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Physics 111 Tuesday, November 2, 2002 • Ch 11: Rotational Dynamics Torque Angular Momentum Rotational Kinetic Energy Tues Phys Nov Announcements 111 .02. • Wednesday, 8 - 9 pm in NSC 118/119 • Sunday, 6:30 - 8 pm in CCLIR 468 Tues Phys Nov Announcements 111 .02. This week’s lab will be on oscillations. There will be a short quiz. Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. Let’s look at a balance scale... L X? 50 g 250 g If I hang a 50 g mass at …where should I hang the end of the left side the 250 g mass to get of the balance... the scale to balance? Worksheet #1 Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. L L/5 50 g 250 g How did we arrive at such an answer? W1 L2 Perhaps we used one of these = formulae W2 L1 m1 L1 = m2 L2 Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. L L/5 50 g 250 g Let’s restate our balance condition so that we get all the subscript 1’s on the same side and all the subscript 2’s on the same side... W1 L2 = L1W1 = L2W2 W2 L1 Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. L L/5 50 g 250 g L1W1 = L2W2 So in the case of this balance, we have... L(50 g)(9.8 m/s ) = ( L / 5)(250 g)(9.8 m/s ) 2 2 Good! The balance 490 gm/s2 L = 490 gm/s2 L is in balance! Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. L L/5 What happens to 250 g our balance if I 25 g change the mass on the left side from 50 g to 25 g? L L/5 25 g 250 g The scale begins to rotate around the pivot point. Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. Let’s look at one more case with our balance... L L X? 100 g 100 g 50 g X = L/2 How did we arrive at Worksheet #2 this answer? Now where should I hang the 100 g mass to get the scale to balance? Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. L L X? 100 g 100 g 50 g First, we examined the left side of the balance: gm L1W1 = L(100 g)(9.8 m/s ) = 980L 2 s2 Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. L L X? 100 g 100 g 50 g Then, we examined the right side: L2W2 + L3W3 = X (100 g)(9.8 m/s 2 ) + L(50 g)(9.8 m/s2 ) gm gm 980 X s 2 + 490L s2 Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. L L X? 100 g 100 g 50 g If the scale is balanced, the two sides should equal one another: gm gm gm 980L s 2 = 980 X s 2 + 490L s2 X = L/2 Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. • In our scale experiments, we’ve been playing around with a physical quantity that we’ve not yet encountered formally. • Our experiments show us that this quantity is related to a force applied at a distance. • The farther away from the pivot point the force is applied, the greater our new quantity is. Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. • When present and unbalanced, our new physical quantity causes the scale to rotate in the direction of the unbalanced force. • We’ve also seen that the quantity is additive. That is, if we have two forces on one side of the pivot at two different distances, the resulting physical quantity is simply the sum of the two force * distance products. • Before we name our new physical quantity, let’s examine one more thought experiment. Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. L 50 g Let’s put a block on one side of our balance and let go. What will eventually happen to this system? (Assume there is SOME friction in the pivot, but that the pivot is free to rotate 360o.) Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. The block eventually comes to rest with the balance aligned vertically. We still have a mass of 50 g at a distance L from the pivot L point, so why has the motion stopped? What has happened to the value of our new physical quantity? 50 g Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. For this case, we notice that the gravitational force on the mass is acting along the length of the balance. In other words, the force is parallel to the radius of length L around the pivot point. L In this case, the magnitude of our new physical quantity must be 50 g Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. Let’s finally give a name to this new physical quantity. How about... | τ |= F⊥ d Where d is the distance from the pivot point (the point of rotation) to the point at which the force is applied and… is the component of the applied F force that is perpendicular to the ⊥ line joining the pivot point to the point at which the force is applied. Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. | τ |= F⊥ d [| τ |] = [F⊥ ][d] [| τ |] = N m Although torque has the same units as energy (J), we generally denote torques as forces acting at a distance, and therefore leave the units in the form N m. Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. 