ch5.1 - centripetal force inertia (velocity) top view centripetal force

Ch5.1 - Centripetal Force

INERTIA (velocity)

Top View

CentripetalForce

There is no force pushing outward! Inertia wants object to fly off tangent to circle.(Centrifugal Force is a fake force we think we feel throwing us outward.)

Centripetal Force – real force acting towards center of circle.

INERTIA (velocity)

Top View

CentripetalForce

Ex 1) What agent exerts the centripetal force in each?

earth

m


earth

mFg Fnet = Fg


earth

mFg Fnet = Fg

m.ac = Fg


FT Fnet = FT

Ex 1) what agent exerts the centripetal force in each?

FT Fnet = FT

m.ac = FT


(Top view of record player)

.

(coin)



.

(coin)

Ff,s

Fnet = Ff,s



.

(coin)

Ff,s

Fnet = Ff,s

m.ac = Ff,s

Centripetal Acceleration

r

vac

2


r

vac

2

Ex 2) The ride “Spin- Out” is a circular room, 5m in diameter. Once it gets to speed, its linear speed is 10 m/s. What is the centripetal acceleration?


r

vac

2

Ex 2) The ride “Spin- Out” is a circular room, 5m in diameter. Once it gets to speed, its linear speed is 10 m/s. What is the centripetal acceleration?

2

2

405.2

10

sm

ms

mac

CENTRIPETAL FORCEcc amF

CENTRIPETAL FORCE

r

vmF

amF

c

cc

2

CENTRIPETAL FORCE

r

vmF

amF

c

cc

2

Ex 3) A 60 kg astronaut stands on a bathroom scale in a 2 km diameter rotating space station that spins every 62.8 sec. What does the scale read?

CENTRIPETAL FORCE

r

vmF

amF

c

cc

2

Ex 3) A 60 kg astronaut stands on a bathroom scale in a 2 km diameter rotating space station that spins ever 62.8 sec. What does the scale read?

Fn

Fnet = FN

CENTRIPETAL FORCE

r

vmF

amF

c

cc

2


Fn

Fnet = FN

m.ac = FN

nFr

vm

2

CENTRIPETAL FORCE

r

vmF

amF

c

cc

2


Fnet = FN

m.ac = FN

NF

Fms

mkg

Fr

vm

N

N

n

6001000

100)60(

2

2

Ex 4) A sport car on a flat track rounds a corner with a radius of 100m, at a maximum speed of 25 m/s before sliding out. What is the coefficient of static friction?

Ch5 HW#1 1 – 5

Ex 4) A sport car on a flat track rounds a corner with a radius of 100m, at a maximum speed of 25 m/s before sliding out. What is the coefficient of static friction?

Ff,s

Fnet = Ff,s

Ch5 HW#1 1 – 5 1. A youngster on a carousel horse 5.0 m from the center revolves at

a constant rate, once around in 15.0 s. What is her acceleration?

2. A beetle standing on the edge of a 12-inch record (r = .152m) whirls around at 33.3 rotations per minute. What is its centripetal accl? What agent exerts the force?

3. A ‘foreign-made’ space station only has a radius of 3 meters. If it rotates around once every 1.15 sec, what is the centripetal accl at the floor? If an astronaut is 1.5 m tall, what is the centripetal accl at his head? Does anyone see any physiological implications of this? How could we remedy this?

r

vac

2

r

vac

2

r

vac

2


At head: At feet:

r

va

r

va

t

rv

t

rv

cc

22

2

2


At head: At feet:

2

22

2

22

900.3

16 45

5.1

8

1615.1

)3(22 8

15.1

)5.1(22

sm

ms

m

r

va

sm

ms

m

r

va

sm

s

m

t

rvs

ms

m

t

rv

cc

4. The Talladega Speedway …1500kg cars complete turns of radius 100mat a speed of 80m/s. What is net force on car? What agent exerts it?

Fnet = Fn

m.ac = Fn

5. 700kg flat track car rounds a corner of radius 30m at a speed of 20m/s, without slipping. What is the minimum coeffiecient of static friction that can accomplish this?

Fnet = Ff,s

m.ac = μ.Fn

Ch5.2 - GravityAll matter is attracted to all other matter.

- More matter = larger force- Distance apart is important

(Inverse square law)Newton’s Law of Universal Gravitation




- Universal gravitational

constant, G = 6.67x 10-11

Ex1) Compute the gravitational attraction between 2 100kg uniform spheres by 1.00m. (Roughly 2 220-lb football players)

Ex2) The mass of the moon is 7.35x 1022 kg and its distance from the Earth is 3.84x 10 3 km. Taking the Earth’s mass to be 5.98x10 24 kg, what force keeps the moon in her orbit?

221

r

mGmF

2

2.

kg

mN




- Universal gravitational constant, G= 6.67x 10-11

Ex2) The mass of the moon is 7.35x 1022 kg and its distance from the Earth is 3.84x 10 3 km. Taking the Earth’s mass to be 5.98x10 24 kg, what force keeps the moon in her orbit?

221

r

mGmF

Ex3) 2 objects both of mass m are a distance r apart and exert a force F on each other. How does the force change if:

a) Both masses are doubled.

b) Instead the distance is doubled?

c) Distance made 3X smaller?

