ics Tuesday, ember 2, 2002 Ch 11: Rotational Dynamics Torque Angular Momentum Rotational Kinetic Energy
Announcements Wednesday, 8-9 pm in NSC 118/119 Sunday, 6:30-8 pm in CCLIR 468
Announcements This week s lab will be on oscillations. There will be a short quiz.
Let s look at a balance scale... L X? 50 g 250 g If I hang a 50 g mass at the end of the left side of the balance... where should I hang the 250 g mass to get the scale to balance? Worksheet #1
L L/5 50 g 250 g How did we arrive at such an answer? W 1 W 2 = L 2 L 1 Perhaps we used one of these formulae m 1 L 1 = m 2 L 2
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... W 1 W 2 = L 2 L 1 L 1 W 1 = L 2 W 2
L L/5 50 g 250 g L 1 W 1 = L 2 W 2 So in the case of this balance, we have... L(50 g)(9.8 m/s 2 ) = (L / 5)(250 g)(9.8 m/s 2 ) 490 gm/s 2 L = 490 gm/s 2 L Good! The balance is in balance!
25 g L L L/5 250 g L/5 What happens to our balance if I change the mass on the left side from 50 g to 25 g? 25 g 250 g The scale begins to rotate around the pivot point.
Let s look at one more case with our balance... L L X? 100 g 100 g 50 g X = L/2 Worksheet #2 How did we arrive at this answer? Now where should I hang the 100 g mass to get the scale to balance?
L L 100 g X? 100 g 50 g First, we examined the left side of the balance: L 1 W 1 = L(100 g)(9.8 m/s 2 ) = 980L gm s 2
L L 100 g X? 100 g 50 g Then, we examined the right side: L 2 W 2 + L 3 W 3 = X (100 g)(9.8 m/s 2 ) + L(50 g)(9.8 m/s 2 ) 980X gm s 2 + 490L gm s 2
L L 100 g X? 100 g 50 g If the scale is balanced, the two sides should equal one another: 980L gm s 2 = 980 X gm s 2 + 490L gm s 2 X = L / 2
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.
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.
L Let s put a block on one side of our balance and let go. 50 g What will eventually happen to this system? (Assume there is SOME friction in the pivot, but that the pivot is free to rotate 360 o.)
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 point, so why has the motion stopped? What has happened to the value of our new physical quantity? L 50 g
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. In this case, the magnitude of our new physical quantity must be L 50 g
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.
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.
0.5 m What is the magnitude of the torque experienced by this unbalanced balance? τ = F d pivot 50 g F g Here, the force of gravity acts perpendicularly to the radius joining the two yellow points above. τ = mgd = (0.05 kg)(9.8 m/s 2 )(0.5 m) = 0.245 Nm
Worksheet #3
Tues 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)
m r F t pinned to table The tangential force results in a tangential acceleration. F t = ma t
m r F t It also creates a torque about the pinned point. τ = pinned to table F r = m a r t t
m r F t pinned to table Recalling that tangential acceleration is related to angular acceleration, we get: τ = m a t r = m( r α ) r = mr 2 α
m r F t pinned to table This expression is valid so long as our connecting rod/string is massless. τ = mr 2 α
If we examine a more general system and plot the relationship, we find τ Note: r is the distance to the axis of rotation. α The slope of this line is known as the moment of inertia, I. For our previous example, I = mr 2
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=1 I = m 1 r 1 2 + m 2 r 2 2 + m 3 r 3 2 +... + m N r N 2 = m i r i 2
So we can rewrite our expression for torque in terms of our new quantity, the moment of inertia... τ = Iα Notice the similarity to our linear motion expression of F = ma
Worksheet #4
Tues 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 = mr 2. In order to impart identical angular accelerations, how large must F 2 be? 1. 0.25 N 2. 0.5 N 3. 1 N 4. 2 N 5. 4 N PI, Mazur (1997)
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 rotated about its end is given by I = 1 3 ML2
τ = F R = Iα net The force creating the torque in this case is that of gravity, acting at the center of the stick. τ = Mg(cosθ) L net 2 = 1 3 ML2 α
