Presentation on theme: "Physics 101: Chapter 9 Today’s lecture will cover Textbook Sections 9.1 - 9.6 1."— Presentation transcript:
1Physics 101: Chapter 9Today’s lecture will cover Textbook Sections1
2Rotation Summary (with comparison to 1-D kinematics) See text: chapter 8Rotation Summary (with comparison to 1-D kinematics)Angular LinearAnd for a point at a distance R from the rotation axis:x = Rv = R a = RSee Table 8.1
3Rotational analog of force See text: chapter 9New concept: TorqueRotational analog of forceTorque = (magnitude of force) x (lever arm)t = F l
4Comment on axes and sign (i.e. what is positive and negative) Whenever we talk about rotation, it is implied that there is a rotation “axis”.This is usually called the “z” axis (we usually omit the z subscript for simplicity).Counter-clockwise (increasing q) is usually called positive.Clockwise (decreasing q) is usually called negative.+wz
5Chapter 9, PreflightThe picture below shows three different ways of using a wrench to loosen a stuck nut. Assume the applied force F is the same in each case.In which of the cases is the torque on the nut the biggest?1. Case Case Case 3CORRECT
6Chapter 9, PreflightThe picture below shows three different ways of using a wrench to loosen a stuck nut. Assume the applied force F is the same in each case.In which of the cases is the torque on the nut the smallest?1. Case Case Case 3CORRECT
9torque about pivot due to gravity: g = mgd Static EquilibriumA system is in static equilibrium if and only if:acm = 0 Fext = 0 = 0 ext = 0 (about any axis)Center of masspivotdW=mgtorque about pivot due togravity:g = mgd(gravity acts at center of mass)This object is NOT in static equilibrium
10Not in equilibrium Equilibrium Torque about pivot 0 Center of masspivotdW=mgTorque about pivot 0Center of masspivotTorque about pivot = 0Not in equilibriumEquilibrium
11Homework Hints Painter is standing to the right of the support B. FA FBMgmgWhat is the maximum distance the painter can move to the right without tipping the board off?
12Homework Hints If its just balancing on “B”, then FA = 0 the only forces on the beam are:xFBMgmgUsing FTOT = 0: FB = Mg + mg This does not tell us x
13t (FB ) = 0 since lever arm is 0 Homework HintsFind net torque around pivot B: (or any other place)FBd1d2Mgmgt (FB ) = 0 since lever arm is 0t (Mg ) = Mgd1Total torque = 0 = Mgd1 -mgd2t (mg ) = -mgd2So d2 = Md1 /m and you can use d1 to find x
14Homework Hints Painter standing at the support B. Find total torque about this axisDFAFBdMgmgt(FA) = - FADt(Mg) = MgdTotal torque = 0 = Mgd -FADt(FB) = 0 (since distance is 0)So FA = Mgd /Dt(mg) = 0 (since distance is 0)
15t (m1g) = 0 since lever arm is 0 MORE EXAMPLES (bar and weights suspended by the string): Find net torque around this (or any other) placexTMgm2gm1gt (m1g) = 0 since lever arm is 0
16t (m1g) = 0 since lever arm is 0 t (Mg ) = -Mg L/2
17t (m1g) = 0 since lever arm is 0 xt (m1g) = 0 since lever arm is 0t (Mg ) = -Mg L/2t (T ) = T x
18t (m1g) = 0 since lever arm is 0 t (Mg ) = -Mg L/2t (T ) = T xt (m2g ) = -m2g LAll torques sum to 0: Tx = MgL/2 + m2gL So x = (MgL/2 + m2gL) / T
26Moments of Inertia of Common Objects Hollow cylinder or hoop about central axisI = MR2Solid cylinder or disk about central axisI = MR2/2Solid sphere about centerI = 2MR2/5Uniform rod about centerI = ML2/12Uniform rod about endI = ML2/3
29Chapter 9, PreflightThe picture below shows two different dumbbell shaped objects. Object A has two balls of mass m separated by a distance 2L, and object B has two balls of mass 2m separated by a distance L. Which of the objects has the largest moment of inertia for rotations around the x-axis?1. A 2. B 3. SamemCORRECT2m2LLx2mmABI = mL2 + mL2= 2mL2I = 2m(L/2)2 + 2m(L/2)2= mL2
32Define Angular Momentum See text: chapters 8-9Define Angular MomentumMomentum Angular Momentump = mV L = Iconserved if Fext = 0 conserved if ext =0Vector Vector!units: kg-m/s units: kg-m2/sSee Table 8.1
33Chapter 9, Pre-flightsYou are sitting on a freely rotating bar-stool with your arms stretched out and a heavy glass mug in each hand. Your friend gives you a twist and you start rotating around a vertical axis though the center of the stool. You can assume that the bearing the stool turns on is frictionless, and that there is no net external torque present once you have started spinning.You now pull your arms and hands (and mugs) close to your body.
34Chapter 9, PreflightWhat happens to your angular momentum as you pull in your arms?1. it increases 2. it decreases 3. it stays the sameL1L2CORRECTThis is like the spinning skater example in the book. Since the net external torque is zero (the movement of the arms and hands involve internal torques), the angular momentum does not change.
35Chapter 9, Preflight w1 w2 I1 I2 L What happens to your angular velocity as you pull in your arms?1. it increases 2. it decreases 3. it stays the samew1w2I2I1LCORRECTas with the skater example given in the book....as you pull your arms in toward the rotational axis, the moment of inertia decreases, and the angular velocity increases.My friends and I spent a good half hour doing this once, and I can say...based on a great deal of nausea, that the angular velocity does increase.
36Chapter 9, Preflight w1 w2 I1 I2 L (using L = I ) What happens to your kinetic energy as you pull in your arms?1. it increases 2. it decreases 3. it stays the samew1w2I2I1LCORRECT(using L = I )Your angular velocity increases and moment of inertia decreases, but angular velocity is squared, so KE will increase with increasing angular velocity
37Spinning disksTwo different spinning disks have the same angular momentum, but disk 2 has a larger moment of inertia than disk 1.Which one has the biggest kinetic energy ?(a) disk (b) disk 2
38(using L = I )If they have the same L, the one with the smallest I will have the biggest kinetic energy.w2disk 2w1disk 1I1 < I2
39Preflights: Turning the bike wheel A student sits on a barstool holding a bike wheel. The wheel is initially spinning CCW in the horizontal plane (as viewed from above). She now turns the bike wheel over. What happens?1. She starts to spin CCW. 2. She starts to spin CW. 3. NothingCORRECT
40Turning the bike wheel... Initially: LINI = LW,I Since there is no net external torque acting on the student-stool system, angular momentum is conserved.Remenber, L has a direction as well as a magnitude!Initially: LINI = LW,IFinally: LFIN = LW,F + LSLSLW,ILW,I = LW,F + LSLW,F
41Rotation Summary (with comparison to 1-d linear motion) See text: chapters 8-9Rotation Summary (with comparison to 1-d linear motion)Angular LinearSee Table 8.1