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Copyright © 2012 Pearson Education Inc. PowerPoint ® Lectures for University Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman Chapter.

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Presentation on theme: "Copyright © 2012 Pearson Education Inc. PowerPoint ® Lectures for University Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman Chapter."— Presentation transcript:

1 Copyright © 2012 Pearson Education Inc. PowerPoint ® Lectures for University Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman Chapter 10 Dynamics of Rotational Motion

2 Copyright © 2012 Pearson Education Inc. Goals for Chapter 10 To learn what is meant by torque To see how torque affects rotational motion To analyze the motion of a body that rotates as it moves through space To use work and power to solve problems for rotating bodies To study angular momentum and how it changes with time

3 Copyright © 2012 Pearson Education Inc. Loosen a bolt Which of the three equal-magnitude forces in the figure is most likely to loosen the bolt? F a ? F b ? F c ?

4 Copyright © 2012 Pearson Education Inc. Loosen a bolt Which of the three equal-magnitude forces in the figure is most likely to loosen the bolt? F b !

5 Copyright © 2012 Pearson Education Inc. Torque Let O be the point around which a solid body will be rotated. O

6 Copyright © 2012 Pearson Education Inc. Torque Let O be the point around which a solid body will be rotated. Apply an external force F on the body at some point. The line of action of a force is the line along which the force vector lies. F applied Line of action O

7 Copyright © 2012 Pearson Education Inc. Torque Let O be the point around which a body will be rotated. The lever arm (or moment arm) for a force is The perpendicular distance from O to the line of action of the force Line of action O L F applied

8 Copyright © 2012 Pearson Education Inc. Torque Line of action O The torque of a force with respect to O is the product of force and its lever arm.  = F L Greek Letter “Tau” = “torque” Larger torque if Larger Force applied Greater Distance from O L F applied

9 Copyright © 2012 Pearson Education Inc. Torque  = F L Larger torque if Larger Force applied Greater Distance from O

10 Copyright © 2012 Pearson Education Inc. Torque Line of action O The torque of a force with respect to O is the product of force and its lever arm.  = F L Units: Newton-Meters But wait, isn’t Nm = Joule?? L F applied

11 Copyright © 2012 Pearson Education Inc. Torque Torque Units: Newton-Meters Use N-m or m-N OR...Ft-Lbs or lb - feet Not… Lb/ft Ft/lb Newtons/meter Meters/Newton

12 Copyright © 2012 Pearson Education Inc. Torque Torque is a ROTATIONAL VECTOR! Directions: “Counterclockwise” (+) “Clockwise” (-) “z-axis”

13 Copyright © 2012 Pearson Education Inc. Torque Torque is a ROTATIONAL VECTOR! Directions: “Counterclockwise” “Clockwise”

14 Copyright © 2012 Pearson Education Inc. Torque Torque is a ROTATIONAL VECTOR! Directions: “Counterclockwise” (+) “Clockwise” (-) “z-axis”

15 Copyright © 2012 Pearson Education Inc. Torque Torque is a ROTATIONAL VECTOR! But if r = 0, no torque!

16 Copyright © 2012 Pearson Education Inc. Torque as a vector Torque can be expressed as a vector using the vector cross product:  = r x F Right Hand Rule for direction of torque.

17 Copyright © 2012 Pearson Education Inc. Torque as a vector Torque can be expressed as a vector using the vector cross product:  = r x F Where r = vector from axis of rotation to point of application of Force  = angle between r and F Magnitude:  = r F sin  F applied Line of action O r Lever arm 

18 Copyright © 2012 Pearson Education Inc. Torque as a vector Torque can be evaluated two ways:  = r x F  = r * (Fsin  ) remember sin  = sin (180-  ) F applied Line of action O r Lever arm  Fsin 

19 Copyright © 2012 Pearson Education Inc. Torque as a vector Torque can be evaluated two ways:  = r x F  = (rsin  ) * F The lever arm L = r sin  F applied Line of action O r Lever arm r sin  

20 Copyright © 2012 Pearson Education Inc. Torque as a vector Torque is zero three ways:  = r x F  = (rsin  ) * F  = r * (Fsin  ) If  is zero (F acts along r) If r is zero (f acts at axis) If net F is zero (other forces!) F applied Line of action O r Lever arm r sin  

21 Copyright © 2012 Pearson Education Inc. Applying a torque Find  if F = 900 N and  = 19 degrees

22 Copyright © 2012 Pearson Education Inc. Applying a torque Find  if F = 900 N and  = 19 degrees

23 Copyright © 2012 Pearson Education Inc. Torque and angular acceleration for a rigid body The rotational analog of Newton’s second law for a rigid body is  z =I  z. What is  for this example?

