rotational equilibrium and rotational dynamics. read introduction page 226 if f is the force acting...
TRANSCRIPT
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-
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 = 0
2. 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 / Σmi
ycg=Σmiyi / Σmi
zcg=Σmizi / Σmi
• Problem solving strategy for objects in equilibrium
1. Diagram system
2. Draw the free body diagram
3. Apply Σ τ = 0, the second condition of equilibrium
4. 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 α
m r2 is called momentum of inertia
• Torque on a rotational object: Στ =(Σ m r2)α
Σ m r2= m1r12+m2r2
2+…
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+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 ω
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
• Problem solving strategy (energy and rotation)
1. Choose two points of interest
2. Identify conservative and nonconservative forces
3. Write the work energy theorem
4. Substitute general expression
5. Use v = r ω
6. 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