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Lecture 15 Rotational Dynamics.

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Presentation on theme: "Lecture 15 Rotational Dynamics."— Presentation transcript:

1 Lecture 15 Rotational Dynamics

2 Reading and Review

3 Moment of Inertia The moment of inertia I:
The total kinetic energy of a rolling object is the sum of its linear and rotational kinetic energies:

4 Rolling Down Two spheres start rolling down a ramp from the same height at the same time. One is made of solid gold, and the other of solid aluminum. Which one reaches the bottom first? a) solid aluminum b) solid gold c) same d) can’t tell without more information Answer: b

5 Mass and radius don’t matter, only the distribution of mass (shape)!
Rolling Down Two spheres start rolling down a ramp from the same height at the same time. One is made of solid gold, and the other of solid aluminum. Which one reaches the bottom first? a) solid aluminum b) solid gold c) same d) can’t tell without more information Answer: b Moment of inertia depends on mass and distance from axis squared. For a sphere: I = 2/5 MR2 But you don’t need to know that! All you need to know is that it depends on MR2 initial PE: mgh final KE: MR2 cancels out! Mass and radius don’t matter, only the distribution of mass (shape)!

6 Moment of Inertia Two spheres start rolling down a ramp at the same time. One is made of solid aluminum, and the other is made from a hollow shell of gold. Which one reaches the bottom first? a) solid aluminum b) hollow gold c) same d) can’t tell without more information

7 Moment of Inertia Two spheres start rolling down a ramp at the same time. One is made of solid aluminum, and the other is made from a hollow shell of gold. Which one reaches the bottom first? a) solid aluminum b) hollow gold c) same d) can’t tell without more information initial PE: mgh A solid sphere has more of its mass close to the center. A shell has all of its mass at a large radius. final KE: A shell has a larger moment of inertia than a solid object of the same mass, radius and shape Larger moment of inertia -> lower velocity for the same energy.

8 Power output of the Crab pulsar
Power output of the Crab pulsar, in radio and X-rays, is about 6 x 1031 W (which is about 150,000 times the power output of our sun). Since the pulsar is out of nuclear fuel, where does all this energy come from ? The angular speed of the pulsar, and so the rotational kinetic energy, is going down over time. This kinetic energy is converted into the energy coming out of that star. calculate the rotational kinetic energy at the beginning and at the end of a second, by taking the moment of inertia to be 1.2x1038 kg-m2 and the initial angular speed to be 190 s-1. Δω over one second is given by the angular acceleration.

9 Power output of the Crab pulsar
Power output of the Crab pulsar, in radio and X-rays, is about 6 x 1031 W (which is about 150,000 times the power output of our sun). Since the pulsar is out of nuclear fuel, where does all this energy come from ? The angular speed of the pulsar, and so the rotational kinetic energy, is going down over time. This kinetic energy is converted into the energy coming out of that star. calculate the rotational kinetic energy at the beginning and at the end of a second, by taking the moment of inertia to be 1.2x1038 kg-m2 and the initial angular speed to be 190 s-1. Δω over one second is given by the angular acceleration.

10 Torque We know that the same force will be much more effective at rotating an object such as a nut or a door if our hand is not too close to the axis. This is why we have long- handled wrenches, and why doorknobs are not next to hinges.

11 The torque increases as the force increases, and also as the distance increases.

12 Only the tangential component of force causes a torque

13 A more general definition of torque:
Fsinθ Fcosθ Right Hand Rule You can think of this as either: - the projection of force on to the tangential direction OR - the perpendicular distance from the axis of rotation to line of the force

14 Torque If the torque causes a counterclockwise angular acceleration, it is positive; if it causes a clockwise angular acceleration, it is negative.

15 e) all are equally effective
Using a Wrench a c d b You are using a wrench to loosen a rusty nut. Which arrangement will be the most effective in tightening the nut? Answer: b e) all are equally effective

16 e) all are equally effective
Using a Wrench a c d b You are using a wrench to loosen a rusty nut. Which arrangement will be the most effective in tightening the nut? Because the forces are all the same, the only difference is the lever arm. The arrangement with the largest lever arm (case #2) will provide the largest torque. e) all are equally effective

17 The gardening tool shown is used to pull weeds. If a 1
The gardening tool shown is used to pull weeds. If a 1.23 N-m torque is required to pull a given weed, what force did the weed exert on the tool? What force was used on the tool?

