Center of Gravity

1. Center of Gravity The center of gravity is the point at which all of an objects weight can be considered to be concentrated. The center of gravity can be considered to be the average position of the weight distribution. Any object will balance if allowed to pivot at its center of gravity.

2. The weight of an object, Fw, behaves as if it is concentrated at its cog. • Fw can be thought of as the resultant of an infinite number of points contributing to an objects weight. For a uniform object, the cog is located at its geometric center. • The cog may be either inside or outside of an object.

3. The location of the center of gravity determines the stability of an object. • If a line drawn from the center of gravity straight down falls inside the base, the object will be stable. • Stable in the sense of not toppling over or rotating about a point.

4. In football, lineman line up in a three point stance to become more difficult to tip over. Whenever the base of an object is made larger with its cog closer to the base, the object becomes more rotationally stable. If a net force is applied at the cog, the object will not rotate but accelerate in a straight line.

5. Torque is Cheap In order to make an object rotate, a force must be applied but more specifically a torque must be applied. Torque is the twisting effect of a force and is given by: Torque is a vector quantity. T = F × d

6. d where F is the force in N, is the perpendicular distance from the axis of rotation, and T has units of N•m. When it comes to objects rotating, think of the hands of a clock. • Tcw Tccw • The hands of a clock, a toy top, or a gyroscope are examples of rotary motion.

7. Torque Problem A bar 6.0 m long has its center of gravity 1.8 m from the heavy end. If it is placed on the edge of a block 1.8 m from the light end and a weight of 650 N is added to the light end, the bar is in rotational equilibrium. What is the weight of the bar? l = 6.0 m F1 = 650 N COG 1.5 m from heavy end

8. ΣT = 0 Tcw = Tccw COG AOR Fup 1.8 m 1.8 m Δ . Fw F1 6.0 m Translational Equilibrium ΣF = 0 Fup = Fw + F1 Rotational Equilibrium

9. ΣT = 0 AOR at the fulcrum. Tcw = Tccw F1 × d1 = Fw × dw 650 N × 1.8 m= Fw × 2.4 m Fw = 490 N

10. Talking About Torque The AOR (axis of rotation) is an arbitrary point that can be placed on or off the object. • All perpendicular distances are measured from this point. The COG (center of gravity) is the point on or off the object that Fw is considered to be acting. • An example of the COG not found on the object is a hallow ring.

11. Remember, the distances measured from the AOR must be perpendicular distances. • For example, if F1 had not been perpendicular to the bar, you would have to resolve F1 to find the component that is perpendicular. You know the object is not symmetrical because the problem torques about a heavy end.

12. To maintain translational equilibrium (motion in a straight line) the ΣF must equal zero. To maintain rotational equilibrium (rotary motion about a point), the ΣT must equal zero. If a force is applied at the COG and the object is free to move, the object will move in a straight line.

13. Another Torque Problem A uniform board weighing 360 N and 8.0 m long rests on a dock with 2.4 m of the board extending out over the water. How far from the dock could a 460 N boy walk out on the board without getting wet? Fwb = 360 N l = 8.0 m Fwboy = 460 N

14. ΣT = 0 Tcw = Tccw Fup 1.6 m 2.4 m Δ . Fwb Fwboy 8.0 m ΣF = 0 Fup = Fwb + Fwboy = 360 N + 460 N = 820 N

15. ΣT = 0 AOR at the fulcrum. Tcw = Tccw Fwboy × dwboy = Fwb × dwb 460 N × d= 360 N × 1.6 m dwboy = 1.3 m Beyond 1.3 m the Tcw > 0 and both the boy and the board will fall into the water.

16. Rotary Motion Rotary motion is an object spinning on its axis. • Examples include the hands of a clock or a bicycle tire spinning on its axis. • Rotary motion is not as intuitive as circular, projectile, or rectilinear motion. • Rotary motion uses the same terms as rectilinear motion but are defined differently.

17. Δθ ω = Δt Analogous Terminology Angular Velocity (ω) The magnitude of angular velocity is given by ω is the angular velocity and Δθ is the angular displacement.