α = 3g 2L (cosθ) To get the linear acceleration of the end of the stick, we simply multiply by the length of the stick. a tan = α L = 1.5g(cosθ) So, if the cos θ > 2/3, the end of the stick will move with a linear acceleration > g. θ < 48 0
Recall in our discussion of force, we discovered F = ma = m Δ v Δt = Δ p Δt where p was the momentum in the system p = mv
We can construct a completely analogous argument to define angular momentum τ = Iα = I Δ ω Δt = Δ L Δt where L is the angular momentum L = Iω
L = Iω [ L] = [I][ ω] [ L] = (kg m 2 )(rad/s) = kg m2 s
When there is no NET external torque on a system, the angular momentum of the system will be conserved. L = Iω = Iω = L i i f f So when examining isolated systems, total energy, linear momentum, and angular momentum are all conserved! Conservation of Angular
We ve studied this quantity already in looking at masses at some fixed distance, r, from an axis of rotation. K = 1 2 mv 2 When looking at an object rotating at a constant angular velocity ω at a distance r from the axis, 1 2 2 1 2 2 2 1 2 K = m( rω) = mr ω = Iω 2
K = 1 2 Iω 2 [K] = [ 1 2 ][I][ω 2 ] [K] = (kg m 2 )(rad/s) 2 [K] = kg m2 s 2 = Nm = J
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): K = K + K tot r t
Worksheet #5
Tues 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)
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 nonconservative forces are acting upon the system. E = K + K + U ( + U ) tot r t g spring
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: F net = 0
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...
Exert a constant force tangential to the circle of motion. Worksheet #6 Top view Holding pencil through the pivot point. Draw a free-body diagram for this system. Assume the system is Oriented in a horizontal plane
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
For an object to be in STATIC equilibrium (i.e., at rest, not moving, not rotating), the following TWO conditions must be met. F net = 0 τ = 0 net (We re free to pick ANY origin about which to computer our torques in this case.)
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 m 2 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
(a) What s the new angular speed? (b) What are the initial and final kinetic energies of this system? Pictorial Representation: Initially: r i Finally: r f ω i ω f m m Knowns: I student = 3.00 kg m 2 m = 3.00 kg r i = 1.00 m r f = 0.30 m ω i = 0.75 s -1 Unknowns: ω f
(a) What s the new angular speed? (b) What are the initial and final kinetic energies of this system? I total = I weights + I student = 2 (mr 2 ) + (3 kg m 2 ) Initially: Finally: I total = 2 (3 kg)(1 m) 2 + (3 kg m 2 ) = 9.00 kg m 2 I total = 2 (3 kg)(0.3 m) 2 + (3 kg m 2 ) = 3.54 kg m 2 I i ω i = I f ω f ω f = I i ω I i = (9kg m 2 ) (0.75rad/s) = 1.91rad/s f (3.54kg m 2 )
(a) What s the new angular speed? (b) What are the initial and final kinetic energies of this system? Initially: K i = 0.5 I i ω i 2 = 0.5 (9 kg m 2 )(0.75 rad/s) 2 = 2.53 J Finally: K f = 0.5 I f ω f 2 = 0.5 (3.54 kg m 2 )(1.91 rad/s) 2 = 6.45 J
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 53 o 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 53 o 5.0 m
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 53 o 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 diagram for the plank f N F c 1.5 m 5.0 m 53 o T W 2.5 m
From the free-body diagram we can write down the torque equation and Newton s 2 nd Law in the x and y directions. f N y x F c 1.5 m 2.5 m 5.0 m 53 o T W Compute the torques about the left end of the plank, where the plank meets the wall: τ net = f (0m) F c (1.5m) W (2.5m) + (T sin53 0 )(5.0m) = 0 0 = 0 (600N)(1.5m) (300N)(2.5m) + (4.0m)T T = 412.5 N
From the free-body diagram we can write down the torque equation and Newton s 2 nd Law in the x and y directions. Newton s 2 nd Law in the x-direction: f N y x F c 1.5 m 2.5 m 5.0 m 53 o T W F net,x = N T cos53 0 = 0 0 = N (412.5N)cos53 0 N = 248 N
From the free-body diagram we can write down the torque equation and Newton s 2 nd Law in the x and y directions. Newton s 2 nd Law in the y-direction: f N y x F c 1.5 m 2.5 m 5.0 m 53 o T W F net, y = f + T sin53 0 F c W = 0 0 = f + (412.5N)sin53 0 600N 300N f = 571 N