18 Force and Angular Acceleration
Consider a mass m rotating around an axis a distance r away. Newton’s second law: a = r α Or equivalently,

19 Torque and Angular Acceleration
Once again, we have analogies between linear and angular motion:

20 The L-shaped object shown below consists of three masses connected by light rods. What torque must be applied to this object to give it an angular acceleration of 1.2 rad/s2 if it is rotated about (a) the x axis, (b) the y axis (c) the z axis (through the origin and perpendicular to the page) (a) (b) (c)

21 Torque Project the force onto the tangential direction
Only the tangential component of force causes a torque Fsinθ Fcosθ Project the force onto the tangential direction

22 Torque and Angular Acceleration
Angular motion is analogous to linear motion

23 The L-shaped object shown below consists of three masses connected by light rods. What torque must be applied to this object to give it an angular acceleration of 1.2 rad/s2 if it is rotated about an axis parallel to the y axis, and through the 2.5kg mass?

24 The L-shaped object shown below consists of three masses connected by light rods. What torque must be applied to this object to give it an angular acceleration of 1.2 rad/s2 if it is rotated about an axis parallel to the y axis, and through the 2.5kg mass?

25 A 2. 85-kg bucket is attached to a disk-shaped pulley of radius 0
A 2.85-kg bucket is attached to a disk-shaped pulley of radius m and mass kg. If the bucket is allowed to fall, (a) what is its linear acceleration? (b) What is the angular acceleration of the pulley? (c) How far does the bucket drop in 1.50 s?

26 A 2. 85-kg bucket is attached to a disk-shaped pulley of radius 0
A 2.85-kg bucket is attached to a disk-shaped pulley of radius m and mass kg. If the bucket is allowed to fall, (a) what is its linear acceleration? (b) What is the angular acceleration of the pulley? (c) How far does the bucket drop in 1.50 s? (a) Pulley spins as bucket falls (b) (c)

27 Static Equilibrium Static equilibrium describes an object at rest – neither rotating nor translating. If the net torque is zero, it doesn’t matter which axis we consider rotation to be around; you choose the axis of rotation This can greatly simplify a problem X

28 Center of Mass and Gravitational Force on an Extended Object
xj mj m1 Fj = mj g ... X axis of rotation center of mass mj m1 xj ... xcm F = Mg X axis of rotation

29 Balance If an extended object is to be balanced, it must be supported through its center of mass.

30 Center of Mass and Balance
This fact can be used to find the center of mass of an object – suspend it from different axes and trace a vertical line. The center of mass is where the lines meet.

31 Balancing Rod 1m 1kg a) ¼ kg
b) ½ kg c) 1 kg d) 2 kg e) 4 kg A 1-kg ball is hung at the end of a rod 1-m long. If the system balances at a point on the rod 0.25 m from the end holding the mass, what is the mass of the rod? Answer: c 1kg 1m

32 Balancing Rod a) ¼ kg b) ½ kg c) 1 kg d) 2 kg e) 4 kg A 1-kg ball is hung at the end of a rod 1-m long. If the system balances at a point on the rod 0.25 m from the end holding the mass, what is the mass of the rod? The total torque about the pivot must be zero !! The CM of the rod is at its center, 0.25 m to the right of the pivot. Because this must balance the ball, which is the same distance to the left of the pivot, the masses must be the same !! 1 kg X CM of rod same distance mROD = 1 kg

33 When you arrive at Duke’s Dude Ranch, you are greeted by the large wooden sign shown below. The left end of the sign is held in place by a bolt, the right end is tied to a rope that makes an angle of 20.0° with the horizontal. If the sign is uniform, 3.20 m long, and has a mass of 16.0 kg, what is (a) the tension in the rope, and (b) the horizontal and vertical components of the force, exerted by the bolt?

