Rotational Equilibrium and
Rotational Dynamics
• Read introduction page 226
• If F is the force acting on an object, and
r is position vector from a chosen point
O to the point of application of the
force, with F perpendicular to r. The
magnitude of the TORQUE σ exerted by
the force F is:
τ=rF
SI unit : Newton-meter (Nm)
• When an applied force causes an object to
rotate clockwise, the torque is positive
• When the forces causes an objet to rotate
counterclockwise, the torque of the object
is negative
• When two or more torques act on an
object at rest the torques are added
• The rate of rotation of an object doesn’t
change, unless the object is acted on
by a torque
• The magnitude of
the torque τ exerted
by the force F is:
τ = F r sinθ
Where rFθThe value of τ depends
on the chosen axis of
rotation
• The direction of σ is given
by the right-hand-rule
• An object in mechanical
equilibrium must satisfy:
1. The net external forces
must be zero:
ΣF=0
2. The net external torque
must be zero:
Στ=0
• We wish to locate the point
of application of the single
force of magnitude w=Fg
where the effect on the
rotation of the object is the
same as that of the
individual particles –
center of gravity
(m1g+m2g+..mng)xcg=
m1gx1+m2gx2+…mn g xn
xcg=Σmixi / Σmi
ycg=Σmiyi / Σmi
zcg=Σmizi / Σmi
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Problem solving strategy for objects in
equilibrium
Diagram system
Draw the free body diagram
Apply Σ τ = 0, the second condition of
equilibrium
Apply Σ F = 0 (on x axis and y axis)
Solve the system of ecuation
• Relationship between
torque and angular
acceleration:
Ft = mat.
Ft r = mat r
at = r α
Ft r = m r2 α
τ = m r2 α
m r2 is called momentum of
inertia
• Torque on a rotational object: Στ =(Σ m r2)α
Σ m r2= m1r12+m2r22+…
The momentum of inertia of the whole body: I= Σ m r2
Στ = I α = I α
The angular acceleration of an extended rigid
object is proportional to the net torque acting on
it
M = m1+m2+…
I= Σ m r2= m1r12+m2r22+…
I = (m1+m2+…) R2
I = MR2
• An object roatating about
some axis with an angular
speed ω has rotational
kinetic energy: ½ I ω2.
v=rω
KEτ= Σ(½ m v2)
= Σ(½ mr2 ω2)
= Σ(½ mr2 )ω2
=½ I ω2.
• Conservation of mechanical energy:
(Ex. a bowling ball rolling down the ramp)
(KEt + KEτ +PE)i = (KEt + KEτ +PE)f
KEt – translational KE
KEτ – rotational KE
PE – gravitational potential energy
Work –Energy of mechanical energy:
Wnc = ΔKEt + Δ KEτ + Δ PE
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Problem solving strategy (energy and
rotation)
Choose two points of interest
Identify conservative and
nonconservative forces
Write the work energy theorem
Substitute general expression
Use v = r ω
Solve the unknown
• Angular momentum:
An object of mass m roatates in
an circular path of radius r,
acted by a net force F, resulting
a net torque τ
Στ= Iα = I (Δω/Δt)
= I(ω –ω0) /Δt
= (Iω –Iω0) /Δt
Angular momentum: L = Iω
Στ=ΔL /Δt = change in angular
momentum / time interval
If Στ= 0, angular momentum is
conserved : Li =Lf