Review of Chapters 1-7

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).

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a = d v dt

⇔ v = d r

dt⇔

r = x(t)ˆ i + y(t) ˆ j + z(t) ˆ k

Projectile motion: x-comp. is constant velocity y-comp. is constant acceleration

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x − x0 = v0x t +12axt

2

vx = v0x+ axtvx

2 = v0x2+ 2ax (x − x0)

Constant acceleration

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x − x0 = vx tConstant velocity

Newton’s 3rd law

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Fr∑ = m v2

R; Ftan∑ = matan ; Fz∑ = maz

Coordinate system is NOT constant, it rotates!!

CH 4-5: Newton’s laws of motion

• 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

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Fx∑ = max ; Fy∑ = may ; Fz∑ = mazFor constant coordinate system

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F on A by B = −

F on B by A

CH 6-7: Work and Energy

Work done by a force is

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Wby F = r 1

r 2∫ F ⋅ d r =

F ⋅ ( r 2 −

r 1) = FΔrcosθFΔr

general constant force

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Wnon−conserv = E2 − E1

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E = KE +UEgrav +UEspring =12m v2+mgy +

12kx 2

Work energy theorem (also contains conservation of energy) Power

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PF =dWdt

= F ⋅ v

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?

FN

mg

T +

+

-

-

FNsinθ

FNcosθ

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x : − FN sinθ +T = 0y : FN cosθ −mg = 0

Write down Newton’s 2nd law

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FN =mgcosθ

T = mgtanθ

Then solve for T and FN

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

m2g

FN2 Ff2

T

FN1

FN2

Ff1=µFN

Ff2

m1g

m1gsinθ

m1gcosθ

!

" Ff 2 = m2gsin#" FN 2 = m2gcos#

!

x1: " Ff 1 "m2gsin# "m1gsin# +T = 0y1: FN1 "m2gcos# "m1gcos# = 0

!

"

x1: # Ff 1 # (m2 +m1)gsin$ +T = 0y1: FN1 # (m2 +m1)gcos$ = 0

!

"

FN1 = (m2 +m1)gcos#T = µ1(m2 +m1)gcos# + (m2 +m1)gsin#

!

x1: " Ff 1 " Ff 2 "m1gsin# +T = 0y1: FN1 " FN 2 "m1gcos# = 0

!

x2 : "m2gsin# + Ff 2 = 0y2 : FN 2 "m2gcos# = 0

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

T

mg

+

+

-

-

Tcosθ

Tsinθ

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⇒ T =mgcosθ

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⇒mgcosθ

sinθ = m 4π2R

T 2

⇒ T =4π 2Rgtanθ

=4π 2(3.00 + 5.00sin30)

(9.8)tan30s

!

r : + T sin" = mv2

R= m

4# 2RT 2

y : T cos" $mg = 0

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

FN

Ff=µFN

mg

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x : − Ff = may : FN −mg = 0

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⇒ − µFN = ma⇒ FN = mg

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⇒ − µg = aKnowing a you can use kinematic eqns.

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v2 f = v20+ 2a(x − x0)0 = v20− 2µg(x − x0)

v0 = 2µg(x − x0) =

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Wnon−conserv = E2 − E1

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E = KE +UEwhere

One can also use the work-energy theorem concepts

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Wnon−conserv = −Ff (x − x0) = −µmg(x − x0)

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E0 =12m v0

2+mgy0 =12m v0

2+ 0

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E f =12m v f

2+mgy f = 0 + 0

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⇒ − µmg(x − x0) = 0 − 12mv0

2

v0 = 2µg(x − x0)

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?

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EA = EB

12m vA

2+mgyA +12kxA

2 =12m vB

2+mgyB +12kxB

2

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0 + 0 + 0 = 0 +mg(−hmax ) +12kxmax

2

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k =2mghmaxxmax2 =

2(91.0)(9.8)(41.0)(11.0)2

= 604 Nm

This way you find the k you need for the spring and then use equilibrium problem set-up

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kx = F ⇒ x =Fk

= 0.69m

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 ?

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Wn−con from B→C = EC − EB

Wn−con from B→C =12m vC

2+mgyC −12m vB

2−mgyB

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−µmg(xA − xB ) = 0 + 0 − 12m vB

2− 0

⇒ µ =12

vB2

g(xA − xB )= 0.39

(a)

(b)

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Wn−con from A→B = EB − EA

Wn−con from A→B =12m vB

2+mgyB −12m vA

2−mgyA

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Wn−con from A→B =12(0.200)(4.80) 2 + 0 − 0 − (0.200)(9.8)(1.60) J

Wn−con from A→B = − 0.83 J