1. Rotational Equilibrium and Rotational Dynamics

2. 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)

3. 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

4. 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

5. 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

6. 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 (m 1 g+m 2 g+..m n g)x cg = m 1 gx 1 +m 2 gx 2 +…m n g x n x cg =Σm i x i / Σmi y cg =Σm i y i / Σmi z cg =Σm i z i / Σmi

7. 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

8. Relationship between torque and angular acceleration: F t = ma t. F t r = ma t r a t = r α F t r = m r 2 α τ = m r 2 α m r 2 is called momentum of inertia

9. Torque on a rotational object: Στ =(Σ m r 2 )α Σ m r 2 = m 1 r 1 2 +m 2 r 2 2 +… The momentum of inertia of the whole body: I= Σ m r 2 Στ = I α = I α The angular acceleration of an extended rigid object is proportional to the net torque acting on it

10. M = m 1 +m 2 +… I= Σ m r 2= m 1 r 1 2 +m 2 r 2 2 +… I = (m 1 +m 2 +…) R 2 I = MR 2

11. An object roatating about some axis with an angular speed ω has rotational kinetic energy: ½ I ω 2. v = r ω KE τ = Σ(½ m v 2 ) = Σ(½ mr 2 ω 2 ) = Σ(½ mr 2 )ω 2 =½ I ω 2.

12. Conservation of mechanical energy: (Ex. a bowling ball rolling down the ramp) (KE t + KE τ +PE) i = (KE t + KE τ +PE) f KE t – translational KE KE τ – rotational KE PE – gravitational potential energy Work –Energy of mechanical energy: W nc = ΔKE t + Δ KE τ + Δ PE

13. 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

14. 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 –ω 0 ) /Δt = (Iω –Iω 0 ) /Δt Angular momentum: L = Iω Στ=ΔL /Δt = change in angular momentum / time interval If Στ= 0, angular momentum is conserved : L i =L f