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Rotational Equilibrium and Rotational Dynamics. Read introduction page 226 - PowerPoint PPT Presentation
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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 FSI 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 sinθWhere r- F- θ-The 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:
Σ F = 02. The net external torque
must be zero: Σ τ = 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 / Σmiycg=Σmiyi / Σmizcg=Σmizi / Σmi
• Problem solving strategy for objects in equilibrium
1. Diagram system2. Draw the free body diagram3. Apply Σ τ = 0, the second condition of
equilibrium4. Apply Σ F = 0 (on x axis and y axis)5. 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 αm r2 is called momentum of
inertia
• Torque on a rotational object: Στ =(Σ m r2)αΣ m r2= m1r1
2+m2r22+…
The momentum of inertia of the whole body: I= Σ m r2
Στ = I α = I αThe angular acceleration of an extended rigid object
is proportional to the net torque acting on it
M = m1+m2+…
I= Σ m r2= m1r12+m2r2
2+…
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 ωKEτ= 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 KEKEτ – rotational KEPE – gravitational potential energyWork –Energy of mechanical energy:Wnc = ΔKEt + Δ KEτ + Δ PE
• Problem solving strategy (energy and rotation)
1. Choose two points of interest2. Identify conservative and
nonconservative forces3. Write the work energy theorem4. Substitute general expression5. Use v = r ω6. 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 = IωΣτ=ΔL /Δt = change in angular
momentum / time intervalIf Στ= 0, angular momentum is
conserved : Li =Lf