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Lecture Ch 08
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Lecture Ch 08

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  • 1. Chapter 8 Rotational Motion
  • 2. Units of Chapter 8
    • Angular Quantities
    • Constant Angular Acceleration
    • Rolling Motion (Without Slipping)
    • Torque
    • Rotational Dynamics; Torque and Rotational Inertia
    • Solving Problems in Rotational Dynamics
    • Rotational Kinetic Energy
    • Angular Momentum and Its Conservation
  • 3. 8-1 Angular Quantities In purely rotational motion, all points on the object move in circles around the axis of rotation (“ O ”). The radius of the circle is r . All points on a straight line drawn through the axis move through the same angle in the same time . When P (at radius r) travels an arc length  , OP sweeps out an angle θ . θ  Angular Displacement of the body
  • 4. 8-1 Angular Quantities
    • θ  Angular Displacement
    • Commonly, measure θ in degrees.
    • Math of rotation: Easier if θ is measured in Radians
    • 1 Radian  Angle swept out
    • when the arc length = radius
    • When   r, θ  1 Radian
    • θ in Radians is defined as:
      • θ = ratio of 2 lengths ( dimensionless )
      • θ MUST be in radians for this to be valid!
    In doing problems in this chapter, need calculators in RADIAN MODE!!!!
  • 5.
    • θ in Radians for a circle of radius r , arc length  is defined as: θ  (  /r)
    • Conversion between radians & degrees:
    • θ for a full circle = 360º = (  /r) radians
    • Arc length  for a full circle = 2 π r
    •  θ for a full circle = 360º = 2 π radians
    • Or 1 radian (rad) = (360/2 π )º  57.3º
    • Or 1º = (2 π /360) rad  0.017 rad
      • In doing problems in this chapter, put your calculators in RADIAN MODE!!!!
    8-1 Angular Quantities
  • 6. θ  3  10 -4 rad = ? º r = 100 m,  = ? a) θ = (3  10 -4 rad)  [(360/2 π )º/rad] = 0.017º For small angles: The chord  arc length b)  = r θ = (100)  (3  10 -4 ) = 0.03 m = 3 cm θ MUST be in radians in part b Example 8-2 :
  • 7. 8-1 Angular Quantities Angular displacement: The average angular velocity is defined as the total angular displacement divided by time : The instantaneous angular velocity: (8-2a) (8-2b) (Units = rad/s ), Valid ONLY if  θ is in rad !
  • 8. 8-1 Angular Quantities The angular acceleration is the rate at which the angular velocity changes with time: The instantaneous acceleration: (8-3a) (8-3b) (Units = rad/s 2 ) Valid ONLY if  θ is in rad &  ω is in rad/s !
  • 9.
    • From Ch. 5 (circular motion): A mass moving in a circle has a linear velocity v & a linear acceleration a .
    • We’ve just seen that it also has an angular velocity and an angular acceleration.
    • Every point on a rotating
    • body has an angular
    • velocity ω and a linear
    • velocity v .
    • There MUST be relations
    • between the linear & the
    • angular quantities!
  • 10. Connection Between Angular & Linear Quantities v = (   /  t),   = r  θ  v = r(  θ /  t) = r ω Radians!  v = r ω  Depends on r ( ω is the same for all points!) v A = r A ω A v B = r B ω B v B > v A since r B > r A Therefore, objects farther from the axis of rotation will move faster .
  • 11. 8-1 Angular Quantities If the angular velocity of a rotating object changes , it has a tangential acceleration: a tan = (  v/  t),  v =r  ω = r (  ω /  t)  a tan = r α Even if the angular velocity is constant, each point on the object has a centripetal acceleration:
  • 12. 8-1 Angular Quantities Here is the correspondence between linear and rotational quantities:
  • 13. 8-1 Angular Quantities The frequency is the number of complete revolutions per second: or ω = 2 π f (angular frequency) Frequencies are measured in hertz . The period is the time one revolution takes:
  • 14. 8-2 Constant Angular Acceleration The equations of motion for constant angular acceleration are the same as those for linear motion, with the substitution of the angular quantities for the linear ones. NOTE: These are ONLY VALID if all angular quantities are in radian units!!
  • 15. 8-3 Rolling Motion (Without Slipping) In (a), a wheel is rolling without slipping. The point P, touching the ground, is instantaneously at rest , and the center moves with velocity v. In (b) the same wheel is seen from a reference frame where C is at rest. Now point P is moving with velocity –v. The linear speed of the wheel is related to its angular speed :
  • 16. Example 8-7
    • Bicycle: v 0 = 8.4 m/s . Comes to rest after 115 m . Diameter of wheel = 0.68 m ( r = 0.34m )
    • a) ω 0 = ( v 0 /r) = 24.7 rad/s
    • b) total θ =(  /r)
    • = (115m)/(0.34m)
    • = 338.2 rad = 53.8 rev
    c) α = ( ω 2 – ω 0 2 )/(2 θ ) Stopped  ω = 0  α = -0.902 rad/s 2 d) t = ( ω – ω 0 )/ α Stopped  ω = 0  t = 27.4 s
  • 17. 8-4 Torque To make an object start rotating, a force is needed; the position and direction of the force matter as well. The perpendicular distance from the axis of rotation to the line along which the force acts is called the lever arm .
