2. What happens when you drive too
fast over a hill!
• http://www.youtube.com/watch?v=
QhjNmXpFnms&feature=related
3. What are we covering today?
• Flying off hills
• Sliding around corners
• Staying on a racetrack
4. At what speed will the car leave the ground?
What are the forces on the car when it
is on the ground?
Up
S (reaction force)
(mv2
)/r (Centripetal force)
Down
mg (weight)
mg
(mv2
)/rS +
When the car is on the ground the forces are balanced
mg = s + (mv2
)/r
If the car is travelling at a speed greater than (V0) it will leave the ground
At this point s = 0 as there is no reaction force from the ground
5. mg
(mv2
)/r
m v0
2
/ r = mg
m = mass
V0 = is the first speed it takes to
the air
r = radius of the hill from the
centre of curvature
Re- arrange to make V0 the centre of attention to find the speed for take off
V0 = g r
Therefore a car would jump a hill of radius 6 metres if it is going a
speed greater than 7.67 m/s
http://www.animations.physics.unsw.edu.au/jw/circular.htm
6. On a roundabout
http://h2physics.org/?cat=63
A vehicle of mass m and speed v in a
circle of radius r around a roundabout
r
The centripetal force is created by the
tyres contact friction on the road
Friction Force F = mv2
r
If F is greater than the grip available from the
tyre and surface the car will slide – this will
happen at a speed v0 and a corresponding force
F0
The grip from the tyre is proportional to the weight of the vehicle and the road
surface and the tyre combine to give the co-efficient of friction µ
Sliding Force F0 = mv0
2
r
http://phys23p.sl.psu.edu/phys_anim/mech/
embederQ2.20150.html
7. The grip from the tyre is proportional to the weight of the vehicle and the road
surface and the tyre combine to give the co-efficient of friction µ
Sliding Force F0 = mv0
2
r
F0 = µmg
The grip from the tyre can be broken down into its component parts
µmg = mv0
2
r
Rearrange to make v0 the centre of attention
v0 = sqrt (µ g r)
8. Banked Track
θ
N1
N2
θ
θ
Car going around a banked track
You can travel faster around a banked corner as a
component of the cars mass keeps the car into the
centre of the trackmg
Resolving the vertical and horizontal components
Horizontal component = (N1 + N2) sin θ
Vertical component = (N1 + N2) cos θ
The horizontal component act as the centripetal force
mv2
/r = (N1 + N2) sin θ
The vertical component balances the weight
(N1 + N2) cos θ = mg
using tan θ = sin θ / cos θ
9. Banked track
(N1 + N2) sin θ
(N1 + N2) cos θ
=
using tan θ = sin θ / cos θ
(N1 + N2) cos θ = mgmv2
/r = (N1 + N2) sin θ
mv2
mgr
Reduces down to ...Reduces down to ...
Tan θ = v2
/g rTan θ = v2
/g r
Making v the focus
V2
= g r tan θ
http://phys23p.sl.psu.edu/phys_anim/mech/
embederQ2.20170.html
