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Rotational Kinetic Energy and Inertia Textbook Sections 8.1 – 8.4 PHYSICS 220 Lecture 13 Rotational Kinetic Energy and Inertia Textbook Sections 8.1 – 8.4 Lecture 13 Purdue University, Physics 220 1 Overview • Last Lecture – Co llis ions an d Exp los ions • Draw “before”, “after” • Define system so that Fext = 0 • Set up coordinate system • Compute ptotal “before” •Compppute ptotal “after” • Set them equal to each other – Center of Mass (Balance Point) GG G mr11+ mr 2 2 rcm = ∑ mi • Today – Rotational Kinetic Energy – Rotational Inertia – Torque – Equilibrium Lecture 13 Purdue University, Physics 220 2 Exam 1 Average: 97.4 out ooff 150150 Std Dev: 28.7 ts tuden ss # of Lecture 13 Purdue University, Physics 220 3 Recall: Rotational Kinematics Angular Linear α = constant a = constant vv=+ at ω =+ωα0 t 0 1 2 1 2 θ =+θω00tt + α xx=+ vtat + 2 002 And for a point at a distance R from the rotation axis: x=Rx = Rθ v=v = ωR a=a = αR Lecture 13 Purdue University, Physics 220 4 Angular Velocity • Since angular velocity is a vector quantity, it must have a direction • If θ increases with time, then ω is positive – Therefore, a counterclockwise rotation corresponds to a positive angular velocity – Clockwise would be negative • When an object is rotating, all the points on the object have the same angular velocity – Makes ω a useful quantity for describing the motion • The linear velocity is not the same for all points – It depends on the distance from the rotational axis Lecture 13 Purdue University, Physics 220 5 Rotations: Axes and Sign When 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). Use the right-hand rule to determine the direction of rottitation. +ω CtCounter-clkilockwise (increas ing θ)i) is usua lly called positive. z Clockwise (decreasing θ) is usually called negative. Lecture 13 Purdue University, Physics 220 6 Angular Acceleration and Centripetal Acceleration • Angular acceleration and centripetal acceleration are different • As an example , assume a particle is moving in a circle with a constant linear velocity – The particle’s angular position increases at a constant rate, there fore its angu lar ve loc ity is cons tan t – Its angular acceleration is 0 – Since it is movinggp in a circle, it experiences a centri petal 2 acceleration of ac = v / r – This is not zero, even though the angular acceleration is zero • The centripetal acceleration refers to the linear motion of the particle • The angular acceleration is concerned with the related angular motion Lecture 13 Purdue University, Physics 220 7 Quiz 1) Where is the center of mass of a baseball bat? A) NththikdNearer the thick end B) Nearer the thin end C) Midway between the two ends 2) Can a body’s center of gravity be outside its volume? A) Yes B) No Lecture 13 Purdue University, Physics 220 8 Rotational Kinetic Energy • Consider a mass M on the end of a stringgg being spun around in a circle with radius r and angular velocity ω – Mass has speed v = ω r M – Mass has kinetic energy KE = ½ M v2 = ½ (M r2) ω2 • Rotational Kinetic Energy is energy due to circular motion of object. Lecture 13 Purdue University, Physics 220 9 An Old Example You and a friend are playing on the merry-go-round at Happy Hollow Park. You stand at the outer edge of the merry-go-rounddd and your fidtdhlfbtfriend stands halfway between the outer edge and the center. Assume the rotation rate of the merry-go-round is constant. Who has greater angular velocity? A) You do B) Your friend does C) Same Because within the same amount of time you and your friend both travel 2π. Lecture 13 Purdue University, Physics 220 10 An Old Example Who has greater tangential velocity? A) You do B) Your friend does C) Same v = r ω and since my r is great...so is my velocity (if I were to fly off!) In one rotation, the person on the outside is covering more distance in the same amount of time as the one on the inside. This means it's a faster speed. Lecture 13 Purdue University, Physics 220 11 Question Who has greater kinetic energy A) You do B) Your friend does C) Same v is greater for you because you are farther from the center so must have the largest kinetic energy Lecture 13 Purdue University, Physics 220 12 Kinetic Energy of Rotating Disk • Consider a disk with radius R and mass M, spinning with angular velocity ω r – Each “pppiece” of disk has speed vi=ωri i – Each “piece” has kinetic energy 2 KEi = ½ mi vi 2 2 = ½ mi ω ri – Combine all the pieces 2 2 ΣKEi = Σ ½ mi ω ri 2 2 = ½ (Σ mi ri ) ω =½I= ½ I ω2 Lecture 13 Purdue University, Physics 220 13 Rotational Inertia • Tells how hard it is to get an object spinning. Just like mass tells you how hard it is to get an object moving. 