221

r

mGmF

Ex4) How fast does an object have to travel, to stay in circle?- what is the direction of the instantaneous velocity?- what would happen if it traveled slower?- what would happen if it traveled faster?

Ch5 HW#2 1 – 4

M

m

Ch5 HW#2 1 – 4 1. Using our equation for the universal law of gravitation:

Explain how the force will change if:a) One mass doubles

b) Both masses cut in half and the distance between them cut in half.

c) The masses stay the same and the distance triples.

d) The masses double and the distance between is cut in half.

221

r

mGmF

Ch5 HW#21. Using our equation for the universal law of gravitation:

Explain how the force will change if:a) Both masses double

b) Both masses cut in half and the distance between them cut in half.

c) The masses stay the same and the distance quadruples.

d) The masses triple and the distance between is cut to one third.

221

r

mGmF

221

221

221 2

))(2(

r

mGm

r

mmG

r

mGmF

221

24

1

2141

22

1

221

121

221

)(

))((

r

mGm

r

mGm

r

mmG

r

mGmF

2. I want you to calculate my weight 2 ways:

(My mass = 75 kg; Earth’s mass = 6x1024 kg; Earth’s radius = 6x106 m)

1) Fg = m.g

2)

(Roughly the same with rounding error.)

22

r

mGMF EG

3. What is the force of gravity on me (75 kg) on the surface of Jupiter?

MJ = 2x1027 kg RJ = 7x107 m

4. What is the gravitational attraction between the earth and the Sun?

Msun = 2x1030 kg DES = 1.5x1011 m



1) Fg = m.g = (75kg)(9.8m/s2) = 748.5N

2)


Nx

xx

r

mGMF EG 833

)106(

)75)(106)(1067.6(26

2411

22


MJ = 2x1027 kg RJ = 7x107 m


Msun = 2x1030 kg DES = 1.5x1011 m



1) Fg = m.g = (75kg)(9.8m/s2) = 748.5N

2)


Nx

xx

r

mGMF EG 833

)106(

)75)(106)(1067.6(26

2411

22


MJ = 2x1027 kg RJ = 7x107 m

Nx

xx

R

mGMF

J

JG 2042

)107(

)75)(102)(1067.6(27

2711

22


Msun = 2x1030 kg DES = 1.5x1011 m

Ch5.3 - Gravity ApplicationsAcceleration of gravity at surface of any object:

Apparent weightlessness-What does the bathroom scale read in each accelerating elevator?

(scale reads FN) (assume m= 100kg)

a = 0 m/s2 a = 10 m/s2

Satellite Orbits

Orbital Speed:

Ex1) A satellite in geostationary orbit must be at a height of 3.6x107m above the earth’s surface. What speed must it travel at to stay directly above good ol’ AV?

Kepler’s 3 Laws of Planetary Motion:1) The planets move in elliptical orbits with the sun as 1 focus.2) Each planet sweeps out equal (areas) in equal time intervals.3) The ratio of the average distance from the Sun cubed

to the period squared is the same constant value.

Ch5 HW#3

Ch5 HW#3 p172 37,44,46,47, + Hard Bonus ProblemHard Bonus Problem. (Year 2000 AP Test FRQ #1)

Prove that the acceleration on the surface of Mars is 38% the acceleration on the surface of the Earth, given MMars = 0.11.MEarth, RMars = 0.53.REarth.

37. The acceleration due to gravity on the surface of mars is 3.7 m/s2. If the planet’s diameter is 6.8x106 m, determine the mass of the planet and compare it to Earth.

Fnet = FG

37. The acceleration due to gravity on the surface of mars is 3.7 m/s2. If the planet’s diameter is 6.8x106 m, determine the mass of the planet and compare it to Earth.

Fnet = FG

26

11

2

)104.3(

)1067.6(7.3

x

Mx

R

mGMam

m

m

m

Mm = 6.4x1023 kg

44. Locate the position of a spaceship on the Earth-Moon center line, such that at that point, the tug of each celestial body exerted on it will cancel and the craft would literally be weightless.

Fnet = FER – FRM

E m

Fnet = FER – FRM

E m


Fnet = FER – FRM

E m

28

22

2

24

22

22

22

)104(

107106

)(

0

ERER

EREM

M

ER

E

RM

RM

ER

RE

RM

RM

ER

RE

rx

x

r

x

rr

M

r

M

r

mGM

r

mGM

r

mGM

r

mGM

RER = 3.46x108m (346,000km)


46. Three very small spheres of mass 2.50 kg, 5.00 kg, and 6.00 kg are located on a strait line in space awayf rom everything else. The first one is a point between the other two, 10.0 cm to the right of the second and 20.0 cm to the left of the third. Compute the net gravitational force on it.

5 62

46. Three very small spheres of mass 2.50 kg, 5.00 kg, and 6.00 kg are located on a strait line in space awayf rom everything else. The first one is a point between the other two, 10.0 cm to the right of the second and 20.0 cm to the left of the third. Compute the net gravitational force on it.

5 62

Nxx

r

mGmFG

82

11

225 103.8

)10.0(

)5.2)(5)(1067.6(

Nxx

r

mGmFG

82

11

226 105.2

)20.0(

)5.2)(6)(1067.6(

47. It is believed that during the gravitational collapse of certain stars, such great densities and pressures will be reached that the atoms themselves will be crushed, leaving only a residual core of neutrons. Such a neutron star is, in some respects, very much like a giant atomic nucleus with a tremendous density of roughly about 3 x 1017 kg/m3. DON’T compute the surface acceleration due to gravity for a one-solar-mass neutron star, just trip out on this unfathomable fact!

