_rod are all given in expressions of omega_disk, therefore,
the kinetic energy of a rod can be determined
as 1.07 omega_disk squared.
Kinetic energy of the disk can be determined as well. Therefore this is the total kinetic energy,
which is an expression with only one unknown, omega_disk.
Now we have analyzed every term in this equation,
the initial total kinetic energy is zero, initial total potential energy,
final total kinetic energy as an expression of omega_disk, and final total potential energy.
Therefore we have one equation, only one unknown, we can solve it.
And if we know the angular velocity of the disk, because the angular velocity of the rod is 0.4 omega_disk, therefore we can determine
the angular velocity of the rod as well.
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