34 (a) the tension in the rope, and
When you arrive at Duke’s Dude Ranch, you are greeted by the large wooden sign shown below. The left end of the sign is held in place by a bolt, the right end is tied to a rope that makes an angle of 20.0° with the horizontal. If the sign is uniform, 3.20 m long, and has a mass of 16.0 kg, what is (a) the tension in the rope, and (b) the horizontal and vertical components of the force exerted by the bolt? Torque, vertical force, and horizontal force are all zero But I don’t know two of the forces! I can get rid of one of them, by choosing my axis of rotation where the force is applied. Choose the bolt as the axis of rotation, then: (b)

35 Dumbbell I a) case (a) b) case (b) c) no difference d) it depends on the rotational inertia of the dumbbell A force is applied to a dumbbell for a certain period of time, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater center-of-mass speed ? Answer: c

36 Dumbbell I a) case (a) b) case (b) c) no difference d) it depends on the rotational inertia of the dumbbell A force is applied to a dumbbell for a certain period of time, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater center-of-mass speed ? Because the same force acts for the same time interval in both cases, the change in momentum must be the same, thus the CM velocity must be the same.

37 F = ma implies Newton’s first law:
without a force, there is no acceleration Now we have Linear momentum was the concept that tied together Newton’s Laws, is there something similar for rotational motion?

38 Angular Momentum Consider a particle moving in a circle of radius r,
I = mr2 L = Iω = mr2ω = rm(rω) = rmvt = rpt

39 Angular Momentum For more general motion (not necessarily circular),
The tangential component of the momentum, times the distance

40 Angular Momentum For an object of constant moment of inertia, consider the rate of change of angular momentum analogous to 2nd Law for Linear Motion

41 Conservation of Angular Momentum
If the net external torque on a system is zero, the angular momentum is conserved. As the moment of inertia decreases, the angular speed increases, so the angular momentum does not change.

42 Conservation of Angular Momentum
Angular momentum is also conserved in rotational collisions

43 Figure Skater a) the same b) larger because she’s rotating faster
A figure skater spins with her arms extended. When she pulls in her arms, she reduces her rotational inertia and spins faster so that her angular momentum is conserved. Compared to her initial rotational kinetic energy, her rotational kinetic energy after she pulls in her arms must be: a) the same b) larger because she’s rotating faster c) smaller because her rotational inertia is smaller

44 Figure Skater a) the same b) larger because she’s rotating faster
A figure skater spins with her arms extended. When she pulls in her arms, she reduces her rotational inertia and spins faster so that her angular momentum is conserved. Compared to her initial rotational kinetic energy, her rotational kinetic energy after she pulls in her arms must be: a) the same b) larger because she’s rotating faster c) smaller because her rotational inertia is smaller KErot = I 2 = L2 / I (used L = I ). Because L is conserved, smaller I means larger KErot. The “extra” energy comes from the work she does on her arms.

45 Rotational Work s = r Δθ W = (r Δθ) F = rF Δθ = τ Δθ τ = r F
A torque acting through an angular displacement does work, just as a force acting through a distance does. Consider a tangential force on a mass in circular motion: τ = r F Work is force times the distance on the arc: s = r Δθ W = s F W = (r Δθ) F = rF Δθ = τ Δθ The work-energy theorem applies as usual.

46 Rotational Work and Power
Power is the rate at which work is done, for rotational motion as well as for translational motion. Again, note the analogy to the linear form (for constant force along motion):

47 The Vector Nature of Rotational Motion
The direction of the angular velocity vector is along the axis of rotation. A right-hand rule gives the sign. Right-hand Rule: your fingers should follow the velocity vector around the circle

48 The Torque Vector Similarly, the right-hand rule gives the direction of the torque vector, which also lies along the assumed axis or rotation Right-hand Rule: your fingers should follow the force vector around the circle

49 The linear momentum of components related to the vector angular momentum of the system

50 Applied tangential force related to the torque vector

51 Applied torque over time related to change in the vector angular momentum.

52 Cassette Player When a tape is played on a cassette deck, there is a tension in the tape that applies a torque to the supply reel. Assuming the tension remains constant during playback, how does this applied torque vary as the supply reel becomes empty? a) torque increases b) torque decreases c) torque remains constant Answer: b

53 Cassette Player When a tape is played on a cassette deck, there is a tension in the tape that applies a torque to the supply reel. Assuming the tension remains constant during playback, how does this applied torque vary as the supply reel becomes empty? a) torque increases b) torque decreases c) torque remains constant a) torque increases b) torque decreases c) torque remains constant Answer: b As the supply reel empties, the lever arm decreases because the radius of the reel (with tape on it) is decreasing. Thus, as the playback continues, the applied torque diminishes.


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