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

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Rotational Equilibrium and Rotational Dynamics

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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 = 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

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