95.141 Feb 6 , 2013 PHYSICS I Lecture 5 - uml.edufaculty.uml.edu/pchowdhury/95.141/Lectures/LECTURE5.pdf · • Draw diagram, choose coordinate system • Knowns and unknowns •

CHOWDHURY 95.141 PHYSICS I SPRING 2013 LECTURE 5

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95.141 Feb 6 , 2013 PHYSICS I Lecture 5

Register your i>clickers online at http://www.iclicker.com/registration

Last Lecture Chapter 3 Vector kinematics Projectile motion

Today Chapter 3 Beyond 1-D Vectors Relative Velocity


From two measurements of time and position

displacement

!x1, t1 !x2, t2!!x = !x2 "

!x1 a vector

Vector Kinematics: 1D

tip-to-tail

!x2 !!x1!

!x = !x2 "!x1

time interval !t = t2 " t1 a scalar

average velocity vavg =!!x!t

a vector


Vector Kinematics •  Kinematics in more than one dimension •  Previously described 1D displacement as Δx, where

motion could only be positive or negative. •  In more than 1D, displacement is a vector

v!r

!r (t) !v(t)


Vector Kinematics : 2D

12 xxx −=Δ

12 rrr −=Δ

x

y

!r1 !r2

!!r

displacement (in unit vector notation): !!r = (x2 " x1)i + (y2 " y1) j + (z2 " z1)k

average velocity = !!r!t


Average to Instantaneous Just like in 1D, let Δt get smaller and smaller….

!v = lim!t" 0

!!r!t

=d!rdt

!!r !

!r

As Δt tends to zero, Δr tends to zero as well, but the ratio of Δr/Δt tends to a finite value Instantaneous velocity


Instantaneous Velocity

Derivative of the position vector with respect to time:

!r = x(t)i + y(t) j + z(t)k

!v = d!rdt=dxdti + dy

dtj + dx

dtk

!v = vxi + vy j + vzk


Acceleration Vector

Average acceleration =!!v!t

=!v2 "!v1

t2 " t1

!a = lim!t" 0

!!v!t

=d!vdt

Instantaneous acceleration

!a = d!vdt

=d 2xdt2

i + d2ydt2

j + d2zdt2

k=dvxdt

i +dvydt

j +dvydt

k


Example Problem Given position as a function of time, find position, instantaneous velocity and instantaneous acceleration at t=3s

!v(t) = (8t)i + (2t)k

!a(t) = 8i + 2k

!r (t) = (4t2 !1)i + 2 j + (t2 )k

Plug in t=3s to get the final answers


By splitting the equations of motion into component form, we can solve problems one direction at a time An object starts from rest at the origin, with a constant acceleration in the +x direction of +2 m/s2 and in the –y direction of 1 m/s2.

What is the displacement of the object at t = 4 s ?

!x = v0 xt +12axt

2

!y = v0 yt +12ayt

2

Example Problem


Motion with constant acceleration

Vector equations shorthand for separate scalar equations in x and y coordinates

tavvtatvrr +=++= 0

221

00

tavvtatvyy

tavvtatvxx

yyyyy

xxxxx

+=++=

+=++=

02

21

00

02

21

00

For projectile motion, ax = 0, ay = -g

gtvvgttvyyvvtvxx

yyy

xxx

−=−+=

=+=

02

21

00

000

Vector Kinematics


Projectile Motion: Clicker Quiz

A helicopter moving at a constant horizontal velocity to the right drops a package when at position A. Which of the marked trajectories is closest to that observed by a stationary person on the ground?


Projectile Motion

Red ball fired horizontally at

blue ball

Red ball fired at an angle towards

blue ball

Gravity OFF Gravity ON


•  Draw diagram, choose coordinate system

•  Knowns and unknowns •  Divide equations into x and y •  Solve, noting that in the x and

y calculations the common parameter is the time interval t

Helicopter flying horizontally at 70m/s wants to drop supplies on mountain top 200m below. How far in advance (horizontal distance) should the package be dropped?

