### Video Transcript

A uniform rod of mass 1.0 kg and
length 2.0 meters is free to rotate about one of its ends, as shown in the
accompanying diagram. If the rod is released from rest at
an angle of 60 degrees above the horizontal, what is the speed of the tip of the rod
as it passes the horizontal position?

As we solve this problem, weβll
assume that the rod can rotate about its axis without any energy loss due to
friction. Weβll also assume that π, the
acceleration due to gravity, is exactly 9.8 meters per second squared. Letβs highlight some of the vital
information given in the statement.

Weβre told the rod has a mass of
1.0 kilograms and a length of 2.0 meters. The initial angle above the
horizontal, that which the rod is released, is 60 degrees. We want to solve for the speed of
the tip of the rod as it passes the horizontal position. Weβll call that speed π£ sub
π‘.

We can approach this problem under
the assumption that the law of conservation of energy holds. Letβs see how that approach works
itself out in this exercise. We can state the law of
conservation of energy in equation form. Given a system of objects, the
initial energy of the objects in that system, πΈ sub π, is equal to the final
energy of the objects in that same system, πΈ sub π.

We can apply this general principle
to our particular scenario. Letβs consider what is the initial
energy πΈ sub π of the rod. At the start, weβre told that the
rod is released from rest, meaning its initial speed is zero. The rod has no kinetic energy
initially, but it does have potential energy due to gravity.

If we establish the zero value of
our height as the dotted horizontal axis, since the center of mass of a rod is above
β equals zero, that means our rod does have potential energy thanks to gravity. Weβre told that the overall length
of a rod, which weβll call capital πΏ, is 2.0 meters. And that the mass of a rod, which
weβll call π, is 1.0 kilograms.

Because this is a uniform rod, the
center of mass of the rod is that its exact center: one meter from either end. To find the initial energy of the
rod, we want to find the height of that center of mass point just as the rod is
being released. By drawing a dotted vertical line
from that point down to the horizontal axis, we create a right triangle where the
length of the hypotenuse of that triangle is capital πΏ over two.

As we look at the corners of this
triangle, we recognize that for a right triangle like this one the sine of the
angle, in this case 60 degrees, is equal to the opposite side, β, divided by the
hypotenuse πΏ over two.

Since πΏ equals 2.0 meters, and
weβre dividing that value by two, we end up with the value on the right-hand side of
our equation of β divided by 1.0 meters. Since any number divided by one is
equal to that original number, we find that the height, β, simply equals the sine of
60 degrees meters.

Now that we know the initial height
of the center of mass of the rod, we can write out the initial energy of our system,
πΈ sub π, in terms of gravitational potential energy. That term will be π, the mass of
the rod, times π, the acceleration due to gravity, times β, which is the sine of 60
degrees.

Now that weβve covered the initial
energy of our system, we want to solve for the final energy of the system. That occurs at the moment when the
rod is in a horizontal position having fallen through an angular extent of 60
degrees. At that instant in time, the rod
will have no gravitational potential energy, because its center of mass will have a
height of zero. However, the rod will have
rotational kinetic energy.

Letβs recall the equation for
rotational kinetic energy. Rotational kinetic energy, which we
abbreviate KE sub π, is equal to one half the moment of inertia of our object
multiplied by its angular speed squared. Letβs apply this equation to our
particular scenario. As we look at this equation, πΌ,
the moment of inertia, applies to a rod rotating about an axis at the end of the
rod.

For an object like that rotating
about an axis like that, the moment of inertia is equal to one third the mass of the
rod times its length squared. When we plug this equation in for
πΌ in our equation for rotational kinetic energy, we find an expression of one half
times one third ππΏ squared multiplied by π squared.

The π or rotational speed noted in
this equation only applies at one instant in time, at the instant when the rod is in
the horizontal position parallel to the horizontal axis. Thatβs why itβs true that we can
take nearly this rotational kinetic energy term and insert that for the final energy
of our system. Because at that moment when the
road is in that position, potential energy due to gravity of the rod is zero.

We can now replace πΈ sub π, the
final energy of our system, with our expression for rotational kinetic energy of the
rod. We end up with this energy balance
equation for our system. The initial gravitational potential
energy of the rod when itβs released is equal to the rotational kinetic energy of
the rod at the moment that the rod is horizontal.

We want to solve this equation for
π, the angular speed of the rod at that moment. If we can solve for π, then we can
use the relationship between π£, the linear speed, and π, the angular speed to
solve for π£ sub π‘.

As we move towards isolating π,
letβs first notice that this problem is mass independent. The mass of the rod appears on both
sides and, therefore, can be canceled out. If we now multiplied both sides of
the equation by six divided by πΏ squared, then the six and πΏ squared terms cancel
out on the right-hand side of our equation.

We can then take the square root of
both sides of our equation and see that the squared term and square root term on the
right side cancel out. Weβre left with an expression for
π; π equals the square root of six divided by πΏ squared times π times the sine
of 60 degrees. πΏ, we know, is 2.0 meters and π
is 9.8 meters per second squared. When we plug those values in to
this equation, then we are now ready to enter these numbers into our calculator to
solve for π. And when we do that, we find that
π is equal to 3.568 radians per second.

Now weβre ready to take advantage
of a relationship between linear velocity and rotational velocity. That relationship is that linear
velocity π£ is defined as π times π, where π is the distance from the point being
considered to the axis of rotation and π is the angular speed. This equation applies to our
situation, because we want to solve for π£ sub π‘, the speed of the tip of the rod
as it passes the horizontal line.

We know π and π is the length of
the rod πΏ. So letβs write this linear to
rotational conversion equation relative to our scenario. π£ sub π‘, the speed of the tip of
the bar, is equal to the length of the bar multiplied by the angular speed with
which the bar rotates.

When we plug in the values we have
for πΏ and π and then multiply these two numbers together, we find that the tip of
this rod as it passes the horizontal line has a linear speed of 7.1 meters per
second. This is the speed at which the end
of the bar is moving at that moment in time.