Wolfram Alpha:

```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 = mac = mv2/r

The acceleration, ac is directed towards the center because the object
is continually changing its direction as it moves around the circle.

Proof:

S = rcosθi + rsinθj

d2S/dt2 = rcosθ(d2θ/dt2)i + rsinθ(d2θ/dt2)j

= ω2(rcosθi + rsinθj)

= rω2 since |rcosθi + rsinθj| = √(r2cos2θ + r2sin2θ) = r

= v2/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)mv2

At C: mv2/R = mg

∴ v2 = Rg

Also,

(1/2)mv2 = 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 = mv2/r

Substituting for T gives,

mgsinθ/cosθ =  mv2/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 (vtangential = 2πr/T)
a                          α
x = vt                     θ = ωt
v = u + at                 ω = ωo + αt
v2 = u2 + 2as              ω2 = ω02 + 2αθ
s = ut + (1/2)at2          ω = ωot + (1/2)αt2
m                          I
F = ma                     τ = Iα
p = mv                     L = Iω
U = Fd                     U = τθ
K = mv2/2                  K = Iω2/2
W = Fd/t                   W = τθ/t

Moment of Inertia of point mass:  I = mr2

Angular Momentum:  L = r x p
= mvr
= mr2ω
= 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

τMass = 2Rm1gsinθ

τRod = Rm2gsinθ

In equilibrium:

τ = TR = 2Rm1gsinθ +  Rm2gsinθ

Angular acceleration of the disc after the string is cut:

ITotal = IDisk + IRod + IMass

α = τ/ITotal

Linear acceleration of m1:

From (3)

a = α2R

Linear velocity of m1 at the horizontal position:

U = m1ghMass +  m2ghRod

KE = (1/2)Iω2

At horizontal:

(1/2)Iω2 =  m1ghMass +  m2ghRod

(1/2)Iω2 =  2Rm1gcosθ +  Rm2gcosθ

solve for ω

From (2)

v = ωR

Example 2.

Consider the above system.  For the vertical pole assume:  I = 0

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)mv2 = mgh - (1/2)Iω2

but ω = v/r

∴ (1/2)mv2 = mgh - (1/2)Iv2/R2

solve for v.

Linear acceleration of cylinder COM:

mgsinθ - f = ma

τ = fR = Iα

From (3)

fR = Ia/R

=> f = Ia/R2

∴ mgsinθ - Ia/R2 = ma

solve for a

OR

we could use v2 = u2 + 2as

=> v2 = 2as

∴ a = v2/2s

=  v2(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 m2 for equilibrium:

m2gr2 = m1gr1

solve for m2

Angular acceleration of cylinders after m2 removed:

m1g - T = m1a

∴ T = m1(g - a)

but a = αr

∴ T = m1(g - αr1)

τ = Tr1

= m1(gr1 - αr12)

= ITotalα

Solve for α

Tension in cable supporting m1:

T = m1(g - αr1)

Linear speed of m1 at the time it has descended h meters:

mgh = (1/2)mv2 + (1/2)Iω2

but ω = v/r

∴ mgh = (1/2)mv2 + (1/2)Iv2/r2

Solve for v

OR

v2 = 2ah

but a = αr1

∴ v = √(2αr1h)

OR

ω2 = 2αθ

but θ = h/r  and v = rω

∴ v = √(2αr1h)

```