Wolfram Alpha:
Search by keyword:
Astronomy
Chemistry
Classical Physics
Climate Change
Cosmology
Finance and Accounting
General Relativity
Lagrangian and Hamiltonian Mechanics
Macroeconomics
Mathematics
Microeconomics
Particle Physics
Probability and Statistics
Programming and Computer Science
Quantum Field Theory
Quantum Mechanics
Semiconductor Reliability
Solid State Electronics
Special Relativity
Statistical Mechanics
String Theory
Superconductivity
Supersymmetry (SUSY) and Grand Unified Theory (GUT)
test
The Standard Model
Topology
Units, Constants and Useful Formulas
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)