LectureNotes03

# Choose equations to use for describing the motion in

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Choose equations to use for describing the motion in the problem. Time connects the x, y, z dimensions and is sometimes to be eliminated via algebra.

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Copyright R. Janow – Spring 2012 pat h f o r g = 0 x m y m θ s What should monkey (a Physics 111 student) do? Where should hunter aim if monkey jumps? Assume: hunter fires at the instant monkey lets go and aims directly at falling monkey
Copyright R. Janow – Spring 2012 What should monkey (a Physics 111 student) do? Where should hunter aim if monkey jumps? When should hunter fire? When should monkey jump? Coordinates, variables: Bullet: x b (t), y b (t) Monkey: s, y m (t) Initial Conditions: Monkey at (s, h) Bullet at (0,0) x y 0 v r θ h s ) sin( v v ) cos( v v y x θ = θ = 0 0 0 0 To hit monkey: y b (t) = y m (t) at same time t 1 when bullet is at s For bullet at s : s t v ) t ( x x b = = 1 0 1 2 1 0 0 2 1 1 0 1 2 1 2 1 t g s v v gt t v ) t ( y x y y b - = - = v s t x 0 1 = For monkey: 2 1 1 2 1 gt h ) t ( y m - = Monkey should wait until hunter fires before jumping Hunter should aim directly at monkey after he jumps Both know Physics, so everybody waits forever s h ) tan( v v x y = θ = 0 0 ) t ( y ) t ( y m b 1 1 = Equate positions:

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Copyright R. Janow – Spring 2012 Example: Find the arrow’s speed An arrow is shot horizontally by a person whose height is not known. It strikes the ground 33 m away horizontally, making an angle of 86 o with the vertical. What was the arrow’s speed as it was released? 86 o R = 33 m v 0 v f R = v 0 t t = R / v 0 t = flight time = time to hit ground Final velocity components: ) sin( v v v o f fx 86 0 = = No acceleration along x ) cos( v gt v v o f y fy 86 0 - = - = V y0 = 0, t as above Divide equations to eliminate v f : gR v gt v ) tan( ) cos( ) sin( v v o o o fy fx 2 0 0 86 86 86 - = - = - = - = Rearrange: 14.3 33 9.8 ) tan( gR v o × × = = 86 2 0 m/s 68 v = 0
Copyright R. Janow – Spring 2012 Uniform Circular Motion A particle in Uniform Circular Motion moves in a circular path with a constant speed barb2right The radius vector is rotating. The velocity vector is always tangent to the particle’s path (as usual). It is perpendicular to the (rotating) radius vector. The particle is accelerating, since the direction of the velocity is constantly changing. r r v r a r a r a r v r v r r r r r The centripetal acceleration vector always points toward the center of the circle. The magnitudes of all three vectors are constant, but their directions are changing All three vectors are rotating with the same constant angular velocity and period. The period of revolution T is: a , v , r r r r v r T π = 2 a , v , r r r r The frequency f is the reciprocal of the period f = 1/T

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Copyright R. Janow – Spring 2012 Centripetal Acceleration Magnitude Formula The position vectors and velocity vectors both change due to changes in their directions: The triangles for Δ r and Δ v are both right isosceles, with the same angle Δθ at their apexes. They are similar: tail to tail v | v | r | r | r r Δ = Δ | r | r v | v | r r Δ = Δ The definition of the (magnitude of) the average acceleration is: t | r | r v t | v | | a | avg Δ Δ = Δ Δ r In the limit Δ t barb2right 0: v t | r | Δ Δ and a | a | avg r v a 2 c = The magnitude of the centripetal acceleration is It always points to the center of the circular motion
Copyright R. Janow – Spring 2012 Example: Find the Earth’s centripetal acceleration in its orbit about the Sun Assume uniform circular motion The radius of the Earth’s orbit is r = 1.5 x 10 11 m r v a 2 c = Need to find the speed v T r 2 period nce circumfere v π = = T = 1 year = 365X24x3600 s = 3.156 x 10 7 s m/s .

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• Fall '09
• DARILPEDIGO
• Acceleration, Velocity, Yi, Copyright R. Janow

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