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Chapter 4 Kinematics in 2 & 3 Dimensions Kinematics in 2 & 3 Dim Projectile Motion Circular Motion Position and Displacement in 3-D Average Velocity Displacement per unit time Dr Dx ˆ Dy ˆ Dz ˆ vavg = = i+ j+ k Dt Dt Dt Dt Instantaneous velocity dr dx ˆ dy ˆ dz ˆ v(t) = = i+ j+ k dt dt dt dt Kinematics in 2 & 3 Dimensions Kinematic equations must be written in vector form – Unit vector form is most advantageous – Takes advantage of properties of Cartesian coordinates; i.e., independence of coordinates from one another r ( t ) = x ( t ) iˆ + y ( t ) j ˆ dr ( t ) dx ( t ) ˆ dy ( t ) ˆ v (t) = = i+ j = vx ( t ) iˆ + vy ( t ) j ˆ dt dt dt dv ( t ) dvx ( t ) ˆ dvy ( t ) ˆ a (t) = = i+ j = ax ( t ) iˆ + ay ( t ) j ˆ dt dt dt In projectile motion the x and y components of motion are independent Use vector notation for position and velocity v v cos x0 0 v y 0 v0 sin ˆ v0 = v0 xiˆ + v0 y j v ( t ) = vx ( t ) iˆ + vy ( t ) j ˆ What is rabbit’s position at t=15 sec x(t ) 0.31t 2 7.2t 28 y (t ) 0.22 t 2 9.1t 30 ˆ r (t ) x(t ) i y (t ) ˆ j a) What is position at t=15 b) Plot the path for 0<t<25 sec Rabbit’s Velocity at t=15 sec dr dx ˆ dy ˆ v (t) i j dt dt dt Acceleration Here the acceleration is constant dv dv x ˆ dv y ˆ a(t) i j dt dt dt A freely falling object and a object with a horizontal velocity fall at the same rate The motion in the x and y directions are independent So we know the EOMs! Projectile Motion Problems Conclusions from previous slide: – Two sets of kinematic equations used One in the x direction Another in the y direction – Special Conditions apply If no air resistance, ax = 0 If no air resistance, ay = g The Range equation – A special case The projectile lands at the same elevation it starts There is no air resistance g is a constant y yo 0 v0 sin t gt 1 2 2 R x x0 v0 cos t The Range equation continued Eliminating time between the two previous equations yields v 2 cos sin 2 x f x0 R 0 g v0 sin 2 2 g Graphical Behavior of the Range equation Note that each value of range has two solutions for a give v0 except 0o,90o and 45o Example Prob 30 a) ywall=y(tf) b) vx(tf), vy(tf) c) Has it passed apogee Uniform Circular Motion Motion in a circle at constant speed – Magnitude of radius is constant – Magnitude of velocity is constant – Direction of velocity CHANGES Velocity vector is tangent to circle Radius if perpendicular to tangent to circle (and velocity) A change in velocity requires an acceleration Assume D is small Ds D r Dv D v Determining the acceleration Using similar triangles Ds Dv SO! r v 1 Ds 1 Dv Dt r Dt v v a r v v2 v2 a r ˆ a r r Result is called radial or centripetal acceleration Uniform circular Motion r (t), v (t), a (t) θ Example 3 A wheel 10 cm in radius rotates at 150 rpm. What is the centripetal acceleration at a point on the edge of the wheel?

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Adobe Acrobat, File Format, Projectile Motion, Dimensional Kinematics, one dimension, physics kinematics, box dimensions, two-dimensional kinematics, dimensional motion, two dimensional

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posted: | 3/26/2011 |

language: | English |

pages: | 20 |

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