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Solving Equations and Circular Motion

Solving Equations and Circular Motion. Physics 2211. Solving Equations. Algebraic v = ∆x/∆t a = ∆v/∆t Cartesian. Area ∫ vdt = position. Word Problems. V. t. Deriving the Equation. a = c most common case v = ∫ adt = gt ∫ + k = g(t 2 – t 1 ) + k If t 1 = 0

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Solving Equations and Circular Motion

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  1. Solving Equations and Circular Motion Physics 2211

  2. Solving Equations • Algebraic • v = ∆x/∆t • a = ∆v/∆t • Cartesian

  3. Area ∫vdt = position Word Problems V t

  4. Deriving the Equation • a = c most common case • v = ∫ adt = gt∫ + k = g(t2 – t1) + k • If t1 = 0 • gt] + k = gt +k • v = gt +v0 • v = ma +b t2 t2 t1 t1 t 0 m = a (x, k )

  5. Finding x from v=gt + v0 t2 x = ∫ vdt = ∫ (gt + v0)dt = ∫ gtdt + ∫ v0dt = (1/2)gt2 +k + v0∫ dt = (1/2)gt2 + k1 + v0t + k2 = (1/2)gt2 + v0t + k3 k3 = x0 xf = (1/2)gt2 + v0t + x0 t1 t 0 t t 0 0 t 0

  6. Formulas xf= (1/2)gt2 + v0t + xi xf=at + v0 xf = xi + (1/2)(vi + vf)t missing a vf = vi2 + 2a(xf – xi) missing t

  7. Example 1 A man throws a ball straight up with the following specifications: vi = 20m/s, a = -10m/s2, xi = 50m At what time does the ball reach its highest point? What is the highest point? When does the ball reach 60m?

  8. Getting Started Use equation xf= (1/2)gt2 + v0t + xi Negative Parabola when slope = 0, (dx/dt) = 0 Slope = 0 - - + + x

  9. Finding t (dx/dt) xf = (dx/dt) [(1/2)(at2) + vit + xi] 0 = at + vi at = - vi t = -vi/a

  10. Finding ball’s highest height -20(m/s) / -10(m/s2) = 2s Therefore, the ball is at its highest when t = 2s Xh = -5(m/s2) *4s + 20(m/s)*2s + 50m = -20m + 4m + 50m = 70m Therefore, the ball’s highest point is at 70m.

  11. Finding the ball at 60m 0 = (1/2)at2 + vit + (xf – xi) 0 = at2 + bt + c t = (2±√2)s ***More than 1 answer!! There is only 1 answer at the top, everywhere else there are 2 roots

  12. Uniform Circular Motion A marker is placed on an old record player. On top, it goes all the way around; from the side it looks like it is going side to side.

  13. Circular Motion Points on a circle are described as: x = cos (wt) y = sin (wt) w  speed of rotation *Remember: cos starts at 1, sin starts at 0

  14. x= sint x = sin(wt) + x0

  15. x = cos(wt)

  16. Therefore v = (d/dt) cos (wt) = -w sin(wt) a = (dv/dt) = -w2cos(wt) = -wx a is oppositely proportional to position Harmonic motion - sin in circular motion velocities and accelerations that vary

  17. Vectors Cartesian unit vectors v = RcosθÎ + RsinθĴ = R (cosθÎ + sinθĴ) Polar Angles can be useful x = Rcosθ y = Rsinθ

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