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Rotational Dynamics
----------------------
Centripetal Force:
Centripetal force is defined as a force which keeps a body moving with a uniform
speed along a circular path and is directed along the radius towards the centre.
The magnitude of the centripetal force on an object of mass m moving at tangential
speed v along a path with radius of curvature r is:
F = ma_{c} = mv^{2}/r
The acceleration, a_{c} is directed towards the center because the object
is continually changing its direction as it moves around the circle.
Proof:
S = rcosθi + rsinθj
d^{2}S/dt^{2} = rcosθ(d^{2}θ/dt^{2})i + rsinθ(d^{2}θ/dt^{2})j
= ω^{2}(rcosθi + rsinθj)
= rω^{2} since |rcosθi + rsinθj| = √(r^{2}cos^{2}θ + r^{2}sin^{2}θ) = r
= v^{2}/r since v =S/t = rθ/t = rω
Example 1.
Consider a rollercoater:
At A: PE = mgh and KE = 0
At B: PE = 0 and KE = (1/2)mv^{2}
At C: mv^{2}/R = mg
∴ v^{2} = Rg
Also,
(1/2)mv^{2} = mg(h - 2R)
substituting we get,
R/2 = h - 2R
∴ h > 2.5R for the car to stay on the track.
Example 2.
Conical pendulum:
Vertical component: Tcosθ = mg
∴ T = mg/cosθ
Horizontal component: Tsinθ = mv^{2}/r
Substituting for T gives,
mgsinθ/cosθ = mv^{2}/r
∴ v = √(rgtanθ)
= r√(g/h)
Now v = 2πr/t
∴ t = T = 2π√(h/g) = 2π√(Lcosθ/g)
for small θ, cosθ = 1 and the T is the same for the simple pendulum.
Linear versus rotational comparison:
Linear Rotational
----- ----------
x θ
v ω (v_{tangential} = 2πr/T)
a α
x = vt θ = ωt
v = u + at ω = ω_{o} + αt
v^{2} = u^{2} + 2as ω^{2} = ω_{0}^{2} + 2αθ
s = ut + (1/2)at^{2} ω = ω_{o}t + (1/2)αt^{2}
m I
F = ma τ = Iα
p = mv L = Iω
U = Fd U = τθ
K = mv^{2}/2 K = Iω^{2}/2
W = Fd/t W = τθ/t
Moment of Inertia of point mass: I = mr^{2}
Angular Momentum: L = r x p
= mvr
= mr^{2}ω
= Iω ... (1)
Angular Velocity: ω = Δθ/Δt = (1/r)Δs/Δt = v/r = 2π/T ... (2)
Angular Acceleration: α = Δω/Δt = (1/r) Δv/Δt = a/r ... (3)
Torque, τ:
τ = Force x lever arm
= FL
The lever arm is defined as the perpendicular distance from the
axis of rotation to the line of action of the force.
L
o ---------
| \
|θ \
| \
| \
v \
F' F
F' = Fcosθ
Example 1.
Consider the disc, rod and mass arrangement as shown. Assume, the
I's are given or can be calculated.
Tension in string:
φ = 90 - θ ∴ cosφ = sinθ
τ_{Mass} = 2Rm_{1}gsinθ
τ_{Rod} = Rm_{2}gsinθ
In equilibrium:
τ = TR = 2Rm_{1}gsinθ + Rm_{2}gsinθ
Angular acceleration of the disc after the string is cut:
I_{Total} = I_{Disk} + I_{Rod} + I_{Mass}
α = τ/I_{Total}
Linear acceleration of m_{1}:
From (3)
a = α2R
Linear velocity of m_{1} at the horizontal position:
U = m_{1}gh_{Mass} + m_{2}gh_{Rod}
KE = (1/2)Iω^{2}
At horizontal:
(1/2)Iω^{2} = m_{1}gh_{Mass} + m_{2}gh_{Rod}
(1/2)Iω^{2} = 2Rm_{1}gcosθ + Rm_{2}gcosθ
solve for ω
From (2)
v = ωR
Example 2.
Consider the above system. For the vertical pole assume: I = 0
and radius is r.
Downward acceleration of M:
Mg - T = Ma
∴ T = Mg - Ma
τ = Iα = Tr
∴ T = Iα/r
so
Iα/r = Mg - Ma
solve for a.
Example 3.
Translation velocity of cylinder at bottom of plane:
(1/2)mv^{2} = mgh - (1/2)Iω^{2}
but ω = v/r
∴ (1/2)mv^{2} = mgh - (1/2)Iv^{2}/R^{2}
solve for v.
Linear acceleration of cylinder COM:
mgsinθ - f = ma
τ = fR = Iα
From (3)
fR = Ia/R
=> f = Ia/R^{2}
∴ mgsinθ - Ia/R^{2} = ma
solve for a
OR
we could use v^{2} = u^{2} + 2as
=> v^{2} = 2as
∴ a = v^{2}/2s
= v^{2}(2h/sinθ)
Minimum μ for cylinder to roll without slipping:
f = μmgcosθ
ma = mgsinθ - f (cylinder just starts to slip when mgsinθ - f = 0
= mgsinθ - μmgcosθ
a = gsinθ - μgcosθ
solve for μ
Example 3.
Determine m_{2} for equilibrium:
m_{2}gr_{2} = m_{1}gr_{1}
solve for m_{2}
Angular acceleration of cylinders after m_{2} removed:
m_{1}g - T = m_{1}a
∴ T = m_{1}(g - a)
but a = αr
∴ T = m_{1}(g - αr_{1})
τ = Tr_{1}
= m_{1}(gr_{1} - αr_{1}^{2})
= I_{Total}α
Solve for α
Tension in cable supporting m_{1}:
T = m_{1}(g - αr_{1})
Linear speed of m_{1} at the time it has descended h meters:
mgh = (1/2)mv^{2} + (1/2)Iω^{2}
but ω = v/r
∴ mgh = (1/2)mv^{2} + (1/2)Iv^{2}/r^{2}
Solve for v
OR
v^{2} = 2ah
but a = αr_{1}
∴ v = √(2αr_{1}h)
OR
ω^{2} = 2αθ
but θ = h/r and v = rω
∴ v = √(2αr_{1}h)