At all points on the circle, Fnet points to the center of the circle.

FN also always points to the center,but its value changes.

Fg only plays a role top and bottom. (At the sides Fg is not in the direction

of Fnet so we disregard it.)

(The scale reads FN)

(The scale reads more on the bottom, less on the top.)

2

3 1

4

v = 14 m/s

v = 14 m/s

v = 11 m/s

v = 20 m/s

Ch5.5 – Roller CoastersEx1) 60kg person sits on bathroom scale on roller coaster that does a loop of radius 10m. What does the scale read at 1-4?

Ch5.4 – Roller CoastersEx1) 60kg person sits on bathroom scale on roller coaster that does a loop of radius 10m. What does the scale read at 1-4?

2

3 1

4

v = 14 m/s

v = 14 m/s

v = 11 m/s

v = 20 m/s

v = 14 m/s

v = 11 m/s

v = 20 m/s

v = 14 m/s1.

3.

2.

4.

Fnet = FN

Fnet = FN

Fnet = FN + Fg

Fnet = FN – Fg

Ex2) A 50kg student sits on a bathroom scale while riding the Revolution. As the coaster enters the loop it begins slowing. At the top of the loop it has a speed of 14 m/s. By the bottom of the loop it has sped up to 20 m/s. If the loop has a radius of 7m, what does the scale read at the top and bottom?

v = 14 m/s

v = 20 m/s

Ch5 HW#4

Ex2) A 50kg student sits on a bathroom scale while riding the Revolution. As the coaster enters the loop it begins slowing. At the top of the loop it has a speed of 14 m/s. By the bottom of the loop it has sped up to 20 m/s. If the loop has a radius of 7m, what does the scale read at the top and bottom?

v = 14 m/s

v = 20 m/s

v = 14 m/s

v = 20 m/s

Ch5 HW#4

Lab5.1 Circular Motion - due at end of period

- Go over Ch5 HW#4

- Ch5 HW#5 due tomorrow

Ch5 HW#4 1 – 4 1. A front-loading clothes washer has a horizontal drum that is thoroughly perforated with small holes. Assuming it to spin dry at 1 rotation per second, have a radius of 40 cm, and contain a 4.5-kg wet teddy bear, what maximum force is exerted by the wall of the bear? What happens to the water?

1. A front-loading clothes washer has a horizontal drum that is thoroughly perforated with small holes. Assuming it to spin dry at 1 rotation per second, have a radius of 40 cm, and contain a 4.5-kg wet teddy bear, what maximum force is exerted by the wall of the bear? What happens to the water?

Fnet = FN – Fg m.ac = FN – mg FN FN = mv2/r + mg

= 71N + 44N = 115N

Fg

2. While whirling around in a vertical circle with a radius of 1.50 m, a 2.00 kg mass is held on a rope attached to a very light spring scale. What does the scale read when the mass is moving at 4.00 m/s at the lowest point in its orbit? What does it read at the top when moving at 2.50 m/s? What does it read on the way down when moving at 3.00 m/s?

Top: Fnet = FN + Fg

Side: Fnet = FN

Bottom: Fnet = FN – Fg

2. While whirling around in a vertical circle with a radius of 1.50 m, a 2.00 kg mass is held on a rope attached to a very light spring scale. What does the scale read when the mass is moving at 4.00 m/s at the lowest point in its orbit? What does it read at the top when moving at 2.50 m/s? What does it read on the way down when moving at 3.00 m/s?

Top: Fnet = FN + Fg mac = FN + mg FN = mv2/r – mg = 1.7N

Side: Fnet = FN mac = FN

FN = mv2/r = 21.3N

Bottom: Fnet = FN – Fg mac = FN – mg

FN = mv2/r + mg = 41N

v

3. A scale is fitted in the seat of a roller coaster car and a person weighing 800 N sits down on it. The car then descends along a path that has the shape of a 100.0 m radius vertical circle with its lowest point at the bottom where the car reaches its greatest speed of 40.0 m/s. What is the maximum reading of the scale?

v = 40 m/s

Fnet = FN + Fg

v = 40 m/s

Fnet = FN – Fg

mac = FN – 800N

NFN 2080800100

)40)(80( 2

v = 40 m/s

3. A scale is fitted in the seat of a roller coaster car and a person weighing 800 N sits down on it. The car then descends along a path that has the shape of a 100.0 m radius vertical circle with its lowest point at the bottom where the car reaches its greatest speed of 40.0 m/s. What is the maximum reading of the scale?

FN

Fg

Ch5 HW#5 p170+ 25,27,3025. What would happen to the weight of an object if its mass were doubled

and its distance to the center of the earth were also doubled?

27. Suppose 2 identical spheres separated by 1.00 m experience a force of 1.00N on each other. Compute masses.

30. Compute g-force between earth n moon, sun n moon.

Ch 5.5 Conservation of Energy and Roller Coasters

FRQ#1) Part of the track of an amusement park roller coaster is shaped as shown. A safety bar is oriented lengthwise along the top of each car. In one roller coaster car, a small 0.10kg ball is suspended from this bar by a short length of light, inextensible string.a) Initially, the car is at rest in point A. i. On the diagram to the right, draw and label

all the forces acting on the 0.10kg ball. ii. Calculate the tension in the string.

