Transcript Motion Along a Straight Line at Constant Acceleration

Book Reference : Pages 26-27
1.
To show that the centripetal force is provided
by real world forces such as tension, gravity
& friction
2.
To consider three particular cases of motion:
•
Over the top of a hill or humped back
bridge
•
Around flat curves (roundabouts)
•
Around banked curves
During the last lesson we saw that an object
moving in a circle has a constantly changing
velocity, it is therefore experiencing acceleration
and hence a force towards the centre of rotation.
We called this the centripetal force: The force
required to keep the object moving in a circle. In
reality this force is provided by another force,
e.g. The tension in a string, friction or the force
of gravity.
Consider a car with mass m and speed v moving
over the top of a hill...
S
mg
r
At the top of the hill, the support force S, is in the
opposite direction to the weight (mg). It is the
resultant between these two forces which keep
the car moving in a circle
mg – S = mv2/r
If the speed of the car increases, there will
eventually be a speed v0 where the car will leave
the ground (the support force S is 0)
mg = mv02/r
v0 = (gr)½
Any faster and the car will leave the ground
On a level road, when a car travels around a
roundabout the centripetal force required to keep
the car moving in a circle is provided by the
friction between the road surface and tyres
Force of Friction F
F = mv2/r
friction
velocity
To avoid skidding or slipping, the force of friction
F0 must be less than the point where friction is
overcome which occurs at speed v0
Friction is proportional to weight and can be given
by the coefficient of friction ():
F  mg
F = mg
At the point of slipping:
F0 = mv02/r
 v0 = (gr)½
 mg = mv02/r
For high speed travel, race tracks etc have banked
corners. In this way a component of the car’s
weight is helping friction keep the car moving in a
circle
N1
N2


Towards centre of rotation
mg

Without any banking the centripetal force is
provided by friction alone. Banked corners allows
greater speeds before friction is overcome
The centripetal force is provided by the horizontal
components of the support forces
(N1 + N2) sin  = mv2/r
and the vertical components balance the weight
(N1 + N2) cos  = mg
Rearranging
sin  = mv2/ (N1 + N2) r
cos  = mg / (N1 + N2)
and since tan  = sin  / cos 
tan  = mv2/ (N1 + N2) r x (N1 + N2) / mg
tan  = mv2 / mgr v2 = gr tan 
Thus there is no sideways frictional force if the
speed v is such that v2 = gr tan 
A car with mass 1200kg passes over a bridge with a radius
of curvature of 15m at a speed of 10 m/s. Calculate:
a. The centripetal acceleration of the car on the bridge
b. The support force on the car when it is at the top
The maximum speed without skidding for a car with mass
750kg on a roundabout of radius 20m is 9m/s. Calculate:
a. The centripetal acceleration of the car on the
roundabout
b. The centripetal force at this speed
A car is racing on a track banked at 25° to the
horizontal on a bend with radius of curvature of
350m
a. Show that the maximum speed at which the
car can take the bend without sideways
friction is 40m/s
b. Explain what will happen if the car takes the
bend at ever increasing speeds