Hello, everyone! I'm Jesse Mason and in this installment of Teach Me we take a look at
banked turns and how an advisory speed limit is determined when the geometry of the turn
is known. Here we're given both the inclination and the radius of the turn. Armed with these
two pieces of information, and a little physics, we're going to determine the rated speed for
the turn.
The rated speed, by the way, is the speed for which friction is unnecessary to maintain
a constant turn radius. In other words, if a vehicle hits a patch of ice at this speed
it will not slide off the road but rather remain on the intended path. As always, we
begin by drawing a picture... Our picture will show the turn as viewed from
above. We'll use our coffee mug here to make sure the turn is nice and circular. Why is
our turn circular? Because the given radius is constant. To make this a little more realistic
we'll depict our vehicle entering this turn from a straight-away and exiting the turn
to a straight-away. Here we'll mark the center of the turn, c. And label the radius of the
turn, r. Finally, we'll draw our vehicle in mid-turn. Below we'll note that r represents
the radius of the turn. At the moment, we have the vehicle traveling upwards. So this
vector arrow will represent the vehicle's tangential velocity -- which is just a geometrically
correct way of describing a body's instantaneous velocity while undergoing uniform circular
motion. Lastly, we'll indicate the direction of the centripetal acceleration of our vehicle.
Oops, forgot an arrow. Now a couple forces are acting on our vehicle
keeping it in uniform circular motion. And that means we need a Free Body Diagram. Since
we're moving clockwise around the turn, the road -- as well as the vehicle on the road
-- will be inclined clockwise as well. Our FBD, of course, shows our vehicle from behind.
To make this view a little clearer, we'll give the vehicle a license plate. And maybe
some tail lights will help. Or maybe it just looks like an angry robot now. ("No disassemble
Number 5!") Anyway, the two forces acting on our vehicle
are its weight and the normal force applied to the vehicle by the road. Now these two
forces are not perpendicular, so it isn't completely obvious which coordinate system
is most convenient to use. Motion analysis -- vis-à-vis Newton's Second Law -- is always
simplified by choosing a coordinate system such that one axis of the coordinate system
-- here it's the x-axis - is aligned with the body's acceleration vector. Since our
vehicle is undergoing uniform circular motion, the acceleration vector is to the right, toward
the center of curvature. Now that we've assigned a coordinate system
to our Free Body Diagram, we can break any oblique force vectors into their x and y components.
So here's y-component of the normal force and the x-component of the normal force.
We'll call this angle theta. And it turns out that the angle between the normal force
and weight vectors is equal to the angle of the incline, so we'll label this theta as
well. So theta is defined as the angle of inclination which is a known value... so we
write "equals fifteen degrees." Now our turn radius is also given to us in
the wording of our problem -- one hundred meters.
The tangential velocity is the rated speed of our turn, which makes it the unknown variable;
our task is to determine its value. Finally, the centripetal acceleration is not given
but its constancy is implied by our definition of rated speed: "MAINTAIN A CONSTANT TURN
RADIUS." To maintain a constant turn radius is to undergo uniform circular motion, which
implies a constant centripetal acceleration. And that does it for our knowns-n-unknowns.
As mentioned previously and forecasted by our use of a Free Body Diagram, we will be
analyzing the vehicle's motion using Newton's Second Law. Given the multidimensional aspect
of its motion, we'll analyze each component of its motion, and its causes, separately.
So here we'll start with N2 in the x-direction. So we have: The sum of the forces in the x-direction
is equal to the product of its mass and its acceleration in the x-direction. Here the
acceleration in the x-direction is equal to its centripetal acceleration, so we'll set
a-sub-x to a-sub-c. Summing our forces in the x-direction we have... just N-sub-x, the
normal force in the x-direction. So we write N-sub-x equals m a-sub-c. Now, trigonometrically,
N-sub-x is related to N by a factor of sine theta, so we can write N sine theta equals
m times - and we'll write the mathematical definition of centripetal acceleration here
- the square of the tangential velocity, v-sub-t, over the turn radius, r. Solving for v-sub-t
we get: v-sub-t equals plus-or-minus the square root of N, r, sin theta over m. Now I didn't
actually write the plus-or-minus symbol because only the positive value for v-sub-t makes
physical sense here. So it's not like I just forgot to write it in. Okay? Okay. Moving
right along, we would like to have an expression for N here, so we'll label this equation (1)
and then go look for that expression using Newton's Second Law in the y-direction.
So we have: The sum of the forces in the y-direction equal the product of the vehicle's mass and
its acceleration in the y-direction. How might acceleration in the y-direction occur in a
vehicle rounding a banked turn? It would appear in the lateral movement of the vehicle -- up
or down the incline. But since our vehicle is maintaining a constant turn radius its
position on the incline is fixed, and therefore is not accelerating in the y-direction. So
we can set a-sub-y to zero. Summing the forces in the y-direction we get:
N-sub-y in the positive y-direction, and the vehicle's weight in the negative y-direction.
Equals zero. Here N-sub-y is related to N by a factor of cosine theta, so we can write:
N cosine theta -- and we're going to move weight over to the other side and use its
mathematical definition -- equals m g. Solving for N, we get N equals m g over cosine theta.
And we'll label this equation number (2). So now we insert equation two into equation
one and we get: v-sub-t equals the square root of m g over cos theta, times r sin theta,
divided by m. The m's divide out, revealing an interesting physical insight about rated
speeds, and the trigonometric functions combine to give us tangent theta. So we're left with
v-sub-t equals the square root of g r tangent theta.
Inserting our values for g, r and theta, we find that the tangential velocity of a body
undergoing uniform circular motion around a banked curve of these specifications -- and
therefore the rated speed for the turn - is 16.2 m/s, which works out to be about 35 miles
per hour. Which -- given our values for r and theta - makes sense.
I'm Jesse Mason. I hope you found this video helpful. If you have any suggestions for future
teach me videos or just want to say hello from your part of the world, please let me
know in the comments section below. And until next, happy learning!