The car is then accelerated horizontally, goes up a 30° incline, and then goes around a vertical circular loop of radius 25 meters. For each of the four situations described in parts (b) to (e), do all three of the following: In each situation, assume that the ball has stopped swinging back and forth.

• Determine the horizontal component Th of the tension in the string

in newtons and record your answer in the space provided.

• Determine the vertical component Tv of the tension in the string

in newtons and record your answer in the space provided.• Show the adjacent diagram the approximate direction of the string

with respect to the vertical. The dashed line shows the vertical line in each situation.b) The car at point B moving horizontally to the right

with an acceleration of 5.0 m/s2.

Th = ____________N Tv = _____________N

c) The car is at point C and is being pulled up the 30° incline with a constant speed of 30 m/s.

Th =____________N Tv =_____________N

d) The car is at point D moving down the 30° incline with an acceleration of 5.0 m/s2.

Th =____________N Tv =_____________N

e) The car is at point E moving upside down with an instantaneous speed of 25 m/s and no tangential acceleration at the top of the vertical loop of radius 25 meters.

Th =____________N Tv =_____________N

Ch5 HW#6 RC FRQ

Ch5 HW#6 Roller Coaster FRQA roller coaster ride at an amusement park lifts a car of mass 700kg to pt A at a height of 90m above the lowest point on the track, as shown. The car starts from rest at point A, rolls with negligible friction down the incline and follows the track around a loop of radius 20m. Point B, the highest point on the loop, is at a height of 50m above the lowest point on the track.a) i. indicate on the figure the point P at which the maximum speed of the car is attained.

ii. Calculate the value vmax of this maximum speed.

b) Calculate the speed vB of the car at point B.

c) i. On the figure of the car, draw and label vectors to represent the forces acting on the car when it is upside down at point B. ii. Calculate the magnitude of all the forces identified in (c-i). d) Now suppose that friction is not negligible. How could the loop be modified to maintain the same speed at the top of the loop as found in (b)? Justify your answer.

Ch5 HW#6 Roller Coaster FRQA roller coaster ride at an amusement park lifts a car of mass 700kg to pt A at a height of 90m above the lowest point on the track, as shown. The car starts from rest at point A, rolls with negligible friction down the incline and follows the track around a loop of radius 20m. Point B, the highest point on the loop, is at a height of 50m above the lowest point on the track.a) i. indicate on the figure the point P at which PE KE the maximum speed of the car is attained. mgh = ½mv2

ii. Calculate the value vmax of this maximum speed. v = 42 m/s


c) i. On the figure of the car, draw and label vectors to represent the forces acting on the car when it is upside down at point B. ii. Calculate the magnitude of all the forces identified in (c-i). d) Now suppose that friction is not negligible. How could the loop be modified to maintain the same speed at the top of the loop as found in (b)? Justify your answer.

a) i. indicate on the figure the point P at which PE KE the maximum speed of the car is attained. mgh = ½mv2

ii. Calculate the value vmax of this maximum speed. v = 42 m/s

b) Calculate the speed vB of the car at point B. mgh = ½mv2

c) i. On the figure of the car, draw and label vectors v = 28 m/s to represent the forces acting on the car when it is upside down at point B. ii. Calculate the magnitude of all the forces identified in (c-i). d) Now suppose that friction is not negligible. How could the loop be modified to maintain the same speed at the top of the loop as found in (b)? Justify your answer.


mgh = ½mv2 v = 28 m/s

c) i. On the figure of the car, draw and label vectors to represent the forces acting on the car when it is upside down at point B. ii. Calculate the magnitude of all the forces identified in (c-i).

d) Now suppose that friction is not negligible. FN

How could the loop be modified to maintain the same speed

at the top of the loop as found in (b)? Justify your answer. Fg

c-ii) Fnet = Fg + FN

NF

FNN

FN

Fmgr

mv

N

N

N

N

440,20

7000440,27

700020

)28)(700(

2

2


mgh = ½mv2 v = 28 m/s

c) i. On the figure of the car, draw and label vectors to represent the forces acting on the car when it is upside down at point B. ii. Calculate the magnitude of all the forces identified in (c-i).

d) Now suppose that friction is not negligible. FN

How could the loop be modified to maintain the same speed

at the top of the loop as found in (b)? Justify your answer. Fg

c-ii) Fnet = Fg + FN

d) Drop the loop lower.More PE needed PE = KE + Wf

NF

FNN

FN

Fmgr

mv

N

N

N

N

440,20

7000440,27

700020

)28)(700(

2

2

Ch 8.1 – Rotational MotionAngular Displacement- distance around in circles

r

Ch 8.1 – Rotational MotionAngular Displacement- distance around in circles

rC 2

2r

C

r

s

rs or r

1 radianwhen s=r

or 6.28 radaround circle2

Ex1) A pendulum of length .40m sweeps out an angle of 30° to cover half its period. What is displacement of the pendulum bob?

o30

Angular speed - how fast you go in circles Linear speed (v)

relates to angular speed (ω)

t

omegaradians

sec

Angular speed - how fast you go in circles

Ex p243) A horse completed a race around a 1 mile track in 92.2 seconds, at an average speed of 17.4 m/s. What was the angular speed?

