### Video Transcript

A particle with a mass π is
suspended from a string of negligible mass and a length of 1.0 meters, as shown in
the diagram. The particle is displaced to a
position where the taut string is at an angle of 30 degrees from the vertical. And the particle is released from
rest at that position. The particle moves through an arc,
where the lowest point of the arc is the point π. What is the instantaneous speed of
the particle at point π? What is the vertically upward
displacement of the particle from point π when its instantaneous speed is 0.81
meters per second?

We can call this instantaneous
particle speed at point π π£. And the vertically upward
displacement of the particle from π, weβll name π. Weβre told that point π on our
diagram is the location of where the mass would be when the string is hanging
straight down. We start off solving for the speed
of the mass when itβs at point π. Because gravity is a conservative
force, we know that the potential energy of the mass π when itβs in its original
position can be converted into kinetic energy of the mass when itβs at point π.

We can write our energy balance
equation to say that the initial kinetic plus potential energy is equal to the final
kinetic plus potential energy of the mass. At the outset, the speed of the
mass is zero. So, its initial kinetic energy is
zero. And if we set the altitude of the
mass at point π to be zero, then its final potential energy will be zero as
well. So, we can say the initial
potential energy of the mass is equal to its final kinetic energy.

Recalling that gravitational
potential energy equals mass times the acceleration due to gravity times height and
that kinetic energy equals one-half an objectβs mass times its speed squared, we can
write that π times π times β is equal to one-half ππ£ squared. We see that the mass of this object
cancels out. And rearranging to solve for the
speed π£, we find itβs equal to the square root of two times π times β.

π, the acceleration due to
gravity, we treat as exactly 9.8 meters per second squared. And we see on our diagram that β is
the height difference between the point π and the original location of the
mass. That height difference is equal to
1.0 meters, the length of the string, minus the length of the string times the cos
of 30 degrees. Plugging this value for β and the
known value for π into our expression for π£, to two significant figures, π£ is 1.6
meters per second. Thatβs how fast the mass is moving
when itβs at its lowest point π.

In part two, we imagine that the
speed of our mass, weβll call it π£ sub π, is 0.81 meters per second. We want to solve for the vertical
distance π of our mass at that speed above point π. In this case, when we write out our
energy balance equation, weβre only able to eliminate the initial potential energy
of our mass because we imagine its initial point is at point π.

At point π, the mass has kinetic
energy. And at the point where its speed is
0.81 meters per second, it will have both kinetic and potential energy. Using the variables weβve decided
on, we can write that one-half ππ£ squared, where π£ is the speed we solved for in
part one, is equal to one-half ππ£ sub π squared plus π times π times π, the
distance we want to solve for. Again, the objectβs mass cancels
out. And we can rearrange to solve for
π. Itβs π£ squared minus π£ sub π
squared all over two times π. When we plug in for these values
and calculate π, to two significant figures, we find itβs 3.3 centimeters. Thatβs the vertical displacement of
the mass above point π.