Rotation in Engineering Physics: Lecture 10

1. Chapter 10 Rotation

2. Engineering Physics: Lecture 10Topics • Rotational Kinematics • Analogy with one-dimensional kinematics • Kinetic energy of a rotating system • Moment of inertia • Discrete particles • Continuous solid objects • Parallel axis theorem

3. Rotation Introduction • Up until now we have gracefully avoided dealing with the rotation of objects. • We have studied objects that slide, not roll. • We have assumed pulleys are without mass. • Rotation is extremely important, however, and we need to understand it! • Most of the equations we will develop are simply rotational analogues of ones we have already learned when studying linear kinematics and dynamics.

4. The Rotational Variables A rigid body is a body that can rotate with all its parts locked together and without any change in its shape. A fixed axis means that the rotation occurs about an axis that does not move. Figure skater Sasha Cohen in motion of pure rotation about a vertical axis. (Elsa/Getty Images, Inc.)

5. The Rotational Variables

6. The Rotational Variables: Angular Position Here s is the length of a circular arc that extends from the x axis (the zero angular position) to the reference line, and r is the radius of the circle. An angle defined in this way is measured in radians (rad).

7. The Rotational Variables: Angular Displacement If a body rotates about the rotation axis as in, changing the angular position of the reference line from q1 to q2, the body undergoes an angular displacement Dq given by An angular displacement in the counterclockwise direction is positive, and one in the clockwise direction is negative.

8. The Rotational Variables: Angular Velocity Suppose that our rotating body is at angular position q1 at time t1 and at angular position q2 at time t2. Then the average angular velocity of the body in the time interval t from t1 to t2 to be. The instantaneous angular velocity w is the limit of the ratio as Dt approaches zero.

9. Relation Between Linear and Angular Quantities • In Cartesian coordinates, we say velocity dx/dt = v. • x = vt • In polar coordinates, angular velocity d/dt = . •  = t •  has units of radians/second. • Displacement s = vt. but s = r = rt, so: y v r s t x v = r

10. QuestionRotations • Bonnie sits on the outer rim of a merry-go-round, and Klyde sits midway between the center and the rim. The merry-go-round makes one complete revolution every two seconds. • Klyde’s angular velocity is: (a)the same as Bonnie’s (b)twice Bonnie’s (c)half Bonnie’s

11. AnswerRotations • The angular velocity w of any point on a solid object rotating about a fixed axis is the same. • Both Bonnie & Klyde go around once (2p radians) every two seconds. (Their “linear” speed v will be different since v = wr). w

12.  Period and Frequency • Recall that 1 revolution = 2 radians • frequency(f) = revolutions / second (a) • angular velocity () = radians / second(b) • By combining(a)and (b) •  = 2 f • Realize that: • period (T) = seconds / revolution • So T = 1 / f = 2/ v R s  = 2 / T = 2f

13. Summary x = R cos()= R cos(t) y = R sin()= R sin(t) = arctan(y/x) =t s=v t s=R=Rt v=R v (x,y) R s t

14. Are Angular Quantities Vectors?

15. The Rotational Variables: Angular Acceleration If the angular velocity of a rotating body is not constant, then the body has an angular acceleration. If w2 and w1 be its angular velocities at times t2 and t1, respectively, then the average angular acceleration of the rotating body in the interval from t1 to t2 is defined as The instantaneous angular acceleration a, is the limit of this quantity as Dt approaches zero. These relations hold for every particle of that body. The unit of angular acceleration is (rad/s2).

16. ^ ^  r Aside: Polar Unit Vectors • We are familiar with the Cartesian unit vectors: i j kNow introduce“polar unit-vectors” r and  : • r points in radial direction •  points in tangential direction ^ ^ ^ ^ (counter clockwise) y R  j x i

17. Centripetal Acceleration • UCM results in acceleration: • Magnitude: a= v2 / R • Direction: - r (toward center of circle) ^ R a 

18. Derivation: We know that and v = R Substituting for v we find that:  a =2R

19. ProblemUniform Circular Motion • A fighter pilot flying in a circular turn will pass out if the centripetal acceleration he experiences is more than about 9 times the acceleration of gravity g. If his F18 is moving with a speed of 300 m/s, what is the approximate diameter of the tightest turn this pilot can make and survive to tell about it ? (a) 500 m (b) 1000 m (c) 2000 m

20. 2km Solution

21. Example: Propeller Tip • The propeller on a stunt plane spins with frequency f = 3500 rpm. The length of each propeller blade is L = 80cm. What centripetal acceleration does a point at the tip of a propeller blade feel? f what is a here? L