0.5 m pivot Fg What is the 50 g magnitude of the torque experienced Here, the force of by this unbalanced gravity acts balance? perpendicularly to the radius joining | τ |= F⊥ d the two yellow points above. | τ |= mgd = (0.05 kg)(9.8 m/s )(0.5 m) = 0.245 Nm 2 Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. Worksheet #3 ≥ Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. Worksheet #3 You are using a wrench and trying to loosen a rusty nut. Which of the arrangements shown is most effective in loosening the nut? List in order of descending efficiency the following arrangements: 1) 1, 2, 3, 4 5) 1, 3, 4, 2 2) 4, 3, 2, 1 6) 4, 2, 3, 1 3) 2, 1, 4, 3 7) 2, 3, 4, 1 4) 2, 4, 1, 3 8) 2, 3, 1, 4 PI, Mazur (1997) Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. m Ft r pinned to table The tangential force results in a tangential acceleration. Ft = mat Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. m Ft r pinned to table It also creates a torque about the pinned point. | τ |=| Ft || r |= m | at || r | Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. m Ft r pinned to table Recalling that tangential acceleration is related to angular acceleration, we get: 2 | τ |= m | at || r |= m(| r || α |) | r |= mr | α | Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. m Ft r pinned to table This expression is valid so 2 long as our connecting τ = mr α rod/string is massless. Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. If we examine a more general system and plot the relationship, we find The slope of this τ line is known as the moment of inertia, I. α For our previous example, Note: r is the distance to the axis of rotation. I = mr 2 Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. 2Imr= [I ] = [m][r ] 2 [I ] = kg m 2 If we have an extended object, we can compute its moment of inertia about some axis from the sum of the moments of each of the individual pieces of that object: N I = m1r12 + m2r22 + m3r32 + ... + mN rN = ∑ mi ri2 2 i=1 Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. So we can rewrite our expression for torque in terms of our new quantity, the moment of τ = Iα inertia... Notice the similarity to our linear motion expression of F = ma Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. Worksheet #4 ≥ Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. Worksheet #4 Two wheels with fixed hubs, each having a mass of 1 kg, start from rest, and forces are applied as shown. Assume the hubs and spokes are massless, so that the rotational inertia is I = mR2. In order to impart identical angular accelerations, how large must F2 be? 1. 0.25 N 2. 0.5 N 3. 1 N 4. 2 N 5. 4 N PI, Mazur (1997) Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. For the ball to fall into the basket, the linear acceleration of the end of the stick must exceed that of gravity. Is that possible? 1) Yes 2) No The moment of inertia of a stick 1 rotated about its end is given by I = ML2 3 Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. τ net = F⊥ R = Iα The force creating the torque in this case is that of gravity, acting at the center of the stick. L 1 2 τ net = Mg(cosθ ) = ML α 2 3 Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. To get the linear 3g acceleration of the end of α= (cosθ ) the stick, we simply multiply 2L by the length of the stick. atan = α L = 1.5g(cosθ ) So, if the cos θ > 2/3, the end of the stick θ < 480 will move with a linear acceleration > g. Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. Recall in our Δv Δp discussion of force, F = ma = m = we discovered Δt Δt where p was the momentum p = mv in the system Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. We can construct a completely analogous argument to define angular momentum Δω ΔL τ = Iα = I = Δt Δt where L is the angular momentum L = Iω Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. L = Iω [ L] = [I ][ω ] kg m 2 [ L] = (kg m )(rad/s) = 2 s Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. When there is no NET external torque on a system, the angular momentum of the system will be conserved. Li = Iω i = Iω f = L f So when examining isolated systems, total energy, linear momentum, and angular momentum are all conserved! Conservation of Angular Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. We’ve studied this quantity already in looking at masses at some fixed distance, r, K = mv 1 2 2 from an axis of rotation. When looking at an object rotating at a constant angular velocity ω at a distance r from the axis, K = m(rω ) = mr ω = Iω 1 2 2 1 2 2 2 1 2 2 Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. K = Iω 1 2 2 [K ] = [ ][I][ω ] 1 2 2 [K ] = (kg m )(rad/s) 2 2 2 kg m [K ] = 2 = Nm = J s Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. So we’ve defined another form of kinetic energy. Before, we studied translational kinetic energy. Now we have rotational kinetic energy, too. So the total kinetic energy of a system now includes two terms (if the system is rotating and translating): Ktot = Kr + Kt Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. Worksheet #5 ≥ Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. Worksheet #5 A figure skater stands on one spot on the ice (assumed frictionless) and spins around with her arms extended. When she pulls in her arms, she reduces her rotational inertia and her angular speed increases so that her angular momentum is conserved. Compared to her initial rotational kinetic energy, her rotational kinetic energy after she has pulled in her arms must be 1. the same. 2. larger because she’s rotating faster. 