24 Copyright © 2012 Pearson Education Inc. Torque and angular acceleration for a rigid body The rotational analog of Newton’s second law for a rigid body is  z =I  z.  = I  and  = Fr so I  = Fr  = Fr/Iand a = r 

25 Copyright © 2012 Pearson Education Inc. Another unwinding cable – Example 10.3 What is acceleration of block and tension in cable as it falls?

26 Copyright © 2012 Pearson Education Inc. Rigid body rotation about a moving axis Motion of rigid body is a combination of translational motion of center of mass and rotation about center of mass Kinetic energy of a rotating & translating rigid body is KE = 1/2 Mv cm 2 + 1/2 I cm  2.

27 Copyright © 2012 Pearson Education Inc. Rolling without slipping The condition for rolling without slipping is v cm = R .

28 Copyright © 2012 Pearson Education Inc. A yo-yo What is the speed of the center of the yo-yo after falling a distance h? Is this less or greater than an object of mass M that doesn’t rotate as it falls?

29 Copyright © 2012 Pearson Education Inc. The race of the rolling bodies Which one wins?? WHY???

30 Copyright © 2012 Pearson Education Inc. Acceleration of a yo-yo We have translation and rotation, so we use Newton’s second law for the acceleration of the center of mass and the rotational analog of Newton’s second law for the angular acceleration about the center of mass. What is a and T for the yo-yo?

31 Copyright © 2012 Pearson Education Inc. Acceleration of a rolling sphere Use Newton’s second law for the motion of the center of mass and the rotation about the center of mass. What are acceleration & magnitude of friction on ball?

32 Copyright © 2012 Pearson Education Inc. Work and power in rotational motion Tangential Force over angle does work Total work done on a body by external torque is equal to the change in rotational kinetic energy of the body Power due to a torque is P =  z  z

33 Copyright © 2012 Pearson Education Inc. Work and power in rotational motion Example 10.8: Calculating Power from Torque Electric Motor provide 10-Nm torque on grindstone, with I = 2.0 kg-m 2 about its shaft. Starting from rest, find work W down by motor in 8 seconds and KE at that time. What is the average power?

34 Copyright © 2012 Pearson Education Inc. Work and power in rotational motion Analogy to 10.8: Calculating Power from Force Electric Motor provide 10-N force on block, with m = 2.0 kg. Starting from rest, find work W down by motor in 8 seconds and KE at that time. What is the average power? W = F x d = 10N x distance travelled Distance = v 0 t + ½ at 2 = ½ at 2 F = ma => a = F/m = 5 m/sec/sec => distance = 160 m Work = 1600 J; Avg. Power = W/t = 200 W

35 Copyright © 2012 Pearson Education Inc. Angular momentum The angular momentum of a rigid body rotating about a symmetry axis is parallel to the angular velocity and is given by L = I . Units = kg m2/sec (“radians” are implied!) Direction = along RHR vector of angular velocity.  

36 Copyright © 2012 Pearson Education Inc. Angular momentum For any system of particles  = dL/dt. For a rigid body rotating about the z-axis  z = I  z. Ex 10.9: Turbine with I = 2.5 kgm 2 ;  = (40 rad/s 3 )t 2 What is L(t) & t = 3.0 seconds; what is  (t)?

37 Copyright © 2012 Pearson Education Inc. Conservation of angular momentum When the net external torque acting on a system is zero, the total angular momentum of the system is constant (conserved). Follow Example using Figure below.

38 Copyright © 2012 Pearson Education Inc. A rotational “collision” Follow Example using Figure at the right.

39 Copyright © 2012 Pearson Education Inc. Angular momentum in a crime bust A bullet (mass = 400 m/s) hits a door (1.00 m wide mass = 15 kg) causing it to swing. What is  of the door?

40 Copyright © 2012 Pearson Education Inc. Angular momentum in a crime bust A bullet (mass = 400 m/s) hits a door (1.00 m wide mass = 15 kg) causing it to swing. What is  of the door? L intial = m bullet v bullet l bullet L final = I  = (I door + I bullet )  I door = Md 2 /3 I bullet = ml 2 L initial = L final so mvl = I  Solve for  and KE final

41 Copyright © 2012 Pearson Education Inc. Gyroscopes and precession For a gyroscope, the axis of rotation changes direction. The motion of this axis is called precession.

42 Copyright © 2012 Pearson Education Inc. Gyroscopes and precession

43 Copyright © 2012 Pearson Education Inc. Gyroscopes and precession For a gyroscope, the axis of rotation changes direction. The motion of this axis is called precession.

44 Copyright © 2012 Pearson Education Inc. A rotating flywheel Magnitude of angular momentum constant, but its direction changes continuously.

45 Copyright © 2012 Pearson Education Inc. A precessing gyroscopic Example 10.13


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