  • 18. 8-4 Torque A longer lever arm is very helpful in rotating objects.
  • 19.
    • Lever Arm  r  =  distance of the axis of rotation from the “line of action” of force F
    • r  = Distance which is  to both the axis of rotation and to an imaginary line drawn along the direction of the force (“line of action”).
    • Find: Angular acceleration
    • α  (force)  (lever arm) = Fr 
    •  Define: TORQUE
    • τ  Fr  τ causes α
    • (Just as in the linear motion case, F causes a )
    Lower case Greek “tau”  8-4 Torque
  • 20. 8-4 Torque Here, the lever arm for F A is the distance from the knob to the hinge ; the lever arm for F D is zero ; and the lever arm for F C is as shown.
  • 21. 8-4 Torque The torque is defined as: F  = F sin θ F  = F cos θ τ = rF sin θ Units of torque: Newton-meters (N m)
  • 22. Example 8-9 r  = r 2 sin60 º τ 2 = -r 2 F 2 sin60 º τ 1 = r 1 F 1 τ = τ 1 + τ 2 = -6.7 m N Always use the following sign convention ! Counterclockwise rotation  + torque Clockwise rotation  - torque
  • 23. Section 8-5: Rotational Dynamics
    • Goal: Newton’s 2 nd Law for rotational motion.
    • We’ve just seen:
    • The angular acceleration is  the net torque or
    • α  τ net = ∑ τ = sum of torques
      • Analogous to Newton’s 2 nd Law :
      • a  F net = ∑ F = sum of forces
      • Recall, Newton’s 2 nd Law: ∑ F = ma
      • a  α , F  τ
    • Question: What plays the role of mass m for rotational problems?
  • 24. Simplest Possible Case A mass m moving in a Circle of radius r , one force F TANGENTIAL to the circle τ = rF Newton’s 2 nd Law + relation (a = r α ) between tangential & angular accelerations  F = ma = mr α So τ = mr 2 α  Newton’s 2 nd Law for Rotations Proportionality constant between τ & α is mr 2 (point mass only!)
  • 25.
    •  Newton’s 2 nd Law, rotational motion is:
    • ∑ τ = I α
    • The net torque = (moment of inertia)  (angular acceleration)
    • Moment of Inertia of the body:
    • I = ∑ (mr 2 )
    • Moment of Inertia I = A measure of the rotational inertia of the body. Analogous to the mass m = A measure of translational inertia of the body.
  • 26.
    •  For one particle, moving in a circle, Newton’s 2 nd Law for rotational motion is:
    • τ = mr 2 α
    • Extrapolate to a rigid body, made of many particles. Newton’s 2 nd Law, rotational motion becomes
    • ∑ τ = ∑ (mr 2 ) α
      • Since α is the same for all points in the body, it comes out of the sum.
    •  Write ∑ τ = I α , where the proportionality constant I = ∑ (mr 2 ) = m 1 r 1 2 + m 2 r 2 2 + m 3 r 3 2 + …
    •  Moment of Inertia of the body
  • 27. 8-5 Rotational Dynamics; Torque and Rotational Inertia The quantity is called the rotational inertia of an object. The distribution of mass matters here – these two objects have the same mass, but the one on the left has a greater rotational inertia , as so much of its mass is far from the axis of rotation.
  • 28. 8-5 Rotational Dynamics; Torque and Rotational Inertia The rotational inertia of an object depends not only on its mass distribution but also the location of the axis of rotation – compare (f) and (g), for example.
  • 29. Example 8-10
    • I = ∑ (mr 2 )
    • a) I = (5)(2) 2 +(7)(2) 2
    • = 48 kg m 2
    • b) I = (5)(0.5) 2
    • +(7)(4.5) 2
    • = 143 kg m 2
  • 30. 8-6 Solving Problems in Rotational Dynamics
    • Draw a diagram.
    • Decide what the system comprises.
    • Draw a free-body diagram for each object under consideration, including all the forces acting on it and where they act.
    • Find the axis of rotation ; calculate the torques around it.