3. smaller because her rotational inertia is smaller. PI, Mazur (1997) Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. So, when we examine the total energy of a system, we must take care to include the rotational kinetic energy term as well. The total energy (including rotational kinetic energy) will be conserved so long as NO non- conservative forces are acting upon the system. Etot = Kr + Kt + U g ( +U spring ) Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. We have studied objects in “equilibrium” before. What is the defining characteristic of an object in equilibrium? And from our earlier studies, the fact that it did not accelerate indicated that: Fnet = 0 Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. But is this a sufficient condition to ensure that in fact an object does not accelerate in some way? This “zero net force” condition only guarantees that the center of mass of the object on which the forces are being exerted does not accelerate. Let’s look at the meter stick again... Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. Worksheet #6 Exert a constant force tangential to the circle of motion. Top view Holding pencil through the pivot point. Draw a free-body Assume the system is diagram for this Oriented in a horizontal plane system. Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. Is the net force in this system zero? So why does the ruler rotate with increasing speed? Because the net torque is NOT zero! We have a new equilibrium condition: τ net = 0 Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. For an object to be in STATIC equilibrium (i.e., at rest, not moving, not rotating), the following TWO conditions must be met. Fnet = 0 τ net = 0 (We’re free to pick ANY origin about which to computer our torques in this case.) Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. A student sits on a rotating stool holding two weights, each of mass 3.00 kg. When his arms are extended horizontally, the weights are each 1.00 m from the axis of rotation, and he rotates at an angular speed of 0.750 rad/s. The moment of inertia of the student + stool is 3.00 kg m2 and is assumed to be constant. The student now pulls the weights horizontally in to a distance of 0.300 m from the rotation axis. (a) What’s the new angular speed? (b) What are the initial and final kinetic energies of this system? Problem #1 Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. (a) What’s the new angular speed? (b) What are the initial and final kinetic energies of this system? Pictorial Representation: Knowns: ωi Istudent = 3.00 kg m2 Initially: m = 3.00 kg ri m ri = 1.00 m rf = 0.30 m ωi = 0.75 s-1 Finally: ωf rf m Unknowns: ωf Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. (a) What’s the new angular speed? (b) What are the initial and final kinetic energies of this system? Itotal = Iweights + Istudent = 2 (mr2) + (3 kg m2) Initially: Itotal = 2 (3 kg)(1 m)2 + (3 kg m2) = 9.00 kg m2 Finally: Itotal = 2 (3 kg)(0.3 m)2 + (3 kg m2) = 3.54 kg m2 Ii ωi = If ωf Ii (9kg m 2 ) ω f = ωi = 2 (0.75rad/s) = 1.91rad/s If (3.54kg m ) Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. (a) What’s the new angular speed? (b) What are the initial and final kinetic energies of this system? Initially: Ki = 0.5 Ii ωi2 = 0.5 (9 kg m2)(0.75 rad/s)2 = 2.53 J Finally: Kf = 0.5 If ωf2 = 0.5 (3.54 kg m2)(1.91 rad/s)2 = 6.45 J Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. A uniform horizontal beam weighs 300 N, is 5.00 m long, and is attached to a wall by a pin connection that allows the beam to rotate. Its far end is supported by a cable that makes an angle of 53o with the horizontal. If a 600-N person stands 1.50 m from the wall, find the tension in the cable and the force exerted by the wall on the beam. 1.5 m Problem #2 53o 5.0 m Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. A uniform horizontal beam weighs 300 N, is 5.00 m long, and is attached to a wall by a pin connection that allows the beam to rotate. Its far end is supported by a cable that makes an angle of 53o with the horizontal. If a 600-N person stands 1.50 m from the wall, find the tension in the cable and the force exerted by the wall on the beam. Free-body T diagram f 5.0 m 53o for the N Fc W plank 1.5 m 2.5 m Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. y From the free-body T f x diagram we can write 5.0 m 53o down the torque equation N Fc W and Newton’s 2nd Law in 1.5 m the x and y directions. 2.5 m Compute the torques about the left end of the plank, where the plank meets the wall: τ net = f (0m) − Fc (1.5m) − W (2.5m) + (T sin53 )(5.0m) = 0 0 0 = 0 − (600N)(1.5m) − (300N)(2.5m) + (4.0m)T T = 412.5 N Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. y From the free-body T f x diagram we can write 5.0 m 53o down the torque equation N Fc W and Newton’s 2nd Law in 1.5 m the x and y directions. 2.5 m Newton’s 2nd Law in the x-direction: Fnet,x = N − T cos53 = 0 0 0 = N − (412.5N)cos53 0 N = 248 N Tues Phys Nov Ch 11: Rotational Dynamics 111 .02. y From the free-body T f x diagram we can write 5.0 m 53o down the torque equation N Fc W and Newton’s 2nd Law in 1.5 m the x and y directions. 2.5 m Newton’s 2nd Law in the y-direction: Fnet, y = f + T sin53 − Fc − W = 0 0 0 = f + (412.5N)sin53 − 600N − 300N 0 f = 571 N