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General Physics (PHY 2130) Lecture 21 • Rotational dynamics rolling angular momentum conservation of angular momentum rotational kinetic energy http://www.physics.wayne.edu/~apetrov/PHY2130/ Lightning Review Last lecture: 1. Rotational dynamics: Equilibrium 2nd Newton’s Law for rotation Review Problem: Two cars, one twice as heavy as the other, are at rest on a horizontal truck. A person pushes each car for 5 s. Ignoring friction and assuming equal force exerted on both cars, the momentum of the light car after the push is 1. smaller than 2. equal to 3. larger than the momentum of the heavy car. Newton’s Second Law for a Rotating Object • Recall last lecture: • The angular acceleration is directly proportional to the net torque • The angular acceleration is inversely proportional to the moment of inertia of the object Στ = Iα • There is a major difference between moment of inertia and mass: the moment of inertia depends on the quantity of matter and its distribution in the rigid object. • The moment of inertia also depends upon the location of the axis of rotation 4 Rolling Objects An object that is rolling combines translational motion (its center of mass moves) and rotational motion (points in the body rotate around the center of mass). For a rolling object: K tot = K T + K rot 1 2 1 2 = mvcm + Iω 2 2 If the object rolls without slipping then vcm = Rω. 5 Example: Two objects (a solid disk and a solid sphere) are rolling down a ramp. Both objects start from rest and from the same height. Which object reaches the bottom of the ramp first? Idea: the object with the largest linear velocity (v) at the bottom of the ramp will win the race! Apply conservation of mechanical energy: h Ei = E f θ Ui + K i = U f + K f Since the objects are also rolling, need to remember that kinetic energy has two parts: 2 1 2 1 2 1 2 1 !v$ mgh + 0 = 0 + mv + Iω = mv + I # & 2 2 2 2 "R% 1! I $ mgh = # m + 2 & v 2 Solving for v: 2" R % v= 2mgh ! I $ #m + 2 & " R % 6 Now let’s figure out which object is faster! From the table 8.1: 1 mR 2 2 2 = mR 2 5 I disk = The moments of inertia: I sphere For the disk: For the sphere: vdisk 4 = gh 3 vsphere Since Vsphere> Vdisk the sphere wins the race. 10 = gh 7 Compare these to a box sliding down the ramp: vbox = 2 gh 7 Rolling Objects: more detail How do objects in the previous example roll? y N FBD: w x Both the normal force and the weight act through the center of mass so Στ = 0. This means that the object cannot rotate when only these two forces are applied. 8 Add friction: y ∑τ = F r = Iα ∑ F = w sin θ − F = ma ∑ F = N − w cosθ = 0 s FBD: N Fs x s cm y θ w Also need acm = αR and x v 2 = v02 + 2aΔx The above system of equations can be solved for v at the bottom of the ramp. The result is the same as when using energy methods. It is static friction that makes an object roll. Example: Consider a flywheel (cylinder pulley) of mass M=5 kg and radius R=0.2 m and weight of 9.8 N hanging from rope wrapped around flywheel. What are forces acting on flywheel and weight? Find acceleration of the weight. mg N Example: I= 1 MR 2 = 0.10 kg ⋅ m 2 2 T Mg Given: masses: M = 5 kg weight: w = 9.8 N radius: R=0.2 m T mg 1. Draw all applicable forces Find: Forces: Forces=? Torques: ∑ F =mg − T = ma τ = T ⋅ R = I ⋅α need T ! α= T ⋅R I Tangential acceleration at the edge of flywheel (a=at): TR 2 at = αR or a t = I or ∑ F =mg − T = ma I 0.10 kg ⋅ m 2 T = 2 at = a = (2.5 kg )at R (0.2 m)2 t a= mg − (2.5 kg )at = ma mg 9.8 N = = 2.8 m s 2 (m + 2.5 kg ) 3.5 kg Use the example: Consider a flywheel (cylinder pulley) of mass M=5 kg and radius R=0.2 m with weight of 9.8 N hanging from rope wrapped around flywheel. What are forces acting on flywheel and weight? Find acceleration of the weight. mg If flywheel initially at rest and then begins to rotate, a torque must be present: τ = I ⋅α ⎛ ω − ω 0 ⎞ = I ⎜ ⎟, since ω = ω 0 + α Δt Δ t ⎝ ⎠ Define physical quantity: