Two-Dimensional Motion Projectile Motion Periodic Motion

Projectile Moion VxVx VxVx VxVx VyVy VyVy VyVy V x = constant V y = varying

VxVx VxVx VxVx VyVy VyVy VyVy Formulas: V x = constant therefore, V x = d/t V y = varying therefore, acceleration v f = v i + at v f 2 = v i 2 + 2ad d = v i + 1/2at 2

Projectile Motion vivi vyvy vxvx V y = sin  (v i ) V x = cos  (v i ) V y controls how long it’s in the air and how high it goes V x controls how far it goes 

Projectile Motion “Range formula” vivi R = v i 2 sin2  /g yiyi yfyf Range formula works only when y i = y f Remember!!!!! v i is the velocity at an angle and the sin2  is the sine of 2 x 

Projectile Motion “Range formula” vivi R = v i 2 sin2  /g yiyi yfyf If v i = 34 m/s and  is 41 o then, R = 1160 m 2 /s 2 (0.99)/9.8 m/s 2 R = 120 m R = (34 m/s) 2 sin82 o /9.8 m/s 2

Projectile Motion “Range formula” vivi  Note that if  becomes the complement of 41 o, that is,  is now 49 o, then, v i = 34 m/s and  is 49 o then, R = 1160 m 2 /s 2 (0.99)/9.8 m/s 2 R = 120 m R = (34 m/s) 2 sin98 o /9.8 m/s 2 So, both 41 o and 49 o yield “R”

Projectile Motion “Range formula” vivi yiyi yfyf OR, If v i = 34 m/s and  is 41 o then, vyvy  v y = sin41 o (34m/s) = 22m/s, and t = v fy - v iy /g = -22m/s - (22m/s)/-9.8m/s 2 = 4.5 s vxvx d x = v x (t) = 26m/s (4.5 s) = 120 m v x = cos    4 m/s) = 26 m/s, and

Circular Motion n When an object travels about a given point at a set distance it is said to be in circular motion

Cause of Circular Motion u 1 st Law…an object in motion stays in motion, in a straight line, at a constant speed unless acted on by an outside force. u 2 nd Law…an outside force causes an object to accelerate…a= F/m u THEREFORE, circular motion is caused by a force that causes an object to travel contrary to its inertial path

Circular Motion Analysis v1v1 v2v2 r r 

v1v1 v2v2 r r v1v1 v2v2  v = v 2 - v 1 or  v = v 2 + (-v 1 ) (-v 1 ) = the opposite of v 1 v1v1 (-v 1 )  

v1v1 v2v2 r r 0  v = v 2 - v 1 or  v = v 2 + (-v 1 ) (-v 1 ) = the opposite of v 1 v1v1 (-v 1 ) v1v1 v2v2 v2v2 vv Note how  v is directed toward the center of the circle  

v1v1 v2v2 r r v1v1 v2v2 v2v2 (-v 1 ) vv  Because the two triangles are similar, the angles are equal and the ratio of the sides are proportional  l  

v1v1 v2v2 r r  v1v1 v2v2 v2v2 (-v 1 ) vv ll  Therefore,  v/v ~  l/rand  v = v  l/r now, if a =  v/t,and  v = v  l/r then, a = v  l/rt,since v =  l/t THEN, a = v 2 /r 

Centripetal Acceleration a c = v 2 /r now, v = d/t and, d = c = 2  r then, v = 2  r/t and, a c = (2  r/t) 2 /r or, a c = 4  2 r 2 /t 2 /r a c = 4  2 r/T 2

The 2 nd Law and Centripetal Acceleration FcFc acac vtvt F = ma a c = v 2 /r = 4  2 r/T 2 therefore, F c = mv 2 /r or, F c = m4  2 r/T 2

Simple Harmonic Motion or S.H.M. Simple Harmonic motion is motion that has force and acceleration always directed toward the equilibrium position and has its maximum values when displacement is maximum. Velocity is maximum at the equilibrium position and zero at maximum displacement Pendulum motion, oscillating springs (objects), and elastic objects are examples

F = max a = max v = 0 F = max a = max v = 0 F = greater a = greater v = less F = 0 a = 0 v = max F = less a = less v = greater Simple Harmonic Motion Force acceleration

Pendulum Motion FwFw FTFT Note that F T (the accelerating force is a component of the weight of the bob that is parallel to motion (tangent to the path at that point).

Pendulum Motion FwFw FTFT Note that as the arc becomes less so does the F T, therefore the force and resulting acceleration also becomes less as the “bob” approaches the equilibrium position.

Pendulum Motion FwFw FTFT a c =  r/T 2 a c = g and r = l g =  l/T 2 T 2 =  l/g T = 2  l/g

Oscillating Elastic Objects F e = max F e = less a = max a = less a and F = 0

FwFw FTFT Note that no part of F w is in the direction on Motion, or F T There, F and a is zero!!!