rotational equilibrium and rotational dynamics
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Rotational Equilibrium and Rotational Dynamics. Read introduction page 226

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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 = r F

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 r-



The value of τdepends 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

Problem solving strategy for objects in equilibrium
  • Diagram system
  • Draw the free body diagram
  • Apply Στ = 0, the second condition of equilibrium
  • Apply ΣF = 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 r2α

τ = m r2α

m r2is called momentum of inertia


Torque on a rotational object: Στ =(Σm r2)α

Σm r2= m1r12+m2r22+…

The momentum of inertia of the whole body: I= Σm 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= Σm 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 = r ω

KEτ= Σ(½ m v2)

= Σ(½ mr2ω2)

= 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

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