Iω = (angular momentum) ≡ L τ= change in ang. momentum ΔL = time interval Δt Angular Momentum • Similarly to the relationship between force and momentum in a linear system, we can show the relationship between torque and angular momentum • Angular momentum is defined as L = I ω ΔL τ= Δt (compare to Δp F= Δt ) • If the net torque is zero, the angular momentum remains constant • Conservation of Angular Momentum states: The angular momentum of a system is conserved when the net external torque acting on the systems is zero. • That is, when Στ = 0, Li = L f or I iω i = I f ω f Return to our example once again: Consider a flywheel (cylinder pulley) of mass M=5 kg and radius R=0.2 m with weight of 9.8 N hanging from rope wrapped around flywheel. What are forces acting on flywheel and weight? Find acceleration of the weight. mg Each small part of the flywheel is moving with some velocity. Therefore, each part and the flywheel as a whole have kinetic energy! ∑ KE pulley i =∑ mass pulley i 2 v 2 pulley i KE pulley = 1 2 Iω 2 1 1 Thus, total KE of the system: KEtot = I pulleyω 2 + mv 2 2 2 14 Angular Momentum Linear Rotational Δp Fnet = lim Δt → 0 Δt p = mv ΔL τ net = lim Δt →0 Δt L = Iω Units of p are kg m/s Units of L are kg m2/s When no net external forces act, the momentum of a system remains constant (pi = pf) Conservation of linear momentum When no net external torques act, the angular momentum of a system remains constant (Li = Lf). Conservation of angular momentum 15 Example: A turntable of mass 5.00 kg has a radius of 0.100 m and spins with a frequency of 0.500 rev/sec. What is the angular momentum? Assume a uniform disk. Given: m = 5.00 kg R = 0.100 m f = 0.500 rev/s Shape: disk (I = MR2/2) Find: Angular momentum L =? Idea: angular momentum is obtained when we know moment of intertia and the angular velocity. Determine angular velocity: rev ⎛ 2π rad ⎞ ω = 0.500 ⎜ ⎟ = 3.14 rad/sec sec ⎝ 1 rev ⎠ Knowing the shape of the turntable (disk), determine angular momentum L: ⎛ 1 ⎞ L = Iω = ⎜ MR 2 ⎟ω = 0.079 kg m 2 /s ⎝ 2 ⎠ 16 The Vector Nature of Angular Momentum Angular momentum is a vector. Its direction is defined with a right-hand rule. Curl the fingers of your right hand so that they curl in the direction a point on the object moves, and your thumb will point in the direction of the angular momentum. 17 Conservation of Angular Momentum Consider a person holding a spinning wheel. When viewed from the front, the wheel spins CCW. Holding the wheel horizontal, he steps on to a platform that is free to rotate about a vertical axis. nothing happens. 18 Conservation of Angular Momentum He then moves the wheel so that it is over his head. As a result, the platform turns CW (when viewed from above). Initially there is no angular momentum about the vertical axis. When the wheel is moved so that it has angular momentum about this axis, the platform must spin in the opposite direction so that the net angular momentum stays zero. Lf = Li 19 20 What will happen when he moves hands closer? 21 Example: A skater is initially spinning at a rate of 10.0 rad/sec with I=2.50 kg m2 when her arms are extended. What is her angular velocity after she pulls her arms in and reduces I to 1.60 kg m2? Idea: Use angular momentum conservation! The skater is on ice, so we can ignore external torques. Li = L f I i ωi = I f ω f ⎛ I i ω f = ⎜⎜ ⎝ I f ⎞ ⎛ 2.50 kg m 2 ⎞ ⎟ωi = ⎜ ⎟⎟(10.0 rad/sec) = 15.6 rad/sec 2 ⎜ ⎟ ⎝ 1.60 kg m ⎠ ⎠ Note on problem solving: • The same basic techniques that were used in linear motion can be applied to rotational motion. • Analogies: F becomes τ , m becomes I and a becomes α , v becomes ω and x becomes θ • Techniques for conservation of energy are the same as for linear systems, as long as you include the rotational kinetic energy • Problems involving angular momentum are essentially the same technique as those with linear momentum • The moment of inertia may change, leading to a change in angular momentum 23