Physics 211 Space - time & space-space diagrams Kinetic Equations of Motion Projectile motion Uniform circular motion Moving coordinate systems Relative.

Physics 211

•Space - time & space-space diagrams•Kinetic Equations of Motion•Projectile motion•Uniform circular motion•Moving coordinate systems•Relative motion

•Galilean Transformation of coordinates

3: Two Dimensional Motion

r(t1)

r(t2)

x

yspace-space diagram

position vectors r t1 ; r t2

r t1 x t1 i y t1 j

r t2 x t

2 i y t2 j

d t r t2 r t1 x t

2 x t1 i y t

2 y t1 j

v t1

dr t1

dtdx t

1 dt

i dy t

1 dt

j

v t1

dr t1 dt

vx t1 i vy t1 j

a t1

dv t1 dt

d2x t

1

dt2i d 2y t

1

dt 2j

a t1

dv t1

dtax t1 i ay t1 j

speed = v t v t dx t dt

2

dy t dt

2

acceleration = a t a t d2x t dt 2

2

d 2y t dt 2

2

average quantities = final value-initial valuetime taken

If a is constant ax and ay being constant

then we can use Kinetic Equations of Motion

d t 12

at2 v 0 t

v t at v 0

•acceleration due force of gravity near the •earths surface is approximately constant •Neglect air resistance•Neglect rotation of earth

Then we can use kinetic equations of motion for projectile motion

i

j

y

x

a g 9. 81 j ms 2

r 0 0

r t 12gt 2

j v x 0 t i v y 0 t j

v x 0 t i v y 0 t 12gt 2

j

x t vx 0 t

y t vy 0 t 12gt 2

y

x

Horizontal and vertical positions

velocity in x directionxt =

dx

dt v

x0

velocity in y direction y t = dy

dt v y 0 9 .81 t

when projectile reaches highest point vy(t)=0 v y 0 9 .81 t 0

thigh

v

y0

9 . 81

x thigh

vx

0 v y0

9. 81;

y thigh

v y 0 2

9. 81-

12

9.81v y 0 9 .81

2

=12

v y 0 2

9 . 81=

v y 0 2

19 . 62

projectile hits ground when y t 0

t vy

0 1

29. 81 t

0

t 0 or t 2 vy0

9 .81 x

max

2 vx

0 v y0

9 . 81

v

v

trajectory angle

tan slope of tangent to path at t0

slope of velocity vector at t 0

v

y 0 vx 0

dy

dtdx

dt

dy

dx

tan 1v

y 0 vx 0

y

x

initial speed = v 0 vx 0 2 vy 0 2

vx 0 v 0 cos

vy 0 v 0 sin

x thigh

v 0 2sin cos g

v 0 2sin 22g

y thigh v 0 2sin 2

2g

and

xrange

2vx 0 vy 0

9.81

2v 0 2 sin cos g

v 0 2 sin 2

g

Maximun height and range can be expressed in terms

of v 0 and

y

x

v(0)

v 0 2 sin 22g

v 0 2 sin (2

g

Uniform circular motion

rv

v t v = constant ; r t r constant

position vector r t = x t , y t r cos t , r sin t r cos t , r sin t rcos t , r sin t

t t d t dt

constant

is angular speed

angular acceleration d t dt

0

distance travelled s = r r t

linear speed = v =dsdt

r

r t rcos t i r sin t j r ˆ r t

v t dr t dt

r sint i r cost j

r sint i cos t j r ˆ v t

a t dv t dt

r 2 cost i 2 sint j r 2 cost i sin t j r 2 ˆ r t

a t r 2 ˆ r t v 2

rˆ r t

a t v 2

r constant

Relative motion

observer 1

observer 2

u(t)=u=constant

r2(t)

r1(t)

r(t)=ut

r1t position vector of object in coordinate system

r2t position vector of object in coordinate system

r1 t r2 t ut

r2t r

1t u t

v2t v

1t u

a2t a

1t

Galilean Transformation 1 2

1

2

Nonuniform curvilinear motion

ˆ v t

ˆ c t

r t r t cos t i sin t j

v t dr t dt

v t ˆ v t

find unit vector, ˆ c t perpendicular to ˆ v t

points to center of curvature of path at this point

ˆ v t ˆ v t 1 d

dtˆ v t ˆ v t 0

d ˆ v t dt

ˆ v t ˆ v t dˆ v t dt

0

ˆ v t d ˆ v t dt

0 thus ˆ c t

d ˆ v t dt

d ˆ v t dt

a t dv t dt

dv t dt

ˆ v t v t dˆ v t dt

a t at t ̂ v t +ar t ̂ c t

ar t v t 2

r t radial (centripetal) acceleration

at t r t t tangential acceleration

r t distance to center of curvature