Download Ch 1-7 Review Lecture

Review of Chapters 1-7
CH 4-5: Newton’s laws of motion
CH1-3: Kinematics: equations of motion
• Time of flight, rotation, etc.
• IF you know acceleration then all motion follows
• General understanding of acceleration: Acceleration
component along/against velocity vector increases/
decreases speed; perpendicular acceleration component
€
changes direction (left or right). IF particle is going along a
circle the radial component is equal to v2/r (due to geometry,
otherwise it spirals in or out).

 dv
a=
dt

 dr

⇔v=
⇔ r = x(t)iˆ + y(t) ˆj + z(t) kˆ
dt
Constant acceleration
Constant velocity
€
x − x 0 = vx t
1
x − x 0 = v 0x t + ax t 2
2
v x = v 0x + ax t
v x 2 = v 0x 2 + 2ax (x − x 0 )
€
Projectile motion:
x-comp. is constant velocity
y-comp. is constant acceleration
For constant coordinate system
∑F
x
= max ; ∑ Fy = may ; ∑ Fz = maz
Newton’s 3 law
Fon A by B = −Fon B by A
rd
• They are the ones from which you find the acceleration of
objects (connection to Ch. 1-3)
• Steps: (1) draw sketch, (2) draw all€forces and label 3rd law pairs,
(3) draw free body diagram for each object, (4) choose
coordinates for each object (if circular motion there is no choice,
one has to be radial –positive towards center- and the other
tangential), (5) decompose forces that are not along axis chosen,
(6) write Newt. 2nd law for EACH object, (7) are there relations
among objects (e.g. same velocity, or one twice the other, etc.),
(8) how many equations and how many unknowns. NOW you are
ready to solve for the question – this is a good time to look back at
the question.
• Force of friction: know distinction between static (no
acceleration) and kinetic (there is motion relative to the surface)
• Circular motion: if moving along a circle sum of forces along the
radial direction MUST add to mv2/r
Coordinate system
is NOT constant, it rotates!!
2
€
∑F
r
=m
v
; ∑ Ftan = matan ; ∑ Fz = maz
R
CH 6-7: Work and Energy
general
Work done by a force is
W by F =
∫

r2

r1
€
constant force
    
F ⋅ dr = F ⋅ ( r2 − r1 ) = FΔr cosθ FΔr
Power
Work energy theorem (also contains conservation of energy)
W non −conserv = E 2 − E1
1
1
€
E = KE + UE grav + UE spring = m v 2 + mgy + kx 2
2
2
€
€
PF =
€
 
