α = p = m v L = I ω Review: Torque Physics 201, Lecture 21 Review: Rotational Dynamics a = Στ = I α

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1 Physics 1, Lecture 1 Today s Topics q Static Equilibrium of Rigid Objects(Ch ) Review: Rotational and Translational Motion Conditions for Translational and Rotational Equilibrium Demos and Exercises q Stable and Instable Equilibrium (Section 7.9) Review: Torque q τ Fsinφ r ( τ = r F ) Torque depends on F, r, and sinφ o torque If F=, or φ= o /18 o, or r = q The torque acting on the object is proportional to its angular acceleration Στ = I α Moment if inertia Hope you have previewed! Review: Rotational Dynamics Στ = I α Rotational Dynamics compared to 1-D Dynamics Angular Linear τ F α = a = I m ω = ω + αt v v + at θ = θ ωt αt ω + αδθ ω = KE = 1 Iω L = I ω = x = x + v 1 + vt at = v + aδx KE = 1 mv p = m v Review: Motion of Rigid Object: Translation + Rotation = + 1

2 Review: Dynamics For Rigid Objects Mechanical Equilibrium (Static) q Translational Motion Σ F external = M a CM q Rotational Motion Σ τ = dl /dt (= Iα for object around fixed axis) Mechanic Equilibrium q Mechanic Equilibrium: Translational acceleration a=, AD Rotational acceleration α= q Conditions of Mechanic Equilibrium: Ø a= The net external force must be zero (think a=f/m) Σ F = F x = F y =, F Z = Ø α= The net external torque must be zero (think α = τ/i) Σ τ = Mechanical Equilibrium (Case Analyses) mg mg center of gravity for board+bottle q Static and Dynamic Equilibrium ote in equilibrium, translational velocity does not have to be Static equilibrium: v= Dynamic equilibrium: v =constant (!=) m 1 g Mg m g

3 Solving Equilibrium Problems q Draw Free Body Diagram (FBD) with exact acting point of each force q Establish convenient coordinate axes for each object. Ø Apply the First Condition of Equilibrium F net =. (ote in D this gives you F net,x = and F net,y = ) q Choose a convenient rotational axis for calculating net torque on object. and then apply τ net =. v ote: In the situation of static equilibrium, the choice of rotational axis is arbitrary. Just choose a convenient one such as a pivot (some choices are better, choose a simple one cleverly.) q Solve the resulting simultaneous equations for all of the unknowns. The Acting point of Gravity: Center of Gravity q The force of gravity acting on an object must be considered in determining equilibrium q In finding the torque produced by the force of gravity, all of the weight of the object can be considered to be concentrated at one point called center of gravity (cg) q Effectively, assuming gravitation field is uniform, the CG of an object is the same as its CM (that is usually true at the Phy13 level) x cg = Σm ix i Σm i and y cg = Σm iy i Σm i See demo: finding CG Example of a Free Body Diagram Another Example of FBD Fig 8.1, p.8 Slide 17 Important: The direction and acting point of each force must be drawn correctly. 3

4 Exercise: Seesaw q Masses of father and daughter are m f =7 kg and m d =45 kg, respectively. The seesaw bar has a mass of M=5 kg. Ø where the father has to sit to balance the bar? Identify 4 forces as in FBD. Choose the pivot point as origin for torque τ daughter = - m d g L τ seesaw = Mg x m = L= m τ = x m = τ father = m f g d Στ = d= (m d /m f )L = 1.3m Ø How much is? ΣF= = m f g+m d g+mg = (7+45+5)g =1617 Exercise: Leaning On the Wall (I) wall at an angle φ=6 o as shown. Assume there is no friction on the wall. Ø What is the friction from ground. A:.8, B:4.9, C:9.8, D: not sure as length is not given Identify forces ( F, W, mg, F friction ), draw FBD Take P as the origin for torque. τ W = W sinφ L out of page τ mg =- mgsin(9 o -φ)l/ into page τ F = τ friction = Στ= W sinφ L mgcosφl/= W =mg/(tanφ) =.83 also: = ΣF x = F friction - W F friction = W =.83 to the right Exercise: Leaning On the Wall() wall at an angle φ=6 o as shown. Assume there is no friction on the wall. Ø What is the friction from ground. Identify forces ( F, W, mg, F friction ), draw FBD Take CM of the ladder as the origin for torque. (Spend 1- min. now, more after class) Repeat steps on previous slide And verify that you get the same answer Exercise: Leaning On the Wall (3) wall at an angle φ as shown. There is no friction on the wall and the µ s between the ladder and the ground is.35. Ø What is the minimum angle φ such that ladder does not sliding down? Identify forces ( F, W, mg, F friction ), draw FBD Take P as the origin for torque. τ W = W sinφ L out of page τ mg =- mgsin(9 o -φ)l/ into page τ F = τ friction = Στ= W sinφ L mgcosφl/= W =mg/(tanφ) also: in y: F -mg= F =mg in x = ΣF x = F friction - W F friction = W = mg/(tanφ) < µ s F tanφ > 1/(µ s ) f > 55 o 4

5 Stable and Unstable Equilibriums (Section. 7.9, conceptual only) q It can be shown that when a system is in equilibrium, the first derivative of its potential energy must be zero. (du/dx=) This is consistent with conditions F=, and τ= q There are class of equilibrium: Stable and Unstable du/dx= and d U/dx > du/dx= but d U/dx < stable equilibrium at x= (lowest energy principle) unstable equilibrium at x= 5