Linear speed (v) relates to angular speed (ω)

t

omegarad

sec

rvt

rv

t

sv

Angular acceleration:Tangential acceleration:

Ex p246) At one moment in a race, a race car moving around a turn of radius 50m had an angular speed of 0.60 rad/s and an angular accerastion of 0.20 rad/s2.a) Linear speed:

b) Centripetal accl:

c) Tangential accl:

t

raT

Ch8 HW#1 p278 7,13,14,19,25,27,35,42

Ch8 HW#1 p278 7,13,14,19,25,27,35,427. A model plane at the end of a control line circles at a constant speed 10.6 times around in 50.0 s. Through how many radians does it fly in 25.0 s?

In 25 sec, completes 5.3 revs.

13. If a ball 30 cm in diam rolls 65 m without slipping, how many revs did it make in the process?

s = 65m C = 2πr r = .15m = 2π(.15m)

=

14. The bob at the end of pendulum 100 cm long swings out an arc 15.0 cm in length. Find the angle in radians and degrees through which it moves. Check your answer by determining the fraction of the complete circle to which this corresponds and then taking that fraction of 2 pi.



5.3 rev 2π rad = 33.3 rad 1 rev


s = 65m C = 2πrr = .15m = 2π(.15m)

=






s = 65m C = 2πr 65 m 1 rev = 69 rev r = .15m = 2π(.15m) .94 m

= .94m/rev






s = 65m C = 2 πr 65 m 1 rev = 69 rev r = .15m = 2 π(.15m) .94 m

= .94m/rev


θ = s/r .15rad 180° = 8.6° = .15m/1.00m π rad

= .15 rad

19. The angular speed of a wheel is to be determined by affixing a tiny mirror to the circumference and recording the returning light bounced off it as it spins past a laser beam. If 100 return pulses are detected in 0.0200s, what is the angular speed of the wheel?

ω = ? ω =

25. Thirty seconds after the start button is pressed on a big electric motor, its shaft is whirling around at 500 rev/s. Determine its average acceleration.

ωi = 0 ωf = 500 rev/s ω = 100 rev 2π rad = 3142 rad/s t = 30 sec 0.02 s 1 revα = ? α = ∆ω =

∆t

27. A steam engine is running at 200 rpm when the engineer shuts it off. The friction of its various parts produces torques that combine to decelerate the machine at 5.0 rad/s . How long will it take to come to rest?

ωf = 0ωi = 200 rpm = 21 rad/s α = - 5 rad/s2


ω = ? ω = 100 rev 2π rad = 3.14x104 rad/s 0.02 s 1 rev


ωi = 0 ωf = 500 rev/s ω = 100 rev 2π rad = 3142 rad/s t = 30 sec 0.02 s 1 revα = ? α = ∆ω =

∆t






ωi = 0 ωf = 500 rev/s ω = 100 rev 2π rad = 3142 rad/s t = 30 sec 0.02 s 1 revα = ? α = ∆ω = 3142 rad/s – 0

rad/s ∆t 30 sec

= 104.7 rad/s2






ωi = 0 ωf = 500 rev/s ω = 100 rev 2π rad = 3142 rad/s t = 30 sec 0.02 s 1 revα = ? α = ∆ω = 3142 rad/s – 0

rad/s ∆t 30 sec

= 104.7 rad/s2


ωf = 0 ∆t = ∆ω = 0 rad/s – 21 rad/s = 4.2 sec

ωi = 200 rpm = 21 rad/s α -5 rad/s2

α = - 5 rad/s2

35. New York City is traveling around at a tangential speed of about 353 m/s (790 mph) as the Earth spins. Assuming the planet is a sphere of radius 6371km, how long is the perpendicular from the city to the spin axis? Compute the city’s latitude (the angle measured above the equator).

vNYC = 353 m/s radius to NYC

t = 86,400 s ω =7.27x10-5 rad/s r = 6.37x106 m θ = 40°

Ch8.2 Angular Equations

Linear Equation Angular Equivalent

1. vf = vi + at 1.



1. vf = vi + at 1. ωf = ωi + αt2. s = ½(vi+ vf)t 2



1. vf = vi + at 1. ωf = ωi + αt2. s = ½(vi+ vf)t 2. θ = ½(ω i+ ω f)t3. s = vit + ½at2 3.



1. vf = vi + at 1. ωf = ωi + αt2. s = ½(vi+ vf)t 2. θ = ½(ωi+ ωf)t3. s = vit + ½at2 3. θ = ωit + ½αt2 4. vf

2 = vi2 + 2as 4.

Ch8.2 Angular EquationsLinear Equation Angular Equivalent

1. vf = vi + at 1. ωf = ωi + αt2. s = ½(vi+ vf)t 2. θ = ½(ωi+ ωf)t3. s = vit + ½at2 3. θ = ωit + ½αt2 4. vf

2 = vi2 + 2as 4. ωf

2 = ωi2 + 2αθ

Ex1) Mounted on a bus engine is a 2.0m diameter flywheel, a massivedisk used to store rotational energy. If it is accl at a const rate of 2rpm per sec, what will be the angular speed of a point on its rim after 5 sec? Thru what angle will it have rotated?

