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A driver makes a turn.
The inertial force prevents them from moving forward and cracking their skull.
This inertial force is fictitious, It seems real, but only because the observerās frame of reference is rotating or accelerating
-
The force of gravity is negligible
-
The force of tension will head to the center
-
The only time this system is undergoing UCM is when the force to keep the system up is enough to counteract the force of gravity
-
The force of gravity would be 0
Top-Down View
Side(Rear) View
a
Acceleration points to center
v
a
Acceleration points to center
a
n
F
g
F
The x-component
nx
F
contributes to acceleration
The y-component
ny
F
counteracts
g
F
Velocity is tangent to the curve
v
If a car wants to make a turn, there is a certain velocity (lets call it v ) that a car needs to be at to successfully make the turn with UCM.
In these cases of UCM, friction does not need to act.
However, if the car moves quicker or slower than , Friction will have to act to assist in the turn.
ideal
v
ideal
ideal
v
assist
Friction
quicker or slower
needs to be at
If v > v
ideal
Centripetal Force is greater
greater
If youāre going faster, you will need to turn quicker to make the turn
Friction acts down the slope. It assists acceleration
If v < v
ideal
Centripetal Force is smaller
smaller
If youāre going slower, you will need to turn slower to make the turn
Friction acts up the slope. It combats acceleration
Friction
Friction
On a banked curve, an object does not need Friction to make a turn. This is because the curve is already on an incline
Instead, what allows the object to turn is the Normal Force
n
F
F
This is because the object is actually accelerating into the ground
accelerating into the ground
Net
F
=
nx
F
=
n
F
Īø
- sin
Īø
Īø
nX
F
v > v
ideal
F
g
F
n
F
f
Īø
Īø
F
fy
F
fx
F
ny
F
nx
Horizontal:
F
net
=
F
fx
F
nx
ma =
F * Ī¼ * cosĪø
n
F * sinĪø
n
Īø
mv
F * Ī¼ * cosĪø
n
F * sinĪø
n
r
2
=
Vertical:
F
ny
=
F
g
F
fy
F * cosĪø
n
=
mg
F * Ī¼ * sinĪø
n
mg
=
F * Ī¼ * sinĪø
n
F * cosĪø
n
Lets divide the 2 equations
mv
r
2
mg
=
F * Ī¼ * cosĪø
n
F * sinĪø
n
F * Ī¼ * sinĪø
n
F * cosĪø
n
v
rg
2
=
Ī¼cosĪø + sinĪø
cosĪø - Ī¼sinĪø
v > v
ideal
F
g
F
f
Īø
F
fy
F
fx
Īø
Horizontal:
F
Net
=
F
nx
F
fx
mv
r
2
=
F * sinĪø
n
F * Ī¼ * cosĪø
n
Vertical:
F
ny
F
fy
=
F
g
F * cosĪø
n
F * Ī¼ * sinĪø
n
F
n
Īø
F
ny
F
nx
=
mg
Divide the 2
mv
r
2
mg
=
F * sinĪø
n
F * Ī¼ * cosĪø
n
F * cosĪø
n
F * Ī¼ * sinĪø
n
v
rg
2
=
sinĪø - Ī¼cosĪø
cosĪø + Ī¼sinĪø