Physic 231 Lecture 4 - Directory | National ...lynch/phy231_2010/lecture4.pdf · Physic 231 Lecture...
Transcript of Physic 231 Lecture 4 - Directory | National ...lynch/phy231_2010/lecture4.pdf · Physic 231 Lecture...
M i i t f t d ’ l t
Physic 231 Lecture 4
• Main points of today’s lecture:• Example: addition of velocities• Trajectories of objects in 2
dimensions:dimensions:
2 downwards m/s 9.8 g =
( )tvvyyy
gtvv
yoy
yy
0
0
21 +=−≡Δ
−=
ygvv
gttvy
y
y
y2
02
20
221
Δ−=−
−=Δ
tvxvv
x
xx
0
0
=Δ=
Example:
A f tb ll l th tt i i th d i b th th• A football player runs the pattern given in the drawing by the three displacement vectors A, B and C. The magnitudes of these vectors are A=5 m, B=15 m, and C=18 m. Using the component method, find the magnitude and direction of the resultant vector A+B+Cmagnitude and direction of the resultant vector A+B+C.
label x-comp y-compA 0 5 mB 15 m 0C 14.7 -10.3mR 29 7 5 3R 29.7m -5.3,m
0x
0
C 18m cos(35 ) 14.7m;
C 18msin(35 ) 10.3m;
= =
= = −R=
yC 18msin(35 ) 10.3m;
2 2R 29.7 5.3 30.2mtan( ) 5 3 / 29 7θ
= + == −
0
tan( ) 5.3 / 29.710.1
θθ = −
quiz
B b t l 3 k d t d th t l 4 k d th R l ti t• Bob travels 3 km due east and then travels 4 km due north. Relative to his starting position what is the magnitude of his total displacement and the angle with respect to due east?
) 7 k 35 d t th th f d t– a) 7 km, 35 degrees to the south of due east.– b) 7 km, 53 degrees to the north of due east– c) 5 km, 35 degrees to the south of due east– d) 5 km, 53 degrees to the north of due east
vector x comp. y comp.
BR
A 3 km 0
B 0 4 km
3 k 4 k
A
R 3 km 4 km
543 2222yx kmRRR =+=+=
east due of north ,53333.1tan o
x
y
RR
===θ
relative velocity problems
L b l h bj t b l tt th t i d f h t it i (f• Label each object by a letter that reminds you of what it is (for example p for paper, g for ground, t for truck).
• Look for phases such as "the velocity of the paper relative to truck" d it th l itand write the velocity as:
pttrucktorelativepaper vv =___
• Take the three velocities and assemble them into an equation such as;pttgpg vvv +=
• Take the three velocities and assemble them into an equation such as;
S l f th l it t N t th t th l iti d t b• Solve for the velocity you want. Note that these velocities need not be parallel. You may need to solve two equations, one in the x direction and another in the y direction.
exercise
Ch k l k h d d G d H H th hi• Chuck looks ahead and sees Grandpa Harper. He throws him a newspaper over the top of the hood to him. The truck is moving 40 km/hr due West. Relative to the truck, the newspaper also moves West with a velocity of 40 km/hr What is the velocity of the newspaperwith a velocity of 40 km/hr. What is the velocity of the newspaper relative to Grandpa Harper? – a) 0 km/hr
b) 80 k /h d t groundtorelativepaperv– b) 80 km/hr due west– c) 80 km/hr due east– d) don’t know 4040
___
___
+=
+=
groundtorelativepaper
trucktorelativepapertruck
groundtorelativepaper
vvv
Westkm/hr 80___
=gp p
vtruck 40 km/hr West
vpaper relative to truck 40 km/hr West
vpaper relative to ground ?
Example
A f b t i t li i di ti 35 1o th f t ith d• A ferry boat is traveling in a direction 35.1o north of east with a speed of 5.12 m/s relative to the water. A passenger is walking with a velocity of 2.71 m/s due east relative to the boat. What is the velocity (magnitude and direction) of the passenger with respect to the water?no
rth
(magnitude and direction) of the passenger with respect to the water?