18. The angular displacement, Δθ, is usually measured in radians. A radian is defined such that length of the intercepted arc of a circle is equal to the radius of the circle. When the length of s equals the length of the radius (s = r), θ is equal to 57.3° or 1 radian. s r

19. Why 57.3°? • 1 revolution = 360° = 2π radians 1 radian = 360°/2/π ≈ 57.3° Keep in mind that a radian is a dimensionless quantity. The direction of the angular velocity is not intuitive either as shown on the following slide.

20. Assume as you look at the wheel above that it is rotating counterclockwise in a horizontal plane. If you curl the fingers of your right hand, your thumb points in the direction of the angular velocity (pointing straight, up perpendicular to the horizontal.

21. Δω Δt A rotating object can have a constant angular velocity of zero or be rotating at a constant rate. • Reminiscent of Newton’s 1st Law. In accordance with Newton’s 2nd Law, a rotating may accelerate by changing its rate of rotation, the direction of the rotation, or both. Angular acceleration (α) is defined by with units of rad/s2. α =

22. Uniformly Accelerated Rotation Equations for uniformly accelerated rotary motion. • ωf = ωi + αΔt • Δθ = ωiΔt + ½αΔt2 • ωf2 = ωi2 + 2αΔθ

23. Δθ ωave = Δt Rotary Motion Problem A circular saw blade completes 1400 revolutions in 60. s while coming to a stop. • Assuming the deceleration is constant, what is the angular deceleration in rad/s2? 2π rad 1400 rev × 1 rev ωave 150 rad/s = = 60. s

24. ωave = (ωi + ωf)/2 ωf = 0; ωi = 2 × ωave = 2 × 150 rad/s = 300 rad/s ωf = ωi + αΔt 0 – 300 rad/s - 5.0 rad/s2 α = = 60. s

25. Its like two people on a merry-go-round. One is close to the axis of rotation and the other on the outer edge but yet they both sweep out 360° or make one revolution in the same amount of time. (d) Do all of the points on the circular saw blade have the same linear speed (v)? No because the further a point is away fromthe center, the faster it must travel to “keepup” with the other points.

26. (b) Assuming the deceleration is constant, what was the initial angular speed? As determined in Part (a), ωi = 300 rad/s. (c) Do all of the points on the circular saw blade have the same angular speed (ω)? Yes, because every point on the blade moves through the same angle in the same time interval.

27. All circles consist of 360° but the larger the circle the larger the circumference. The relationship between linear velocity and angular velocity is given by v = ωr and radians are dimensionless units so m/s = 1/s × m = m/s

28. More Rotary Motion Problems A ball rolls 5.7 m along a horizontal surface before coming to a stop. If the ball makes 12 revolutions, what is the diameter of the ball? 2 × π × r × 12 rev = 75 r 5.7 m = 1 rev r = 7.6 x 10-2 m D = 2 × r = 1.5 x 10-1 m

29. An electric saw blade is spinning at 1750 rev/min. If the blade has a diameter of 0.45 m: • What is its angular velocity in rad/s? rev 2πrad 1 min × × = ω 1750 min 1 rev 60 s = 180 rad/s ω

30. (b) What is the linear speed of a point on the edge of the saw blade? v = ωr = 180 rad/s × 0.45 m = 81 m/s (c)What is the acceleration of a point on the edge of a saw blade? The tangential acceleration (aT) is given by aT = rα. The angular velocity is constant, therefore the tangential acceleration (aT) is zero.

31. v2 (180 rad/s)2 aR = = r = 7.2 x 104 m/s2 0.45 m The centripetal (radial) acceleration is given by The total linear acceleration is equal to the vector sum of a = aT + aR. v2 (wr)2 = ω2r aR = = r r The centripetal (radial) acceleration is given by

32. Moment of Inertia The moment of inertia is an interesting concept and a little more involved than just plain old “inertia” found in Newton’s 1st Law. As with Newton’s 1st Law, a non-rotating wheel will remain at rest unless a torque is applied. A wheel rotating at a constant angular velocity will continue to do so until a net torque is applied.

33. To explain the inertia of a rotating object not only is the mass important but also the distribution of mass with respect to the axis of rotation. A familiar example is that of a figure skater spinning or rotating in place. While a figure skater has her arms extended, she rotates rather slowly but she rotates much faster when she pulls her arms into her body.

34. Her mass does not change but the distribution of her mass changes when she pulls her arms into her body. The close the mass is to the axis of rotation, the less rotational inertia an object has.

