3.6 Further Mechanics and Thermodynamics
3.6.1 Periodic Motion
3.6.1.1 Circular Motion
● Angular frequency is the rate of change of angle (radians per second)
θ
ω= t
= 2πf
● Angular velocity = how fast the object is rotating
|v| = ωr (where r = radius)
Centripetal Acceleration
● Centripetal force is the force which acts on the body towards the center of
the circle of its motion
● Centripetal acceleration is the acceleration of an object moving in circular
motion
● Called centripetal acceleration because the acceleration is always towards
the center of the circle of motion
v2
ac = r
= ω2
mv 2
F Centripetal = r
= mω 2 r
3.6.1.2 Simple Harmonic Motion (SHM)
● A body will undergo simple harmonic motion when the force that tries to
restore the object to its rest position is proportional to the displacement of
the object
● F = − kx
● This motion can be shown by a sine wave (or cosine wave)
x = Acos(ωt)
v = − ω Asin(ωt)
a = − ω 2 Acos(ωt) = − ω 2 x
M ax value of cos(ωt) and sin(ωt) = 1
⇒ M ax speed = ωA
⇒ M ax acceleration = ω 2 A
,3.6.1.3 Simple Harmonic Systems
● A spring is an example of a simple harmonic system
● In a spring system:
√
m
T = 2π k (where k is the spring constant)
● For a pendulum:
√
l
T = 2π g (where l is the length of the pendulum)
● Pendulum equation only works for small angles (see below for reason)
Derivation for Time Period of a Pendulum
F = − kx
− mg sinθ = − k x
mg sinθ
k = x
Arc length s = θr = θl (radius = length of pendulum)
x ≈ s ⇒ x = θl
mg sinθ
k = θl
S M ALL AN GLE AP P ROXIM AT ION S − θ ≈ sinθ, so cancel
mg
k = l
√
m
T = 2π k
m
T = 2π
√ mg
l
√
ml
T = 2π mg
√
l
T = 2π g
Aside
, F = kx
ma = kx
m (ω 2 Acos(ωt)) = k(Acos(ωt))
k
m = ω2
SHM and Energy
E T otal = E P otential + E Kinetic
● As the angle of the pendulum increases, kinetic energy decreases and
potential energy decreases
● As the angle decreases, kinetic energy increases and kinetic energy
decreases
● At the origin position ( x = 0 ), kinetic energy (and so velocity) is at a
maximum (no potential energy)
1 2
ET = 2
kx + 12 mv 2
ET = 1
2
k(A2 cos2 (ωt)) + 12 ( ωk2 )(A2 ω 2 sin2 (ωt))
ET = 1
2
kA2 (cos2 (ωt) + sin2 (ωt))
ET = 1
2
kA2
● From this we can find a formula for v :
1
2
kA2 = 12 kx2 + 1
2
mv 2
mv 2 = kA2 − kx2
( ωk2 )v 2 = k(A2 − x2 )
v 2 = ω 2 (A2 − x2 )
v = ± ω √A2 − x2
v max at x = 0
⇒ v max = ± ω A
3.6.1 Periodic Motion
3.6.1.1 Circular Motion
● Angular frequency is the rate of change of angle (radians per second)
θ
ω= t
= 2πf
● Angular velocity = how fast the object is rotating
|v| = ωr (where r = radius)
Centripetal Acceleration
● Centripetal force is the force which acts on the body towards the center of
the circle of its motion
● Centripetal acceleration is the acceleration of an object moving in circular
motion
● Called centripetal acceleration because the acceleration is always towards
the center of the circle of motion
v2
ac = r
= ω2
mv 2
F Centripetal = r
= mω 2 r
3.6.1.2 Simple Harmonic Motion (SHM)
● A body will undergo simple harmonic motion when the force that tries to
restore the object to its rest position is proportional to the displacement of
the object
● F = − kx
● This motion can be shown by a sine wave (or cosine wave)
x = Acos(ωt)
v = − ω Asin(ωt)
a = − ω 2 Acos(ωt) = − ω 2 x
M ax value of cos(ωt) and sin(ωt) = 1
⇒ M ax speed = ωA
⇒ M ax acceleration = ω 2 A
,3.6.1.3 Simple Harmonic Systems
● A spring is an example of a simple harmonic system
● In a spring system:
√
m
T = 2π k (where k is the spring constant)
● For a pendulum:
√
l
T = 2π g (where l is the length of the pendulum)
● Pendulum equation only works for small angles (see below for reason)
Derivation for Time Period of a Pendulum
F = − kx
− mg sinθ = − k x
mg sinθ
k = x
Arc length s = θr = θl (radius = length of pendulum)
x ≈ s ⇒ x = θl
mg sinθ
k = θl
S M ALL AN GLE AP P ROXIM AT ION S − θ ≈ sinθ, so cancel
mg
k = l
√
m
T = 2π k
m
T = 2π
√ mg
l
√
ml
T = 2π mg
√
l
T = 2π g
Aside
, F = kx
ma = kx
m (ω 2 Acos(ωt)) = k(Acos(ωt))
k
m = ω2
SHM and Energy
E T otal = E P otential + E Kinetic
● As the angle of the pendulum increases, kinetic energy decreases and
potential energy decreases
● As the angle decreases, kinetic energy increases and kinetic energy
decreases
● At the origin position ( x = 0 ), kinetic energy (and so velocity) is at a
maximum (no potential energy)
1 2
ET = 2
kx + 12 mv 2
ET = 1
2
k(A2 cos2 (ωt)) + 12 ( ωk2 )(A2 ω 2 sin2 (ωt))
ET = 1
2
kA2 (cos2 (ωt) + sin2 (ωt))
ET = 1
2
kA2
● From this we can find a formula for v :
1
2
kA2 = 12 kx2 + 1
2
mv 2
mv 2 = kA2 − kx2
( ωk2 )v 2 = k(A2 − x2 )
v 2 = ω 2 (A2 − x2 )
v = ± ω √A2 − x2
v max at x = 0
⇒ v max = ± ω A