eastx
y
n
pw pb bwv v v= +
bw xv 5.12cos(35.1 ) 4.19m / s= ° =
vpb= 2.71 m/s
east bw,x ( )
bw,yv 5.12sin(35.1 ) 2.94m / s= ° =
x y mag θ
vpb 2.71 0 2.71 0
vbw 4.19 2.94 5.12 35.1°
v 6 9 2 94 7 5 23°
35.1oθ
vpw 6.9 2.94 7.5 23°
2 2pwv 6.9 2.94 7.5m / s= + =
( )1 2 94vpb= 2.71 m/s ( )1 2.94tan 23 north of east6.9θ −= = °
Projectile motion in two dimensionsgtvtavv yyyy −=+= 00 00 =+= vtavv xxxx
( )
gttvtatvy
tvvyyy yoy
−=+=Δ
+=−≡Δ
1121
22
0 ( )1
21
2
00
=+=Δ
=+=−≡Δ
tvtatvx
tvtvvxxx xxox
ygyavv
gttvtatvy
yy
yyy
y Δ−=Δ=−
=+=Δ
2222
20
2
00
022
20
2
00
=Δ=−
=+=Δ
xavv
tvtatvx
xx
xxx
x
hi h h f h l i i h• This means that the x-component of the velocity remains constant. The y-component reflects the gravitational acceleration.– This is true; contrary to the presentation of Jason below:
Projectile MotionThe horizontal motion isThe horizontal motion is constant; the vertical motion is free fall:
The horizontal and vertical components of the motion are independent.
Slide 3-37
example
A b ll t i fi d f ifl th t i h ld 1 6 b th d i• A bullet is fired from a rifle that is held 1.6 m above the ground in a horizontal position. The initial speed of the bullet is 1100 m/s. Find (a) the time it takes for the bullet to strike the ground and (b) the horizontal distance traveled by the bullethorizontal distance traveled by the bullet.
If upward is the direction of positive y:
vx0 1100 m/s
vy0 0
direction of positive y:
Δy -1.6 m
xΔ
yΔ
clicker question• The diagram below shows two
successive positions of a particle; it'ssuccessive positions of a particle; it's a segment of a full motion diagram. Which of the following vectors best represents the acceleration betweenrepresents the acceleration between
)
i fv and v
a)
b)
c)
d)d)
example
A ll i h i f th ili A ifl i i d di tl t th• A small can is hanging from the ceiling. A rifle is aimed directly at the can, as the figure illustrates. At the instant the gun if fired, the can is released. Ignore air resistance and show that the bullet will always strike the can regardless of the initial speed of the bullet Assume thatstrike the can, regardless of the initial speed of the bullet. Assume that the bullet strikes the can before the can reaches the ground.
Δygravity Δycan
θ
Reading Quiz 2. The acceleration vector of a particle in projectile motion2. The acceleration vector of a particle in projectile motion
A. points along the path of the particle.B. is directed horizontally.C i h t th ti l ’ hi h t i tC. vanishes at the particle’s highest point.D. is directed down at all times.E. is zero.
Slide 3-9
example• A motorcycle daredevil is attempting to jump across as many buses asA motorcycle daredevil is attempting to jump across as many buses as
possible (see the drawing). The takeoff ramp makes an angle of 18.0°above the horizontal, and the landing ramp is identical to the takeofframp. The buses are parked side by side, and each bus is 2.74 m wide.p p y ,The cyclist leaves the ramp with a speed of 33.5 m/s. What is themaximum number of buses over which the cyclist can jump?
33 5 m/s33.5 m/s _ _
2.74m
v0 33.5 m/s gtvv yy −= a) 0y0v 33.5sin(18 ) 10.35m / s= ° =
21
θ 18o ( )
gttvy
tvvyyy yoy
−=Δ
+=−≡Δ
1c)
21 b)
20
0
2y0
1y v t gt 02
Δ = − =
2y0
1v t gt2
= y02vt 2.11s
g
Which equation to use?ygvv
gttvy
y
y
y Δ−=−
Δ
2 d)2
c)
20
2
0x0x v t 33.5cos(18 )(2.11s) 67.25mΔ = = ° =
busesbus
x 67.25mN 24.5 24busesw 2.74mΔ= = =
Concept problem• A battleship simultaneously fires two shells at enemy ships. If
the shells follow the parabolic trajectories shown which shipthe shells follow the parabolic trajectories shown, which ship gets hit first?
– a) Ab) both at the same time :Bfor velocity verticalinitialthe
exceedsA for velocity verticalinitial
– b) both at the same time– c). B– d) need more information
ByAyA
topy
vvhhghv
ghvv
,0,020
20
2,
;2
02y
>>=
=−=
B
ytottottotytot
yy
gv
tgttvy 020
,,
2210 =−==Δ
BtotAtotByAy ttvv ,,,0,0
:iestrajectortwothecomparing>>