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Ch. 6: Circular Motion & Other Applications of Newton’s Laws

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

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Ch. 6: Circular Motion & Other Applications of Newton’s Laws

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  1. Ch. 6: Circular Motion & Other Applications of Newton’s Laws

  2. 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 valways ac= (v2/r)

  3. 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.

  4. A particle moving in uniform circular motion, radius r(speed v = constant). The acceleration: ac = (v2/r), ac v always!! ac is radially inwardalways! • Newton’s 1st Law: There mustbe 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) • NOTa 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)

  5. 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 ALWAYSinward(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.

  6. 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)], orv = (rg tanθ)½ From trig, r = L sinθ so, v = (Lg sinθtanθ)½ (Reminder: ½ power means the square root)

  7. 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 fsbetween the tires & the road. Calculate the maximum speedvmaxfor 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 vmaxis the solution to μsmg = m[(vmax)2/r] Or, vmax = (μsgr)½ Putting in numbers gives: vmax = 13.4 m/s

  8. 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 θof 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°

  9. 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”.

  10. 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)

  11. Example

  12. 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

  13. Example n n

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