dW
= F⋅ v
dt
Problem 5.15
A horizontal wire holds a solid uniform ball of mass m in place on a tilted ramp that rises 35.0 degrees
above the horizontal. The surface of this ramp is perfectly smooth, and the wire is directed away from
the center of the ball (the figure ). How hard does the surface of the ramp push on the ball and what is
the tension in the wire?
FNsinθ
Write down Newton’s 2nd law
+
x : − FN sin θ + T = 0
y : FN cosθ − mg = 0
FN FNcosθ
T
-
Then solve for T and FN
+
mg
cosθ
T = mgtan θ
FN =
€
mg
-
€
Problem 5.33
You are taking up two boxes, one on top of the other, up the ramp shown in the figure by pulling on a
rope parallel to the surface of the ramp. Both boxes move together at a constant speed of 20.0 cm/s.
The coefficient of kinetic friction between the ramp and the lower box is 0.438, and the coefficient of
static friction between the two boxes is 0.766. What is T? What is the friction on the upper box?
Write down Newton’s 2nd law for each
x1: " F f 1 " F f 2 " m1gsin # + T = 0
y1: FN1 " FN 2 " m1gcos # = 0
x2 : " m2 gsin # + F f 2 = 0 " F f 2 = m2 gsin #
!
FN2 Ff2
m2 g
y2 : FN 2 " m2 gcos# = 0
T
FN2
x1: " F f 1 " m!2 gsin # " m1gsin # + T = 0
!
y1: FN1 " m2 gcos# " m1gcos # = 0
FN1
"
x1: # F f 1 # (m2 + m1 )gsin $ + T = 0
Ff2
Ff1=µFN
m1gcosθ
!
y1: FN1 # (m2 + m1 )gcos $ = 0
"
m1 g
m1gsinθ
" FN 2 = m2 gcos #
!
FN1 = (m2 + m1 )gcos#
T = µ1 (m2 + m1 )gcos# + (m2 + m1 )gsin #
Problem 5.52
The "Giant Swing" at a county fair consists of a vertical central shaft with a number of horizontal arms
attached at its upper end. Each arm supports a seat suspended from a cable 5.00 m long, the upper
end of the cable being fastened to the arm at a point 3.00 m from the central shaft. Find the time of one
revolution of the swing if the cable supporting a seat makes an angle of 30 degrees with the vertical.
Write down Newton’s 2nd law for each
v2
4# 2 R
r : + T sin" = m = m 2
R
T mg
y : T cos" $ mg = 0 ⇒ T =
cosθ
+
Tsinθ
+
T
Tcosθ
-
!
mg
4π 2 R
⇒
sin θ = m 2
€ T
cosθ
⇒T =
mg
€
4π 2R
=
gtan θ
4 π 2 (3.00 + 5.00sin 30)
s
(9.8)tan 30
Problem 5.80
You are called as an expert witness in the trial of a traffic violation. The facts are these: A driver
slammed on his brakes and came to a stop with constant acceleration (hitting his physics professor in
the process). Measurements of his tires and the skid marks on the pavement indicate that he locked his
car's wheels, the car traveled 192 ft before stopping, and the coefficient of kinetic friction between the
road and his tires was 0.750. The charge is that he was speeding in a 45 mph zone. He pleads
innocent and claims the professor, as usual, was not paying attention. The professor calls you as an
impartial witness (although telling you that if you get the problem wrong you will fail the course)
First you convert to mks units: 192 ft=58.5 m and 45 mph=20.1 m/s
⇒ − µFN = ma ⇒ − µg = a
Knowing a you can
y : FN − mg = 0 ⇒ FN = mg
x : − F f = ma
use kinematic eqns.
v 2 f = v 2 0 + 2a(x − x 0 )
0 = v 2 0 − 2 µg(x − x 0 )
v 0 = 2 µg(x − x 0 ) =
FN
€
€ Ff=µFN
€
mg
One can also use the work-energy theorem concepts
W non −conserv = E 2 − E1
where
W non −conserv = −F f (x − x 0 ) = −µmg(x − x 0 )
€
€
1
1
E 0 = m v 0 2 + mgy 0 = m v 0 2 + 0
2
2
€
1
E f = m v f 2 + mgy f = 0 + 0
2
E = KE + UE
1
⇒ − µmg(x − x 0 ) = 0 − mv 20
2
v 0 = 2 µg(x − x 0 )
Speaking about collisions (Ch 8)
Problem 7.51: Life and death (and marriage)
A bungee cord is 30.0 m long and, when stretched a distance , it exerts a restoring force of magnitude
kx. Your father-in-law (mass 91.0 Kg) stands on a platform 45.0 m above the ground, and one end of
the cord is tied securely to his ankle and the other end to the platform. You have promised him that
when he steps off the platform he will fall a maximum distance of only 41.0 before the cord stops him.
You had several bungee cords to select from, and you tested them by stretching them out, tying one
end to a tree, and pulling on the other end with a force of 420 N. When you do this, what distance will
the bungee cord that you should select have stretched?
EA = EB
1
1
1
1
m v A 2 + mgy A + kx A2 = m v B 2 + mgy B + kx B2
2
2
2
2
1 2
0 + 0 + 0 = 0 + mg(−hmax ) + kx max
2
€
k=
2mghmax 2(91.0)(9.8)(41.0)
N
=
=
604
2
x max
(11.0) 2
m
This way you find the k you need for the spring and then use equilibrium problem set-up
€
€
kx = F
€
⇒
x=
F
= 0.69m
k
Problem 7.65: post office job
In a truck-loading station at a post office, a small 0.200 Kg package is released from rest at point A on a
track that is one-quarter of a circle with radius 1.60 m (the figure ). The size of the package is much
less than 1.60 m, so the package can be treated as a particle. It slides down the track and reaches
point B with a speed of 4.80 m/s. From point B , it slides on a level surface a distance of 3.00 m to point
C, where it comes to rest. (a) What is the coefficient of kinetic friction on the horizontal surface? (b)
How much work is done on the package by friction as it slides down the circular arc from to ?
W n −con from B →C = E C − E B
(a)
1
1
W n −con from B →C = m vC 2 + mgyC − m v B 2 − mgy B
2
2
1
−µmg(x A − x B ) = 0 + 0 − m v B 2 − 0
2
2
1
vB
⇒µ=
= 0.39
2 g(x A − x B )
€
(b)
W n −con from A → B = E B − E A
€
€
€
1
1
W n −con from A → B = m v B 2 + mgy B − m v A 2 − mgy A
2
2
1
W n −con from A → B = (0.200)(4.80) 2 + 0 − 0 − (0.200)(9.8)(1.60) J
2
W n −con from A → B = − 0.83 J