12032985
Descripción
Mapa Mental por Meri perkins, actualizado hace más de 1 año
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Creado por Meri perkins
hace casi 7 años
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Motion in 2 and 3 dimensions: Ch. 4
- Projectile Motion
- Horizontal Motion
- x Speed never changes
- Vix=vx
- Vix=vx
- x Speed never changes
- Vertical motion
- vy=vy+ay(t)
- viy-gt
- viy-gt
- vy=vy+ay(t)
- How do we find Range?
- Ex. vi= 55m/s
- h= 500m
- angle= tan^-1(h/x)
- angle is the top left of RT with 90 degree angle on left
- x= vix(t)+1/2(ax)(t^2)
- x=vix(t)
- ax=0
- deltay=viyt +_ 1/2 ay t^2
- -h=sqrt(2h/g)
- 10.15
- This goes into x= vix(t)
- x=(55m/s)(10.15)
- angle = tan^-1 (555.5/500)
- 45 deg
- After Being shot from a cannon soured over 3 Ferris Wheels into a net
- Known values
- vi=26.5m/s
- angle theta= 53deg
- ay=-g
- -9.81m/s^2
- Distance in x To first wheel: 23 m
- h=delta h-H
- Time to cross Ferris wheel
- delta y = viy(t)-1/2gt^2
- delta x/vocosthetai
- delta x/vocosthetai
- delta y=visintheta(deltax/vicosthetai)-1/2 g (deltax/vicosthetai)^2
- =deltax tan theta-1/2g (delta x^2)/(vi^2cos^2 Thetai)
- If Reached max height, How much did rider Clear it?
- Here we are finding range
- R-= vi^2/g * sin 2thetai
- How Far should Net Be? away from initial starting point
- How Far should Net Be? away from initial starting point
- R-= vi^2/g * sin 2thetai
- Use KInematic equations the most
- Here we are finding range
- a) this is calculating clearence over first wheel
- If Reached max height, How much did rider Clear it?
- =deltax tan theta-1/2g (delta x^2)/(vi^2cos^2 Thetai)
- delta y = viy(t)-1/2gt^2
- Time to cross Ferris wheel
- h=delta h-H
- -9.81m/s^2
- ay=-g
- angle theta= 53deg
- vi=26.5m/s
- unkown
- vix=vicostheta
- viy=viysin theta
- viy=viysin theta
- vix=vicostheta
- Known values
- After Being shot from a cannon soured over 3 Ferris Wheels into a net
- 45 deg
- angle = tan^-1 (555.5/500)
- x=(55m/s)(10.15)
- This goes into x= vix(t)
- 10.15
- -h=sqrt(2h/g)
- ax=0
- x=vix(t)
- x= vix(t)+1/2(ax)(t^2)
- angle is the top left of RT with 90 degree angle on left
- angle= tan^-1(h/x)
- h= 500m
- Horizontal Motion
- Uniform Circular motion
- Velocity must be constant for a=0
- To Obtain UCM
- Travels around a circle or circular arc
- Must be at constant speed
- since Velocity Constantly changes The particle is accellerating
- Veloc and accel. Both have
- 1) Constant Magnitude
- 2) Changing direction
- Ex. We have a circle
- v (mag)=-vxi(hat)+vyj(hat)
- Look at where theta is on the circle
- =-vsini(hat)+vcosj(hat)
- costheta=y/x
- In a particular triagle
- In a particular triagle
- Velocity Vector
- v(mag)=-v(y/r)i(hat) + v (x/r)j(hat)
- a=-v/r (dy/dt)i (hat) + v/r (dx/dt) j(hat)
- =-v^2/rvcos i (hat) + v/r * vsin theta j(hat)
- a(mag)= v^2/r * costhetai(hat) + v^2/r * sinthetaj(hat)
- absvalue a = sqrt ((-v^2/r* costheta)^2+(v^2/r* sintheta)^2)
- absvalue a = sqrt ((-v^2/r* costheta)^2+(v^2/r* sintheta)^2)
- a(mag)= v^2/r * costhetai(hat) + v^2/r * sinthetaj(hat)
- =-v^2/rvcos i (hat) + v/r * vsin theta j(hat)
- a=-v/r (dy/dt)i (hat) + v/r (dx/dt) j(hat)
- costheta=y/x
- Look at where theta is on the circle
- v (mag)=-vxi(hat)+vyj(hat)
- Ex. We have a circle
- 2) Changing direction
- 1) Constant Magnitude
- Veloc and accel. Both have
- Speed and veloc not same!
- since Velocity Constantly changes The particle is accellerating
- Circular Motion:
- Centripital Acceleration
- change in speed over change in time
- a=v^2/r
- Period of Revolution
- Time is takes to go around path once
- T=2pir/v
- T=2pir/v
- Time is takes to go around path once
- Period of Revolution
- a=v^2/r
- change in speed over change in time
- Centripital Acceleration
- Must be at constant speed
- Travels around a circle or circular arc
- To Obtain UCM
- Velocity must be constant for a=0
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