Ex2) A cyclist traveling at 5.0 m/s uniformly accl up to 10 m/s in 2.0 sec.A small pebble is caught in the tread of one tire, that has a radius of 35cm. a. What is the angular accl of the pebble in those 2 secs? b. Thru what angle does the pebble revolve? c. How far does the pebble travel?

Ch8 HW#2 p280 43,44,49,53,55

Ch8 HW#2 p280 43,44,49,53,5543. A videodisc revolves at 1800rpm beneath a laser read-out head. If the beam is 12cm from the center of the disc, how many metersof data pass beneath it in 0.10sec?

Ch8 HW#2 p280 43,44,49,53,5543. A videodisc revolves at 1800rpm beneath a laser read-out head. If the beam is 12cm from the center of the disc, how many metersof data pass beneath it in 0.10sec? ω = 1800 rpm r = 0.12m θ = ω.t t = 0.10s = (188.5 rad/s)(0.10s) s = ? = 18.85 rad

s = r . θ = (0.12m)(18.85 rad) = 2.26m

44. A variable speed electric drill motor turning at 100 rev/sis uniformly accl at 50 rev/s2 up to 200 rev/s.How many turns does it make in the process?

44. A variable speed electric drill motor turning at 100 rev/sis uniformly accl at 50 rev/s2 up to 200 rev/s.How many turns does it make in the process?

ωi = 100 rev/s ωf2 = ωi

2 + 2αθ ωf = 200 rev/s 2002 = 1002 + 2(50) θ α = 50 rev/s2 θ = ? θ = 300 rev

49. A wheel is revolving at 20 rad/s when a brake is engaged and the wheel is brought to a stop in 15.92 rev. How much time elapsed and what was the angular deceleration?

ωi = 20 rad/s ωf = 0 rad/s θ = 15.92 rev 100 rad α = ? t = ?

53. A chimp sitting on a unicycle with a wheel diameter of 20in (.508m)is pedaling away at 100 rpm. How fast does he travel?

r = 0.254m ω = 100 rad/s 10.5 rad/s vi = ?

53. A chimp sitting on a unicycle with a wheel diameter of 20in (.508m)is pedaling away at 100 rpm. How fast does he travel?

r = 0.254m ω = 100 rad/s 10.5 rad/s vi = ? v = r . ω

= (0.254m)(10.5 rad/s) = 2.6 m/s

55. An electric circular saw reaches an operating speed of 1500rpmin the process of revolving thru 200 turns.Find the angular accl and time elapsed.

ωi = 0 rpm 0 rad/s ωf2 = ωi

2 + 2αθ ωf = 1500 rpm 157 rad/s θ = 200 revs 1257 rads α =? t = ? θ = ½(ωi+ ωf)t

Ch8.2B – More Rotation EquationsEx1) A record player is turned on and reaches a speed of 33 1/3 rpm

in 6.0 sec. a. What is the ave ang accl? b. Thru what angle is the record displaced?

Ex2) A bicycle is most efficient at a wheel rotation of 50rpm.If the wheel has a diameter of 700mm, a. How fast is the bike moving? b. If the bike’s brakes can decelerate it at 0.5 rad/s2, thru

what angle does it spin? c. What linear distance does it move?

HW#30) The motor is mounted with a 20cm diameter pulley andrevolves at 100rpm. It is used to drive an attic fan with 4 1m long blades.the ends of the blades cannot exceed 7.0 m/s. What is the size of the pulley?

Gear ratio equation:

r.ω = r.ω

Ch8 HW#3 p280+ 30,45,46,47,50

Ch8 HW#3 p280+ 30,45,46,47,5045. A wheel is released from rest on an incline and rolls for 30.0suntil it reaches a speed of 10.0 rad/s. If accl is const, thru what angle did it rotate?

46. A large motor-driven grindstone is spinning at 4 rad/s, whenthe power is turned off. It rotates thru 100 rad as it uniformly comes to rest. What was its ang accl?

47. A ring-shaped space satellite is revolving at 100 rpm when its despin rockets are fired and it decelerates at a constant 2.0 rad/s2. How much time elapses to bring it to 0 rpm’s?Thru what angle does it spin in this process?

ωi = 100 rpm 10.5 rad/s ωf = ωi + αt ωf = 0 rpm 0 rad/s α = –2 rad/s2 t = ? θ = ? θ = ½(ωi+ ωf)t

50. A bicycle with a 24in (0.6m) diameter wheel is traveling at 10 mph (4.47 m/s). What is the angular speed?How much time elapses to complete 1 revolution?

r = 0.12in = 0.30m

v = 4.47 m/s v = r.ω ω = ? t = ?

Ch8.3 Torque and Rotational Equilibrium

Torque - a force that causes an object to rotate.

(must be )

2nd Condition of Equilibrium:

Ex1) A hand exerts a force of 200 N at the end of a lever 1.0 m long. A return spring attached at the midpoint of the lever pulls back with what force, if the lever isn’t moving?

rF

0net

Ex2) A 30kg boy wishes to play on a seesaw with his 10kg dog. When the dog sits 3m, from the pivot, where should the boy sit?

Center of Mass (Center of Gravity) - point on an object where we can pretend all the mass is located

(there is no net torque from that point)

Where is the center of mass on a broom? on a human male?

female?irregular chalkboard?earth-moon system?