22. Example: • First calculate the angular velocity of the propeller: • so 3500 rpm means  = 367 s-1 • Now calculate the acceleration. • a = 2R = (367s-1)2 x (0.8m) = 1.1 x 105 m/s2 = 11,000 g • direction of a points at the propeller hub (-r ). ^

23. Example: Newton & the Moon • What is the acceleration of the Moon due to its motion around the Earth? • What we know (Newton knew this also): • T = 27.3 days = 2.36 x 106 s (period ~ 1 month) • R = 3.84 x 108 m (distance to moon) • RE= 6.35 x 106 m (radius of earth) R RE

24. Moon... • Calculate angular velocity: • So  = 2.66 x 10-6 s-1. • Now calculate the acceleration. • a = 2R= 0.00272 m/s2 = 0.000278 g • direction of a points at the center of the Earth (-r). ^

25. amoon g R RE Moon... • So we find that amoon/g= 0.000278 • Newton noticed that RE2 / R2= 0.000273 • This inspired him to propose that FMm 1 / R2 • (more on gravity later)

26. Space ShuttleCentripetal Acceleration • The Space Shuttle is in Low Earth Orbit (LEO) about 300 km above the surface. The period of the orbit is about 91 min. What is the acceleration of an astronaut in the Shuttle in the reference frame of the Earth? (The radius of the Earth is 6.4 x 106 m.) (a) 0 m/s2 (b) 8.9 m/s2 (c) 9.8 m/s2

27. Space ShuttleCentripetal Acceleration • First calculate the angular frequency : • Realize that: RO 300 km • RO = RE + 300 km • = 6.4 x 106 m + 0.3 x 106 m • = 6.7 x 106 m RE

28. a = 2R a =(0.00115 s-1)2 x6.7 x 106 m a = 8.9 m/s2 Space ShuttleCentripetal Acceleration • Now calculate the acceleration:

29. Relating Linear and Angular Acceleration: Differentiating the linear velocity with respect to time -with r held constant-leads to: at =dv/dt= d(rω)dt = rdω/dt Note that dv/dt =at(tangential acceleration) represents only the part of the linear acceleration that is responsible for changes in the magnitude v of the linear velocity. Like v, that part of the linear acceleration is tangent to the path of the point in question. Also, the radial part of the acceleration is the centripetal acceleration given by:

30. Rotational Quantities: • Rotation about a fixed axis: • Consider a disk rotating aboutan axis through its center: • First, recall what we learned aboutUniform Circular Motion: (Analogous to )  

31. constant Rotational Kinematics Equations • Now suppose  can change as a function of time: • We define the angular acceleration:  • Consider the case when is constant. • We can integrate this to find  and  as a function of time:  

32. Rotational Kinematics Equations (α = const.) • Recall also that for a point at a distance R away from the axis of rotation: • x = r • v = r And taking the derivative of this we find: • a = r v x r   

33. Summary (with comparison to 1-D kinematics) Angular Linear And for a point at a distance r from the rotation axis: • x = rv = r a = r

34. Rotation with Constant Angular Acceleration Just as in the basic equations for constant linear acceleration, the basic equations for constant angular acceleration can be derived in a similar manner. The constant angular acceleration equations are similar to the constant linear acceleration equations.

35. A ball is spinning about an axis that passes through its center with a constant angular acceleration of  rad/s2. During a time interval from t1 to t2, the angular displacement of the ball is  radians. At time t2, the angular velocity of the ball is 2 rad/s. What is the ball’s angular velocity at time t1? a) 6.28 rad/s b) 3.14 rad/s c) 2.22 rad/s d) 1.00 rad/s e) zero rad/s

36. A ball is spinning about an axis that passes through its center with a constant angular acceleration of  rad/s2. During a time interval from t1 to t2, the angular displacement of the ball is  radians. At time t2, the angular velocity of the ball is 2 rad/s. What is the ball’s angular velocity at time t1? a) 6.28 rad/s b) 3.14 rad/s c) 2.22 rad/s d) 1.00 rad/s e) zero rad/s

37. The original Ferris wheel had a radius of 38 m and completed a full revolution (2 radians) every two minutes when operating at its maximum speed. If the wheel were uniformly slowed from its maximum speed to a stop in 35 seconds, what would be the magnitude of the instantaneous tangential speed at the outer rim of the wheel 15 seconds after it begins its deceleration? a) 0.295 m/s b) 1.12 m/s c) 1.50 m/s d) 1.77 m/s e) 2.03 m/s

