Physics 1A
Lecture 8A
"One machine can do the work of fifty ordinary
men. No machine can do the work of one
extraordinary man."
--Elbert Hubbard
Vertical Circular Motion
You must worry about gravity with centripetal
acceleration in the vertical direction.
In this situation you also
must consider that the
tangential velocity will
be different at different
locations of the loop.
For example, vbot will be
greater than vtop (due to
conservation of energy).
The normal force of the
track on the coaster will
also vary with location.
Vertical Circular Motion
Example
In a 1901 circus performance, Allo Diavolo
introduced the stunt of riding a bicycle in a
vertical loop-the-loop. Assuming that the loop is
a circle with radius R=2.7m, what is the minimum
speed Diavolo could have at the top of the loop
to remain in contact with it there?
Answer
First, you must define
a coordinate system.
Let’s say that the
bottom of the loop is
y=0 and down is the
positive y-direction.
Centripetal
Force
Answer
Next, we draw a free-body diagram for the
bicycle at the top of the loop:
bicycle
Fgravity, Earth on bicycle
Fnormal, loop on bicycle
Next, apply Newton’s 2nd Law in the y-direction.
Since this bicycle is undergoing centripetal acceleration.
Centripetal
Force
Answer
So we can say:
If the bicycle is to have the least speed and
remain in contact with the loop, then it is on the
verge of losing contact, such that FN = 0:
Gravitational Force
Earlier we mentioned that Fg = mg only works near
the surface of the Earth.
In general, you can calculate the gravitational force
by:
If there is more than one gravitational force
present, merely sum all of the forces present.
∑Fon1 = F2on1 + F3on1 +...
Remember that these forces are vectors, so you
will need to take their direction into account
when summing.
Gravitational Force
Gravitational Potential Energy also has a similar
constraint.
Near the surface of the Earth: PEgrav = mgh.
In general you can calculate gravitational
potential energy by:
Remember that r is measured from center to center
of the two objects.
Please remember the negative sign, it is important.
Gravitational Force
Unfortunately this equation
assumes that zero potential energy
occurs when you are infinitely far
away from the other object.
You don’t get to define where zero
is with this equation. Your
coordinate system is pre-defined.
As you get closer to the other
object, PE will become negative
and, thus, decrease.
Again ΔPE is much more important
than absolute PE.
Potential Energy
Example
A 5.00kg stone is shot directly upward into outer
space. What minimum speed does it need to be
given in order to escape Earth’s gravitational pull
(i.e. PEgrav = 0)?
Answer
First, you must define a coordinate system.
Let’s say that the center of the Earth is y=0 and
up is the positive y-direction (in reality we didn’t
choose, the PE equation chose for us).
Potential
Energy
Answer
Use conservation of mechanical energy here,
initially:
Finally, the energy will be:
where KEf is zero since we want the minimum speed
that is just large enough for the object to escape.
Energy is conserved here, so: Emec,i = Emec,f.
Potential
Energy
Answer
<-this is how you
calculate escape speed.
Did this escape speed depend on the mass of the stone?
No, it only depended on the mass and radius of the Earth.
Torque
When applying force to rotating objects, it is very
important to understand where the force is being
applied on the object.
We turn to torque, or turning force, τ. The
magnitude of torque is given by:
where F is the applied
force, r is the radius from
the axis of rotation (or
pivot point) and θ is the
angle between r and F
Torque
When the force is parallel to r, then the
magnitude of torque is zero.
When the force is at some angle to r, the
perpendicular component of F is what causes the
rotation.
The SI unit for torque is the [Nm] (not a Joule).
Just as force causes linear accelerations on an object,
torque will cause angular accelerations on an object.
Torque
Torque is a vector quantity. The direction is
perpendicular to the plane determined by r and F.
The direction is determined by the Right Hand Rule.
Point your fingers in the direction of r.
Curl the fingers toward F.
Your thumb points in the direction of
torque.
We use the same conventions as with
rotational variables. A counterclockwise
torque is considered positive and a
clockwise torque is considered negative.
For Next Time (FNT)
Keep working on the Homework for
Chapter 7.
Start Reading Chapter 8