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Lecture radial direction

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					Ch. 6: Circular Motion & Other
Applications of Newton’s Laws
              Recall From Ch. 4:
Acceleration of Mass Moving in Circle (Const. Speed)


                 Particle moving in a circle, radius r,
                 speed v (= constant). The velocity is
                 tangent to the circle. The centripetal
                 acceleration, a = ac is radially inward.

                          ac  v always

                            ac = (v2/r)
Newton’s Laws + Circular Motion
                   ac = (v2/r)  v
                  Newton’s 1st Law:
            There must be a force acting.
                  Newton’s 2nd Law:
                   ∑F = ma = mac
                      = m(v2/r) (magnitude)
         Direction: The total force must
               be radially inward.
• A particle moving in uniform circular motion, radius
  r (speed v = constant). The acceleration: ac = (v2/r), ac 
  v always!! ac is radially inward always!
• Newton’s 1st Law: There must be a force acting!
• Newton’s 2nd Law:
              ∑F = ma  Fr = mac= m(v2/r)
  The total force must be radially inward always!
 The Force entering 2nd Law  Centripetal Force Fr
                     (Center directed force)
• NOT a new kind of force. Could be string tension,
  gravity, etc. The right side of ∑F = ma, not the left
  side! (The form of ma, above, for circular motion)
Example: A ball twirled on a string in a
circle at constant speed. The centripetal
force Fr is the tension in the string.

         MISCONCEPTION!!
The force on the ball is NEVER outward
(“centrifugal force”). The force on the ball
is ALWAYS inward (centripetal force).
An outward force (“centrifugal”) is NOT
a valid concept! The force ON THE
BALL is inward (centripetal).
     What happens when the ball is
      released? (Fr = 0). Newton’s 1st
 Law says it should move off in
      a straight line at constant v.
        Example 6.1: Conical Pendulum
                                 A ball, mass m, is suspended
                                 from a string of length L. It
                                 revolves with constant speed v
                                 in a horizontal circle of radius r.
                                 The angle L makes with the
                                 horizontal is θ. Find an expression
                                 for v.
T ≡ tension in the string. Fig. (b) shows horizontal & vertical
components of T:         Tx = Tsinθ, Ty = Tcosθ.
Newton’s 2nd Law: ∑Fx = Tsinθ = mac= m(v2/r)                    (1)
 ∑Fy = Tcosθ – mg = 0; Tcosθ = mg                               (2)
Dividing (1) by (2) gives: tanθ = [v2/(rg)] , or v = (rg tanθ)½
From trig, r = L sinθ so, v = (Lg sinθ tanθ)½
       (Reminder: ½ power means the square root)
Example 6.2: Car Around a Curve
        Curve radius: r = 35 m. Static friction coefficient
        between tires & road: μs = 0.523. The centripetal
        force that keeps the car on the road is the static
        friction force fs between the tires & the road.
        Calculate the maximum speed vmax for the car to
        stay on the curve. Free body diagram is (b).
        Newton’s 2nd Law (let + x be to left) is:
                 ∑Fx = fs = mac = m(v2/r) (1)
                 ∑Fy = 0 = n – mg; n = mg (2)
        The maximum static friction force is (using (2)) :
                   fs(max) = μsn = μsmg                (3)
         If       m(v2/r) > fs(max), so vmax is the
            solution to μsmg = m[(vmax)2/r]
        Or,             vmax = (μsgr)½
        Putting in numbers gives: vmax = 13.4 m/s
Example 6.4: Banked Curves
       Engineers design curves which are banked (tilted
       towards the inside of the curve) to keep cars on the
       road. If r = 35 m & we need v = 13.4 m/s,
       calculate the angle θ of banking needed (without
       friction). From free body diagram, the horizontal
       (radial) & vertical components of the force n
       normal to the surface are:
                  nx = n sinθ, ny = n cosθ,
                     Newton’s 2nd Law
                ∑Fx = n sinθ = m(v2/r)               (1)
       ∑Fy = 0 = n cosθ – mg; n cosθ = mg (2)
       Dividing (1) by (2) gives: tanθ = [(v2)/(gr)]
       Putting in numbers gives: tanθ = 0.523 or
                         θ = 27.6°
           Example 6.5: “Loop-the-Loop”!
                                  A pilot, mass m, in a jet does a “loop-the-
                                  loop. The plane, Fig. (a), moves in a
                                  vertical circle, radius r = 2.7 km = 2,700 m
                                  at a constant speed v = 225 m/s.
                                  a) Calculate the force, nbot (normal force),
                                      exerted by the seat on the pilot at the
                                      bottom of the circle, Fig. (b).
                                  b) Calculate this force, ntop, at the top of the
                                      circle, Fig. (c).
TOP: Fig. (b). Newton’s 2nd Law in the radial (y) direction (up is “+”).
  ∑Fy = nbot – mg = m(v2/r) so nbot = m(v2/r) + mg or
  nbot = mg[1 + (v2/rg)] = 2.91 mg (putting in numbers) he feels “heavier”.
BOTTOM: Fig. (c). Newton’s 2nd Law in the radial (y) direction (down is “+”).
  ∑Fy = ntop + mg = m(v2/r) so ntop = m(v2/r) - mg or
  ntop = mg[(v2/rg) - 1] = 0.913 mg (putting in numbers) he feels “lighter”.
        Example (Estimate)




m = 0.15 kg, r = 0.6 m, f = 2 rev/s  T = 0.5 s
Assumption: Circular path is  in horizontal
plane, so θ  0  cos(θ)  1
    ∑F = ma           FTx = max= mac = m(v2/r)
           v =(2πr/T) = 7.54 m/s
           FTx = 14 N (tension)
Example
                        Problem
r = 0.72 m, v = 4 m/s
       m = 0.3 kg
• Use: ∑F = mac
• Top of circle:
  Vertical forces:
  (down is positive!)
  FT1 + mg = m(v2/r)
  FT1 = 3.73 N
• Bottom of circle:
  Vertical forces:
  (up is positive)
  FT2 - mg = m(v2/r)
• FT2 = 9.61 N
Example
     n




 n

				
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posted:3/26/2011
language:English
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