Ex3) The bicep muscle is attached 5cm from elbow (pivot). If the fore arm and hand have a mass of 3.85 kg, and a 2 kg ball is in the hand, what kind of force does the bicep exert to keep the 34 cm fore arm horizontal?

http://www.laboratorium.dist.unige.it/~piero/Teaching/Gait/SOUTAS-LITTLE%20Motion%20Analysis%20and%20Biomechanics_files/sout-f26.jpg

Ex4) What does the scale read?mmeterstick = 100g, m1 = 1000g, m2 = 500g

m1 m2

Ch8 HW#4 p282 63,64,67,68

Ch8 HW#4 p282 63,64,67,6863. While working with precision components, especially out in space, it is often necessary to use a torque wrench, a devise that allows the user to exert only a preset amount of torque. Having dialed in a value of 35.0 Nm, what maximum perpendicular force should be exerted on the handle of a wrench 25.0 cm from the bolt?

Τ = 35 Nm Τ = F┴.r

F┴ = ? r = .25m

64. Harry, who weighs 320 N, and 200 N Gretchen are about to play on a 5.00 m long seesaw. He sits at one end and she sits at the other. Where should the pivot be located if they are to be balanced? Neglecting the weight of the seesaw beam, what is the reaction force exerted by the force on it?

Tnet = TH – T G 0 = F.r – F.r

r 5 – r

F┴


Τ = 35 Nm Τ = F┴.r

F┴ = ? r = .25m F┴ = Τ = 35Nm =

140N r .25m


Tnet = TH – T G 0 = F.r – F.r r 5 – r

F┴


Τ = 35 Nm Τ = F┴.r

F┴ = ? r = .25m F┴ = Τ = 35Nm =

140N r .25m


Tnet = TH – T G 0 = F.r – F.r = (320N)(r) – (200N)(5 – r) r = 1.9m from Harry

r 5 – r

F┴

67. Two campers carry their gear (90.72 kg) on a light, rigid horizontal pole whose ends they support on their shoulders 1.829 m apart. If Selma experiences a compressive force of 533.8 N, where is the load hung on the pole, and what will Rocko feel?

Tnet = TS – TR 0 = F.r – F.r

= (533.8N)(r) – (F)(1.829 – r)

r =

r 1.829 - r

67. Two campers carry their gear (90.72 kg) on a light, rigid horizontal pole whose ends they support on their shoulders 1.829 m apart. If Selma experiences a compressive force of 533.8 N, where is the load hung on the pole, and what will Rocko feel? r 1.829-5 Tnet = TS – TR

0 = F.r – F.r = (533.8N)(r) – (F)(1.829 – r)

68. (next slide)

68. The bridge has a uniform weight of 20.0 kN. Calculate forces a, b.

Fg20000

a b

Fg8000 Fg8000

15m 35m 15m 20m

http://www.ecotoys.com.au/product-pics/wooden-toy-car.jpg


68. The bridge has a uniform weight of 20.0 kN. Calculate forces a, b. FgB = ?

15m 35m 15m 20m

Pick a fulcrum, I picked A

Fg8000 Fg8000

Fg20000

Tnet = Fg8000.r + FgB

.r + (-Fg20000.r) + (-Fg8000

.r)

0 = (8000N)(15m) + FgB.(70m) – (20000)(35) – (8000N)

(50m)FgB = 14,000NFgA = 36,000N – 14,000N = 22,000N

a b



Ch8.3b – Net Torque EquationsEx1) What does the scale read? (mass of meterstick= 75g)

2kg

Ex2) What do both scales read? (mass of meterstick 100g)

0.5kg 1kg

Ex3) A symmetric bridge weighs 30,000N is 200M long. A truck weighing 7500N is 50m from the west side, a 3500N car is 80m from the west side. What is the force on each landing?

Ch8 HW#5 p282 69,72, +2 more problems


Ch8 HW#5 p282 69,72, +2 Bonus Questions69. The beam is of negligible mass, what value does the scale read?

2 1

72-a. Find the tension in the rope. The beam is 2m and the rope is attached at 1.50m. Beam is negligible mass.

Tnet = FTy.r + (-Fg

.r)

400N

50°

72-a. Find the tension in the rope. The beam is 2m and the rope is attached at 1.50m. Beam is negligible mass.

Tnet = FTy.r + (-Fg

.r)

0 = (FTy .sinθ )(1.50m) -

(400N)(2m) FT

FT = 686N 50°

Fg

400N

72-b. Find the tension in the rope. The beam is 2m and the rope is attached at 1.50m. Beam is 200N.

Tnet = FTy.r + (-Fg

.r) + (-Fgmeterstick.r)

FT

50°

Fgmeterstick

Fg

400N

1. Find the reading on scale 2 and the unknown mass m2.m1 = 15N, mass of meterstick = 2N, Scale 1 reads 20N

m2m1

Fs1 = 20N FS2 = ?

2. What do the scales read? m1 = 10N, m2 = 5N, massmeterstick = 2N

Tnet = F S1.r + FS2

.r + (-Fg1.r) + (-Fg2

.r) + (-Fms.r)

5N10N

Fs1 = ? FS2 = ?

Lab8.1 Mass of a Meterstick

- Ch8 HW#5 p282 69,72, +2 Bonus Questionsdue at beginning of period. Go over it before lab.