Example (Rescue Helicopter)


t = 6.39s

x = 447m

x0 = 0 y0 = 0v0 x = 70 m/s v0 y = 0

ax = 0 ay = !9.8 m/s2

x = ? y = !200 m

y = yo + voyt +12ayt

2

common parameter is time t Solve y-equations to find t

Plug t into x-equations to find x

Example (Rescue Helicopter)

!200 = 0+ 0+ 12(!9.8)t2

x = xo + voxt +12axt

2

x = 0+ 70(6.39)+ 0


Relative Motion: Clicker Quiz

A helicopter moving at a constant horizontal velocity to the right drops a package when at position A. Which of the marked trajectories is closest to that observed by a person on the helicopter?


A golf ball is hit with initial velocity v0 at an angle θ0 above the horizontal. Find –  Equations of motion

•  Draw diagram and choose coordinate system •  Fill in knowns

x

y

v0

v0x

v0y θ0

x = xo + voxt +12axt

2

x0 = 0, y0 = 0v0 x = v0 cos!0, v0 y = v0 sin!0

ax = 0, ay = !g

Again, common parameter is time t

x = (vo cos!0 )t

y = yo + voyt +12ayt

2 y = (vo sin!0 )t !12gt2


Example (Golf Ball)

A golf ball is hit with initial velocity v0 at an angle θ0 above the horizontal. –  The time of flight (how long the ball is in the air) This depends only on the y-component equations, as the acceleration due to gravity is only in the y-direction.

0 = voyt !12gt2y = yo + voyt +

12ayt

2 Since both y0 and y are zero

Two solutions for t where y=0

The second is the time of flight t =2v0 yg

0 = t(voy !gt2)

t = 0 and 2v0 y

g


Example (Golf Ball)

R = 2vo2 sin!0 cos!o

g=vo2 sin2!0g

A golf ball is hit with initial velocity v0 at an angle θ0 above the horizontal. Find –  Range (how far does ball travel on flat ground)

Use constant x-velocity to calculate how far ball travels horizontally during time of flight (Range)

t =2v0 yg

=2v0 sin!

gtime of flight v0 x = v0 cos!0Constant

Range

Maximum Range occurs when sin(2θ0)=1 or θ0=45° Rmax =vo2

g


Example (Golf Ball)

A golf ball is hit with initial velocity v0 at an angle θ0 above the horizontal. –  Trajectory (height y as a function of position x)

Since common parameter is time t, eliminate t to get y(x)

x = v0 xt y = voyt !12gt2

t = xv0 x

y =v0 yv0 x

!

"#

$

%&x '

g2v0 x

2

!

"#

$

%&x2

Equation of parabola y = Ax !Bx2


Example (Golf Ball)

The Speed Bus In the movie, a bus tries to jump a gap in the highway.


The Speed Bus

So we know: !vo,!x

1)  DRAW DIAGRAM!! 2)  Determine knowns 3)  Pick equations

t = !xv0 x

ov

y = ! 12gt2

= !12(9.8)*(0.5)2m

=15.24m31.3m / s

! 0.5s

!1.2m

Δx = 50m

Δy

= 31.3m/s

Bad news!

Is this even close to possible?


But….


The Speed Bus

•  It looks like the bus is magically launched at an angle of 30º! •  Now can the bus make it?

But….

30º


The Speed Bus

Speed Bus with Magic Launch

30º

ov

R = vo2 sin2!0g

Range

=(31.3)2 sin60°

9.81! 85m


Let’s check!


The Speed Bus

Summary

Relative Velocity Vector kinematics Projectile motion


Documents

95.141 Feb 6 , 2013 PHYSICS I Lecture 5 - uml.edufaculty.uml.edu/pchowdhury/95.141/Lectures/LECTURE5.pdf · • Draw diagram, choose coordinate system • Knowns and unknowns •