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Chapter Eight

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  1. Chapter Eight Dynamics II Motion in a Plane

  2. Uniform Circular Motion Uniform Circular Motion: Movement at constant speed around a circle of radius, r. Period (T) – the time it takes to go around the circle once.

  3. Uniform Circular Motion CCW > 0 CW < 0

  4. Angular Velocity

  5. vtvsω vt = ds/dt s = rθ ds/dt = r dθ/dt vt = r dθ/dt ω = dθ/dt vt = ωr (with ω in rad/s)

  6. Centripetal Acceleration vf= v r v0 = v  r By similar triangles as θ => 0 v t ac=vf- v0 r r  v0  vf

  7. Kinematics of UCM ac = ar

  8. Dynamics of UCM

  9. Test Your Understanding Rank in order, from largest to smallest, the centripetal accelerations (ac)ato (ac)eof particles a to e.

  10. Test Your Understanding Rank in order, from largest to smallest, the centripetal accelerations (ac)ato (ac)eof particles a to e. (ac)b> (ac)e> (ac)a= (ac)c> (ac)d

  11. Acceleration for UCM We have shown that: ac = vt2/r and vt = ωr Substituting gives ac = ω2r2/r=ω2r

  12. Nonuniform Circular Motion If the particle velocity changes, there is tangential acceleration,at. If at is constant, 1D kinematic equations may be applied.

  13. Nonuniform Circular Motion For non-uniform circular motion: • the acceleration vector no longer points towards the center of the circle. • at can be positive (ccw) or negative (cw) but ac is always positive.

  14. Angular vs Tangential Acceleration at = dv/dt where v = vt vt = r ω at = r dω/dt dω/dt =angular acceleration: α = dω/dt, in rad/s2 at = r α or α= at/r

  15. Angular Acceleration

  16. Angular Acceleration if α is constant, θ and ω can be found with the angular equivalents of the kinematic equations.

  17. 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

  18. UCM Dynamics AP Physics C

  19. UCM Dynamics If a 25 kg object on a 2 m rope is moving in a circle at uniform speed and the tension in the rope is 100 N. What is the angular velocity? Known: Mass, radius, Fnet Unknown: ω Plan: Find v, then ω AP Physics C

  20. Problem 1 Find V: Find ω: What if given v or ω and asked for Fr? AP Physics C

  21. Problem 2 A car is going around a curve which has a 50 m radius. If the mass of the car is 1500 kg, and the coefficient of friction is 1, how fast can the car make the curve without sliding? The road is unbanked. Ff Fc a=v2/r W=mg AP Physics C

  22. Problem 2 con’t AP Physics C

  23. Car going around a turn What are the forces that cause this radial acceleration? Fc a=v2/r Car is heading into the screen. W=mg

  24. Car going around a turn The fastest the car can go around the turn without sliding is when the friction is maximum: Ff = mFc. Fc Ff mFc a=v2/r W=mg

  25. Max Velocity on Curve Apply N2L- SFx= Ff = ma = mv2/r For max speed, we need max force of friction: Ff = mFc = mmg Substituting: mmg= mv2/r Solving for v: Fc Ff= mFc a=v2/r W=mg

  26. Banked turn Fc a=v2/r Ff q W=mg GSP

  27. Banked turn SFx= Fc sin(q) + Ffcos(q) = mv2/r SFy = Fccos(q) - Ff sin(q) - mg = 0 Ff= mFc. Fc = mg / [cos(q) - m sin(q)] Fc Fccosθ θ Fc Fc sin θ θ Ffcosθ a=v2/r Ff sin θ Ff Ff q mg W=mg

  28. Banked turn SFx= Fc sin(q) + Ffcos(q) = mv2/r SFy = Fccos(q) - Ff sin(q) - mg = 0 Ff= mFc. Fc = mg / [cos(q) - m sin(q)] Fc Fccosθ θ Fc Fc sin θ θ Ffcosθ a=v2/r Ff sin θ Ff Ff q mg W=mg

  29. Banked turn SFx= Fc sin(q) + Ffcos(q) = mv2/r SFy = Fccos(q) - Ff sin(q) - mg = 0 Ff= mFc. Fc = mg / [cos(q) - m sin(q)] Fc Fccosθ θ Fc Fc sin θ θ Ffcosθ a=v2/r Ff sin θ Ff Ff q mg W=mg

  30. Banked turn - Minimum Speed Find an equation for theminimum speed necessary? What changes in what we did formax speed? Fc Ff a=v2/r q W=mg

  31. Minimum Speed Con’t Fc Ff a=v2/r q AP Physics C W=mg

  32. Banked Curve • For a car traveling with speed v around a curve of radius r, determine a formula for the angle at which a road should be banked so that no friction is required. ∑Fy = may FNcosθ – mg = 0 FN = mg/cosθ FN sin θ – mv2/r mg(sin θ)/cosθ = mv2/r tan θ = v2/gr θ = tan-1(v2/gr) AP Physics C

  33. Problem #1 • What is the maximum speed with which a 1200 kg car can round a turn of radius 80.0 m is on a flat road, if the coefficient of friction between tires and road 0.65? AP Physics C

  34. Problem #2 • Highway curves are marked with a suggested speed. If this speed is based on what would be safe in wet weather, μ = .7, estimate the radius of curvature for a curve marked 50 km/h curve. See the free-body diagram, which assumes the center of the curve is to the right in the diagram. AP Physics C

  35. Problem #3 • What is the minimum speed that a roller coaster must be traveling when upside down at the top of a circular loop so that the passengers do not fall out? Assume a radius of curvature of 7.6 m. At minimum speed FN = 0 AP Physics C

  36. Problem #4 You swing a pail of water in a vertical circle of radius r. The speed of the pail is vt at the top of the circle. (a) Find the force exerted on the water by the pail at the top of the circle. (b) Find the minimum value of vt for the water to remain in the pail. (c) Find the force exerted by the pail on the water at the bottom of thecircle, where the pail’s speed is vb. Two forces act on the water: FP and mg. The ac is –v2/r. (b) AP Physics C

  37. Problem #4 Con’t Similar to top except that force of pail is positive What would be the minimum speed to swing a bucket containing 3kg of water in a vertical circle with a 1 m radius without spilling the water? AP Physics C