Transcript Document

Circular Motion
Centripetal Force
video
Angles can be measured in
RADIANS
The angle in radians is
s/r which is the length
of the arc divided by
the radius
One full circle = 2𝜋
radians i.e. 360o = 2𝜋
radians
Angular velocity (𝝎) is the rate of
change of angle during circular
motion in radians /sec (𝝷/t)
𝝅
𝝎=2
𝑻
For a full circle
( T is the
time period of the rotation
1/T = the frequency of rotation.
f = 1/T
𝝎 = 2𝜋f
Linear speed (v) and (𝝎) are linked
by the equation (v = r𝝎)
The speed is always constant in
the direction to the tangent at any
point on the circumference but he
direction is always changing
towards the centre of the circle.
The acceleration is always towards
the centre of the circle
Centripetal acceleration
towards the centre of the
circle = v2/r or 𝝎2r
According to Newton’s
second law F = ma
therefore centripetal
force = mv2/r
or m𝝎2r
centripetal force
2
= mv /r
m = mass of turning object
v = velocity
r = radius of the turning circle
Example
situation
A vehicle of mass 1200kg is
cornering at a velocity of 30 m/s
around a level corner of radius
110m. Calculate the centripetal
force exerted on the car by friction
between the tyres and the road.
centripetal force
= mv2/r
Mass = 1200kg
v = 30m/s
r = 110m
Centripetal force on vehicle
= (1200 x 302) /110
=9818N
The vehicle is on the limit of
adhesion at this velocity.
Calculate the coefficient of
friction between the tyres and
the road (take g as 9.8m/s2).
Force (centripetal) = μ(coefficient of
friction) x weight of vehicle (mass
x 9.81)
Weight = 1200 x 9.81
=11772N
μ = Force/weight
= 9818/11772
0.83
The corner now becomes banked
at an angle of 12o to the
horizontal. Calculate the maximum
cornering velocity of the vehicle on
this part of the curve.
F = mv2/r
F (centripetal) = μ(coefficient of friction) x weight
of vehicle (mass x 9.81)
μ x m x 9.81 = mv2/r
(this calculation gives us the centripetal force for cornering on a
flat road)
The extra centripetal force due to the 12o banking
Is found by the equation m x 9.81 x tan 12o (= v2/r)
v2 = (r x μ x m x 9.81)/m + (r x m x 9.81 x tan12o)/m
= (r x μ x 9.81) + ( r x 9,81 x tan 12o)
m’s
cancel
out
v2 = (r x μ x 9.81) + ( r x 9.81 x tan 12o)
= (110 x 0.83 x 9.81) + (110 x 9.81x tan 12o)
= 1143.8
V = √1143.8
33.8m/s
The crankshaft of a 125cc two-stroke engine
carries a single piston of bore 51.5 mm and
stroke 60mm. The effective rotating mass of
the connecting rod and piston assembly is
200g and is out of balance with the line of the
crankshaft by an eccentricity of half the stroke.
The engine is running at 900 rpm. Show that
this is equivalent to an angular velocity of
94.2 rad/s and calculate the centripetal force
on the crankshaft bearings due to this out of
balance load.
ω = angular velocity in
radians/second
360o (full circle)= 2π radians
ω = 2π/t (full circle)
1/t = frequency (f)
ω = 2πf
If the engine is running at 900rpm
ω = 900 x 2π/60
= 94.2 rad/sec
v = ωr
F = mv2/r = mω2r
In the question m = 200g =0.2kg
r = half of stroke =30 mm
F = 0.2 x 94.22 x .03
= 53.24 N
The out-of-balance force on the
crank is to be balanced using two
identical counterweights either side
of the connecting rod at an
eccentricity of 0.02m. Calculate the
mass of one counterweight.
a)
m1
2
ωr
1=
2m2
2
ωr
2
m2 = m1r1/2r2
m1 = mass of connecting rod and piston
assembly, 0.2kg
r1 = half stroke, 0.03m
r2 = eccentricity of counter weights,
0.02m
m2 =(0.2 x 0.03)/(2 x 0.02)
= 0.15 kg
Centrifugal
clutch
Centrifugal clutch
Rotating bobs
cylinder
springs
Rotating bobs
cylinder
springs
When the angular velocity of the shaft
increases the centripetal force
Increases causing the springs to extend and
the clearance between the bobs and the
cylinder to decrease. When the centripetal
force is high enough the bobs will engage
with the clutch cylinder.
Rotating bobs
cylinder
0.1m
0.01m
springs
A centrifugal clutch similar to the one
described has 6 rotating bobs each of mass
150g and a radius of 100mm. The spring
strengths are 5kN/m
Calculate the angular velocity (ω) as the bobs
engage with the clutch cylinder. (The bob
clearance is 0.01m and the spring length is
Rotating bobs
0.1m
0.01m
r = 0.1 +
0.01 = 0.11
springs
cylinder
The springs needs to extend by 0.01 m
Force required = 5kN/m x 0.01
= 50N
mrω2 = 50N
ω2 50/(m x r)
ω2 = 50/(0.15 x 0.11)
ω2 = 50/(0.15 x 0.11)
= 3030.3
ω = 55rad/sec
ω = 2πf
f = ω/2π
=55/2π
= 8.75 rps
8.75 x 60 rpm
525 rpm
Calculate the force on the cylinder at a
velocity of 1200rpm.
ω = 2πf
=7539.8 rad/min
=125.7 rad/sec
F = mrω2
0.15 x 0.11 x 125.72
261N
Net force = Centripetal force - spring force
= 261 – 50 = 211N
Explain how the coefficient
of friction between the bob
and the clutch face will
govern the power
transmitted
As ω increases the net force
between the bobs and the
cylinder but will always be 50N
less than the centripetal force
after engagement. Power
transmitted will be affected by
the force between the bobs and
the cylinder ( F = μ x R(reaction force))
And the velocity of the clutch.
As the velocity doubles the
power will multiply by 4 until
slip occurs