35. Rotational inertia is called the moment of inertia and is symbolized by I (capital eye). The geometric shape and the location of the axis of rotation determines the moment of inertia. Consider the following three objects all having the same mass and their moments of inertia: • A solid sphere, I = 2/5mr2, a solid cylinder, I = ½mr2, and a hollow cylinder, I = mr2.

36. What happens when they are released at the same time on the same incline from the same height? • The sphere reaches the bottom first, followed by the solid cylinder, and lastly the hollow cylinder. • The hollow cylinder because of its largest moment of inertia will be the hardest to start and stop rotating. • The units for moment of inertia are kg•m2.

37. Newton’s 2nd Law Newton’s 2nd Law applied to rotating objects is given by T = Iα where T is the torque discussed in the beginning slides, I is the moment of inertia, and α is the angular acceleration.

38. A Moment of Inertia Problem A uniform cylinder has a mass of 0.680 kg and a radius of 9.40 cm. • Calculate its moment of inertia. m = 0.680 kg r = 9.40 cm I = ½mr2 I = ½ × 0.680 kg × (9.40 cm × 1 m/102 cm)2

39. rev min I = 3.00 x 10-3 kg•m2 (b) Calculate the torque needed to accelerate it from rest to 1700 rpm in 8.00 s. ωi = 0 ωf = 1700 rev/min Δt = 8.00 s ωf = ωi + αΔt 1 min 2πrad = 1700 × × rev 60 s 0 + α × 8.00 s

40. α = 22 rad/s2 T = Iα = 3.00 x 10-3 kg•m2 × 22 rad/s2 T = 6.6 x 10-2 m•N

41. Angular Impulse and Momentum Recalling impulse and the change of momentum for rectilinear motion: FnetΔt = Δmv = mΔv an analogous equation can be written for rotary motion: TnetΔt = IΔω

42. TnetΔt = IΔω where Tnet is the net torque in N•m, Δt is the time in s that the Tnet is acting, I is the moment of inertia in kg•m2, and ω is the angular velocity in rad/s. The quantity TnetΔt is the angular impulse expressed in N•m•s and IΔω is the change in angular momentum expressed in kg•m2/s.

43. Just as there is the conservation of linear momentum given by: Δp = 0 pi = pf m1v1 + m2v2 = m1v1’ + m2v2’ there is a conservation of angular momentum given by: ΔL = 0 Li = Lf I1ω1 = I2ω2

44. The example of a figure skater has already been discussed. Have you ever wondered why it is so much easier to balance on a moving bike than a stationary one? • Stationary or not, the location of your center of gravity does not change. • Stationary or not, the size of your base does not change unless your fall over.

45. An object spinning on its axis has a high degree of stability. A large torque is needed to change the angular momentum. • Remember, the direction of the angular momentum is given by curling the fingers of your right hand in the direction of rotation.

46. Angular Momentum Problem A merry-go-round has a mass of 200. kg and a radius of 30. m. A boy whose mass is 50. kg walks from the outer edge toward the center. If the angular velocity of the merry-go-round is 4.0 rad/s when the boy is at the outer edge, what will be the angular velocity of the boy when he is 10. m from the center? Assume I = ½mr2 for the merry-go-round and I = mr2 for the boy.

47. mmgr = 200. kg mb = 20. kg rb = 10. m rmgr = 30. m ω1 = 4.0 rad/s ω2 = ? ΔL = 0 Li = Lf I1ω1 = I2ω2 ((½mr2)mgr + (mr2)b) × ω12 = ((½mr2)mgr + (mr2)b) × ω22

48. (½ × 200. kg × (30. m)2 + 20. kg × (30. m)2) × 4.0 rad/s = (½ × 200. kg × (30. m)2 + 20. kg × (10. m)2) × ω2 ω2 = 4.7 rad/s

49. Counter Steering orCounter Intuitive Physics A rotating wheel will resist changes in its state of rotation when a force is applied to it • A rotating wheel is a gyroscope • Newton’s 2nd Law applied to rotational dynamics • The resistance comes as a result of the I (moment of inertia)

50. The applied force produces an unbalanced torque Gyroscopic precession results when a force is applied to a rotating mass and the force is redirected 90° in the direction of the rotation Gyroscopic precession is best illustrated when going more than ≈ 20 mi/h on a motorcycle • Not noticeable on a bicycle. Why?