2 KEtran = ½ m v Linear Motion 2 KErot = ½ I ω Rotational Motion 2 2 •I =Σ miri (units kg m ) • Note! Rotational Inertia depends on what you are spinning about (basically the ri in the equation). Lecture 13 Purdue University, Physics 220 14 Rotational Inertia Table For objects with finite number of masses, use I = Σ m r2. For “continuous” objects, use table below. Lecture 13 Purdue University, Physics 220 15 Example TbthTwo batons have equa l mass an dlthd length. Which will be “easier” to spin about the center? A) Mass on ends B) Same C) Mass in center I = Σ m r2 Further mass is from axis of rotation, greater moment of itiinertia (hd(harder to spi)in) Lecture 13 Purdue University, Physics 220 16 Merry Go Round Four kids (mass m) are riding on a (light) merry-go-round rotating with angular velocity ω=3 rad/s. In case A the kids are near the cen ter (r= 1. 5 m ), in case B they are near the e dge (r=3 m). Compare the kinetic energy of the kids on the two rides. A B A) KEA > KEB B) KEA = KEB C) KEA < KEB KE = 4 x ½ m ω r2 = ½ I ω2 Where I = 4 m r2 Further mass is from axis of rotation, greater KE it has. Lecture 13 Purdue University, Physics 220 17 Massless Pulley Example Consider the two masses connected byyp a pulley as shown. Use conservation of energy to calculate the speed of the blocks after m2 has dropped a distance h. Assume the pulley is massless (ignore friction). ∑WKEPENC = Δ+Δ Note: Tension does positive work on 1 and negative work on 2. Net work (on 1 and 2) by tension is ZERO. 1122 0 =−mgh212++ mv mv 22 2mgh v = 2 22 mm12+ 2mg212h = mv+ mv Lecture 13 Purdue University, Physics 220 18 Massive Pulley Consider the two masses connected byyp a pulley as shown. If the p ulle y is massive, after m2 drops a distance h, the blocks will be moving A) faster than B) the same speed as Slower because some energy goes into spinning pulley! C) slower than if it was a massless pulley 111 mgh=+ mv222 + mv + Mv 212224 0 =ΔPEKE +Δ 111222 0 =−mgh212 + mv + mv + Iω 222 2mgh2 2 v = 111122⎛⎞⎛⎞ 2v mmM12+ + /2 mg212hMRh =+mv + mv + ⎜⎟⎜⎟MR 2222⎝⎠⎝⎠R Lecture 13 Purdue University, Physics 220 19 Torque • Torque is the product of an applied force and the distance it is applied from the support point – Denoted τ – The poi nt P i s call e d the p ivo t po in t • Since the object can rotate around that point • Only forces with a component perpendicular to the rod can contribute to the angular acceleration • Newton’s Second Law for rotational motion is written as Στ = Ι α – I is the moment of inertia Lecture 13 Purdue University, Physics 220 20 Torque • Rotational effect of force. Tells how effective fittitittibjtforce is at twisting or rotating an object. • τ = ± r Fppperpendicular = r F sin θ τ = ±rperpendicular F rperpendicular = lever arm – Units N m – Sign, CCW rotation is positive F⊥ Lecture 13 Purdue University, Physics 220 21 Torque Example A person raises one leg to an angle of 30 degrees. An ankle weight (89 N) is attached a distance of 0.84 m from her hip. What is the torque due to this weight? 1) Draw Free-BdBody Diagram 2) τ = F r sin θ = F r sin(90 – 30) =65Nm= 65 N m 30 F=89 N If she raises her leg higher, the torque due to the weight will A) Increase B) Same C) Decrease Lecture 13 Purdue University, Physics 220 22 Work Done by Torque • Recall: W = F d cos θ • For a wheel – WFW = Ftangential d = Ftangential 2 π r θ / (2 π) (θ in radians) = Ftangential r θ = τ θ F tangential to the wheel – PW/tP = W/t = /t τ θ = F (to the radius) = τ ω ⊥ Lecture 13 Purdue University, Physics 220 23 Questions • A rod is lying on a table and has two equal but opposite forces acting on it. What is the net force on the rod? A) Up B) Down C) Zero y-direction: Σ Fy = may +F – F = 0 • Will the rod move? A) Yes B) No y F Yittt!Yes, it rotates! x F Lecture 13 Purdue University, Physics 220 24 Rotational Equilibrium • Equilibrium may include rotational equilibrium • An object can be in equilibrium with regard to both its translation and its rotational motion • Its linear acceleration must be zero and its angular acceleration must be zero • The total force being zero is not sufficient to ensure both accelerations are zero • For an object to be in complete equilibrium, the angular acceleration is required to be zero • Στ = 0 • This is a necessary condition for rotational equilibrium – All the torques will be considered to refer to a single axis of rotation – The same ideas can also be applied to multiple axes Lecture 13 Purdue University, Physics 220 25 Rotational Equilibrium, Lever • Use rotational equilibrium to find the force needed to just lift the rock – We can assume that the acceleration is zero – Also ignore the mass of the lever • The force exerted by the person can be less than the weight of the rock • The lever will amplify the force exerted by the person – If Lperson > Lrock Lecture 13 Purdue University, Physics 220 26 Equilibrium • Conditions for Equilibrium Σ F = 0 Translational EQ (Center of Mass) Στ= 0 Rotational EQ • Can choose any axis of rotation….
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