38. The original Ferris wheel had a radius of 38 m and completed a full revolution (2 radians) every two minutes when operating at its maximum speed. If the wheel were uniformly slowed from its maximum speed to a stop in 35 seconds, what would be the magnitude of the instantaneous tangential speed at the outer rim of the wheel 15 seconds after it begins its deceleration? a) 0.295 m/s b) 1.12 m/s c) 1.50 m/s d) 1.77 m/s e) 2.03 m/s

39. Sample problem, Angular Velocity and Acceleration

40. Sample problem, Angular Velocity and Acceleration

41. a R Example: Wheel And Rope • A wheel with radius R = 0.4 m rotates freely about a fixed axle. There is a rope wound around the wheel. Starting from rest at t = 0, the rope is pulled such that it has a constant acceleration a = 4 m/s2. How many revolutions has the wheel made after 10 seconds? (One revolution = 2 radians)

42. = 0 + 0(10) + (10)(10)2 = 500 rad Wheel And Rope... • Use a = Rto find : = a / R = 4 m/s2 / 0.4 m = 10 rad/s2 • Now use the equations we derived above just as you would use the kinematic equations from the beginning of the semester. a  R

43. Sample problem • Consider an induction roller coaster (which can be accelerated by magnetic forces even on a horizontal track). Each passenger is to leave the loading point with acceleration g along the horizontal track. • That first section of track forms a circular arc (Fig. 10-10), so that the passenger also experiences a centripetal acceleration. As the passenger accelerates along the arc, the magnitude of this centripetal acceleration increases alarmingly. When the magnitude a of the net acceleration reaches 4g at some point P and angle qP along the arc, the passenger moves in a straight line, along a tangent to the arc. • What angle qP should the arc subtend so that a is 4g at point P? • Calculations: • Substituting wo=0, and qo=0, and we find: • But • which gives: This leads us to a total acceleration: Substituting for ar, and solving for q lead to: When a reaches the design value of 4g, angle q is the angle qP . Substituting a =4g, q= qP, and at= g, we find:

44. Sample problem, cont. (b) What is the magnitude a of the passenger’s net acceleration at point P and after point P? Reasoning: At P, a has the design value of 4g. Just after P is reached, the passenger moves in a straight line and no longer has centripetal acceleration. Thus, the passenger has only the acceleration magnitude g along the track. Hence, a =4g at P and a =g after P. (Answer) Roller-coaster headache can occur when a passenger’s head undergoes an abrupt change in acceleration, with the acceleration magnitude large before or after the change. The reason is that the change can cause the brain to move relative to the skull, tearing the veins that bridge the brain and skull. Our design to increase the acceleration from g to 4g along the path to P might harm the passenger, but the abrupt change in acceleration as the passenger passes through point P is more likely to cause roller-coaster headache.

45. m4 m1 r1  r4 m3 r2 r3 m2 Rotation & Kinetic Energy • Consider the simple rotating system shown below. (Assume the masses are attached to the rotation axis by massless rigid rods). • The kinetic energy of this system will be the sum of the kinetic energy of each piece:

46. which we write as: v1 m4 v4 m1 r1  r4 v2 m3 r2 r3 m2 Define the moment of inertia about the rotation axis v3 Rotation & Kinetic Energy... • So: butvi = ri I has units of kg m2.

47. Rotation & Kinetic Energy... • The kinetic energy of a rotating system looks similar to that of a point particle:Point Particle Rotating System v is “linear” velocity m is the mass.  is angular velocity I is the moment of inertia about the rotation axis.

48. Moment of Inertia • So where • Notice that the moment of inertia I depends on the distribution of mass in the system. • The further the mass is from the rotation axis, the bigger the moment of inertia. • For a given object, the moment of inertia will depend on where we choose the rotation axis (unlike the center of mass). • We will see that in rotational dynamics, the moment of inertia I appears in the same way that mass m does when we study linear dynamics!

49. Calculating Moment of Inertia • We have shown that for N discrete point masses distributed about a fixed axis, the moment of inertia is: where r is the distance from the mass to the axis of rotation. Example: Calculate the moment of inertia of four point masses (m) on the corners of a square whose sides have length L, about a perpendicular axis through the center of the square: m m L m m

50. I = 2mL2 Calculating Moment of Inertia... • The squared distance from each point mass to the axis is: Using the Pythagorean Theorem so L/2 m m r L m m