  • 31. 5. Apply Newton’s second law for rotation . If the rotational inertia is not provided, you need to find it before proceeding with this step. 6. Apply Newton’s second law for translation and other laws and principles as needed. 7. Solve . 8. Check your answer for units and correct order of magnitude. 8-6 Solving Problems in Rotational Dynamics
  • 32. Example 8-11
    • M = 4 kg, F T = 15 N
    • Frictional torque:
    • τ fr = 1.1 m N, t = 0, ω 0 =0
    • t=3 s, ω =30 rad/s, I = ?
    • Compute angular accel. :
    • ω = ω 0 + α t,
    • α = ( ω - ω 0 )/t
    • = 10 rad/s 2
    • ∑ τ = I α
    • F T R - τ fr = I α
    •  I = [(15)(0.33) -1.1]/10
    • I = 0.385 kg m 2
  • 33. 8-7 Rotational Kinetic Energy The kinetic energy of a rotating object is given by By substituting the rotational quantities, we find that the rotational kinetic energy can be written: An object that has both translational and rotational motion also has both translational and rotational kinetic energy : (8-15) (8-16)
  • 34. Example 8-13:
    • Sphere rolls down incline
    • (no slipping or sliding).
    • KE+PE conservation :
    • (½)Mv 2 + (½)I ω 2 +MgH
    • = constant
    • or
    • (KE) 1 +(PE) 1 = (KE) 2 + (PE) 2
    • where KE has 2 parts :
    • (KE) trans = (½)Mv 2
    • (KE) rot = (½)I ω 2
  • 35. 8-7 Rotational Kinetic Energy When using conservation of energy, both rotational and translational kinetic energy must be taken into account. All these objects have the same potential energy at the top, but the time it takes them to get down the incline depends on how much rotational inertia they have.
  • 36. 8-7 Rotational Kinetic Energy The torque does work as it moves the wheel through an angle θ : (8-17) Torque: τ = Fr Work: W = F   = Fr  = τ 
  • 37. 8-8 Angular Momentum and Its Conservation In analogy with linear momentum, we can define angular momentum L :
    • Similar to momentum case, can show, rotational version of Newton’s 2 nd Law:
    • ∑ τ = (  L/  t) (= I α )
    •  if ∑τ = 0, (  L/  t) = 0 & L = constant
    • Angular Momentum is conserved if total torque = 0
    • Example: t = 0 , angular velocity ω 0 time t , angular velocity ω
    • L = L 0 or I ω = I 0 ω 0
  • 38. 8-8 Angular Momentum and Its Conservation Therefore, systems that can change their rotational inertia through internal forces will also change their rate of rotation:
  • 39. Example 8-15
    • v 1 = 2.4 m/s
    • r 1 = 0.8m
    • r 2 = 0.48 m
    • v 2 = ?
    • I 1 ω 1 = I 2 ω 2
    • m(r 1 ) 2 ω 1 = m(r 2 ) 2 ω 2
    • ω 2 = ω 1 (r 1 ) 2 /(r 2 ) 2
    • ω 1 = (v 1 /r 1 )
    • v 2 = r 2 ω 2 = r 2 (v 1 /r 1 )[(r 2 ) 2 /(r 1 ) 2 ] = v 1 (r 1 /r 2 )
    • v 2 = 4.0 m/s
  • 40. Translation-Rotation
    • Translation Rotation
    • Displacement x θ
    • Velocity v ω
    • Acceleration a α
    • Force (Torque) F τ
    • Mass (moment of inertia) m I
    • Newton’s 2 nd Law ∑ F = ma ∑τ = I α
    • Kinetic Energy (KE) (½)mv 2 (½)I ω 2
    • Work (constant F, τ ) Fd τ  θ
    • Momentum mv I ω
    • CONNECTIONS : v = r ω a tan = r α
    • a R = (v 2 /r) = ω 2 r τ = r  F
    • I = ∑ (mr 2 )
  • 41. Summary of Chapter 8
    • Angles are measured in radians; a whole circle is 2 π radians.
    • Angular velocity is the rate of change of angular position.
    • Angular acceleration is the rate of change of angular velocity.
    • The angular velocity and acceleration can be related to the linear velocity and acceleration.
    • The frequency is the number of full revolutions per second; the period is the inverse of the frequency.
  • 42. Summary of Chapter 8, cont.
    • The equations for rotational motion with constant angular acceleration have the same form as those for linear motion with constant acceleration.
    • Torque is the product of force and lever arm.
    • The rotational inertia depends not only on the mass of an object but also on the way its mass is distributed around the axis of rotation.
    • The angular acceleration is proportional to the torque and inversely proportional to the rotational inertia.
  • 43. Summary of Chapter 8, cont.
    • An object that is rotating has rotational kinetic energy. If it is translating as well, the translational kinetic energy must be added to the rotational to find the total kinetic energy.
    • Angular momentum is
    • If the net torque on an object is zero, its angular momentum does not change.