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# Beginner's Question about Banked Turns

## Beginner's Question about Banked Turns - Physics Forum

### Beginner's Question about Banked Turns - Physics Forum. Discuss and ask physics questions, kinematics and other physics problems.

#1
07-22-2004, 01:12 PM
 Anebt Guest Posts: n/a

When a bank is curved at the certain angle theta to prevent a car in uniform
circular motion from skidding irrespective of any frictional force, the
horizontal component of the normal force equals mv^2/r, or the centripetal
force.

I understand the concept of this, but I have a question:

Where does the horizontal component of the car's weight figure into all
this? It seems that the horizontal component of the weight of the car would
be keeping the car from skidding off the track (thanks to the gravitational
force), yet textbooks only say that the horizontal component of the normal
force does this. I know that mathematically the two are related, but I am
having trouble understanding the concept.

#2
07-22-2004, 02:20 PM
 N:dlzc D:aol T:com \(dlzc\) Guest Posts: n/a

Dear Anebt:

"Anebt" <[Only registered users see links. ]> wrote in message
news:m3PLc.9157\$[Only registered users see links. ].atl.earthli nk.net...
uniform
centripetal
would
gravitational
normal

I think the problem you are hosting comes from the use of the word
"normal". In physics, this word means "perpendicular to" something. In
the context of this problem, the track surface can have a meaningful normal
formed to it.

The component of gravity normal to the surface ( g * cos( theta ) ) simply
describes the supporting force of the track. The component of gravity
parallel to the surface ( g* sin( theta ) ) describes the radial
acceleration of the vehicle around the track.

Imagine if the track were around the Earth, then ramped up to 90 degrees.
The only way to keep the same relationship with the centerline of the track
would be to be in orbit at that altitude.

I hope this helps.

David A. Smith

#3
07-22-2004, 03:04 PM
 AJW Guest Posts: n/a

>
You simply want the "weight vector" of the car to be normal to the road
surface. You already know the downward (vertical) force of a stationary car.
You can compute what the skidding (force vector parallel to the roadway in the
downhill direction) and now you simply have to offset that with a force vector
brought on by radial acceleration in the opposite (uphill) direction. Don't
worry about naming the forces, that can lead to confusion especially on this
newgroup. Just decide what the objective is -- in htis case it's to have the
total forces acting on the car be normal to the raod surface -- then calculate
the circumstances that'll creat that set of circumstances.

#4
07-22-2004, 03:42 PM
 sam1967@hetnet.nl Guest Posts: n/a

On Thu, 22 Jul 2004 13:12:50 GMT, "Anebt" <[Only registered users see links. ]> wrote:

i had trouble understanding this due to ill-informed teachers but
centrifugal force has a part to play in this.

i suggest you read this entire article at NASA (who surely must know

[Only registered users see links. ]

#5
07-22-2004, 06:02 PM
 thomas guild Guest Posts: n/a

"Anebt" <[Only registered users see links. ]> wrote in message
news:m3PLc.9157\$[Only registered users see links. ].atl.earthli nk.net...
uniform
would
gravitational

Of the two forces acting on the car (weight (W) and normal force (N), only N
has a horizontal component. W acts vertically downwards and so its
horizontal component is zero.

#6
07-22-2004, 06:35 PM
 sam1967@hetnet.nl Guest Posts: n/a

On Thu, 22 Jul 2004 18:02:00 GMT, "thomas guild"
<[Only registered users see links. ].uk> wrote:

that is true from the point of view of a linear inertial
(non-accelerating ) reference frame. in the circular or rotational
reference frame things look slightly differently.

there are two main horizontal forces at play in this reference frame.

there is an outward centre fleeing force acting on the car pushing it
outwards away from the centre in a radial direction. this is the
centrifugal force and is equal to (m*v*v )/ r .

where v is the velocity of the car in the tangential direction.

obvisouly as v increases so does the centrifugal force on the car.

the other horizontal force is provided by the friction of the tyres on
the road and acts directly towards the centre (ie it is a centripetal
force).

while the car stays on the bank these two forces are equal and
opposite and the car stays on the bank.

however the friction provided by the tyres has a maximum level beyonw
which it cannot go (determined by the material and the surface area)

as soon as the velocity - v - of the car increases to a certain level
the centrifugal force will overcome the centripetal frictional force
and the car will fly off the bank.

this can easily be seen by sitting a draughts (checkers) piece on a
turntable. the piece will stay on the turntable until the rpms
(revolutions per minute ) of the turntable exceeds some level (eg 70
rpm). at that point the centrifugal force (m*v*v/ r) will overcome the
frictional centripetal force and the piece will fly off the turntable

#7
07-22-2004, 09:10 PM

"Anebt" <[Only registered users see links. ]> wrote in message
news:m3PLc.9157\$[Only registered users see links. ].atl.earthli nk.net...
uniform

Make that "the horizontal component of the force equals mv^2/r."

The word "normal" in physics (as in analytical geometry) means
"perpendicular to the surface."

There are two forces acting from the vehicle onto the ground here: the
weight (W = m*g) due to gravity acts vertically downward and the inertial
force (F = m*v*2/r, sometimes called "centrifugal") that results from the
changing direction of the velocity and acts horizontally (unless the vehicle
is simultaneously going uphill or downhill).

Each of these forces is a vector - meaning it has both magnitude AND
direction. When you add them together, the *net* force N is also a vector.
It often helps to draw a diagram:

CAR
#------F=m*v^2/r
|\
| \
| \
| \
| \
| \
W=m*g N=[F]+[W]

Ideally, you would like the *net* force N to be perpendicular to the ground.

A little trigonometry will show you that the angle theta of the pavement
should equal the angle between the weight vector W and the *net* force
vector N.

gravitational

"Weight" has *no* horizontal component. It acts straight down by definition
of "down."

Hope this helps...

Tom Davidson
Richmond, VA

#8
07-23-2004, 04:03 AM
 Anebt Guest Posts: n/a

news:[Only registered users see links. ]...

definition

I was thinking about the examples with inclined planes found in textbooks.
The horizontal component of the weight provides the force that pulls a
sled down the plane, etc. In these cases the horizontal component of weight
pulls something downward.

I guess that a web site I just found answered my question, at least in
theory:

"Since the car is not sliding down the bank, only components of the normal
are examined."

Source for quote:
[Only registered users see links. ]
centripetal/lessoncentripetalforce.asp

#9
07-23-2004, 09:09 AM
 thomas guild Guest Posts: n/a

"Anebt" <[Only registered users see links. ]> wrote in message
news:r60Mc.10200\$[Only registered users see links. ].atl.earthl ink.net...
weight

In such cases the force acting downhill isn't the -horizontal- component of
the weight (which is zero after all!), but the component acting parallel to
the slope.

#10
07-23-2004, 09:20 AM
 thomas guild Guest Posts: n/a

<[Only registered users see links. ]> wrote in message
news:[Only registered users see links. ]...
the
centripetal
normal
am
only N

You seem to be analysing the case of a car cornering on a horizontal
surface. In the case of a banked surface, friction isn't the only
horizontal force: The horizontal component of the normal force provides an
additional sideways force, which allows the car the possibility of cornering
at a higher speed than it could on the flat.

TG

 Tags banked , beginner , question , turns

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