- Ch8 HW#6 p278 12,48,66 due tomorrow

- Lab8.1 due @ end of period

Ch8 HW#6 p278 12,48,6612. An ant positioned on the very edge of a Beatles record that is 26cmin diam revolves around 100° as the disk turns. What distance traveled?

48. Record player at 78rpm. Brake brings to stop in 1 sec. How many radians does it turn?

66. Weightless ruler, find mass M and scale 1.

2

M1kg

1

2kg

Ch8 HW#6 p278 12,48,6612. An ant positioned on the very edge of a Beatles record that is 26cmin diam revolves around 100° as the disk turns. What distance traveled? θ = 100° π rad s = rθ

180° = 1.75 rad = (.13m)(1.75rad) = .23m



2

M1kg

1

2kg

Ch8 HW#6 p278 12,48,6612. An ant positioned on the very edge of a Beatles record that is 26cmin diam revolves around 100° as the disk turns. What distance traveled? θ = 100° π rad s = rθ

180° = 1.75 rad = (.13m)(1.75rad) = .23m


ωi = 78rpm = 8.2rad/s θ= ½(ωi+ωf)t ωf = 0 = 4.1rad t = 1s θ=?


scale2 = 4.17kg2

M1kg

1

2kg

Pivot at mass M:netT = T2 + T1 – Ts1

Pivot at scale 1:netT = T1 + T2 + TM – Ts2

Ch8.4 Rotational Motion

To keep an object from moving, there must be no net __________.


To keep an object from moving, there must be no net _Force_.(It can be moving at const speed w no net force.)

To keep an object from spinning, there must be no net __________.



To keep an object from spinning, there must be no net _torque_.(It can be spinning at const speed w no net torque.)

If you apple a force to an object’s center of mass, it will accelerate linearly.

F Fnet = m.a.



To keep an object from spinning, there must be no net _torque_.(It can be spinning at const speed w no net torque.)

If you apple a force to an object’s center of mass, it will accelerate linearly.

F Fnet = m.a

If you apply a force to an object outside its CoM, it will rotate (angular accl)

F

moment angular of inertia accl

.

.

Inet

What about gravity?

To keep something from spinning, there must be no net torque.

What about gravity?

To keep something from spinning, there must be no net torque.The center of gravity must be over the support.

1st Law of Rotation - A body at rest stays at rest, a body in rotation tends to keep that rotation unless acted upon by a net torque.

- Spinning objects have rotational inertia (movement of inertia)

- just like moving objects have inertia, which depends on mass, spinning objects have rotational inertia, which depends on distribution of mass (distance from the CoM)

- each type of structure has its own value for . (examples on ppts, p261, and internet)

- more mass, further from center=larger I

- masses spin around their CoM (aka center of gravity)

I

Ex1) A lab group sets up a torque experiment as shown. When they let go of the 100g meterstick, will it fall? If so, what direction and with what angular acceleration?

1N 2.5N


1N 2.5N

Tnet = T1 + Tms – T2.5 I.α = F1r + Fmsr – F2.5r ((1/12)ml2).α = (1N)(.4m) + (1N)(0m) – (2.5N)(.3m) ((1/12)(.1)(1)2.α = -.35

α = 42 rad/s2

(Not completely true, doesn’t rotate around total CoM)


1N 2.5N

Ch8 HW#7 1 – 9

Ch8 HW#7 1 – 91. Inertia depends on mass, rotational inertia depends on mass and what?2. Compare the effects of a force exerted on an object

and torque exerted on an object.3. How do clockwise and counterclockwise torques compare

when a system is balanced?

Ch8 HW#7 1 – 91. Inertia depends on mass, rotational inertia depends on mass and what?2. Compare the effects of a force exerted on an object

and torque exerted on an object.3. How do clockwise and counterclockwise torques compare

when a system is balanced?4. A rock has a mass of 1 kg. What is the mass of the measuring stick

if it is balanced by a support force at the one-quarter mark?

5. Find the rotational inertia (moment of inertia) of a solid cylinder, with a mass of 5 kg and a radius of .25 meters.

I = ½mR2

6. Find the moment of inertia of a ring rotating about its normal axis, with a mass of 150g and a diameter of 10cm.

I = mr2

7. A bicycle wheel has a radius of 0.33m. The bike is flipped over and the wheel is spun by a torque of 68 N.m. If the rim and tire together have a mass of 1.46 kg, determine the angular acceleration of the wheel. Ignore the contribution of the spokes.

8. Determine the angular acceleration that would result when a torque of 0.60 N.m is applied about the central spin axis of a hoop of mass 2.0 kg and radius 0.50 m.

7. A bicycle wheel has a radius of 0.33m. The bike is flipped over and the wheel is spun by a torque of 68 N.m. If the rim and tire together have a mass of 1.46 kg, determine the angular acceleration of the wheel. Ignore the contribution of the spokes.

Tnet = 68 N.m I.α = 68 N.m α = 423 rad/s2 mr2.α = 68 N.m

(1.46kg)(.33m)2.α = 68 N.m8. Determine the angular acceleration that would result when a torque of 0.60 N.m is applied about the central spin axis of a hoop of mass 2.0 kg and radius 0.50 m.

9. A 10kg solid steel cylinder with a 10cm radius is mounted on bearings so that it rotates freely about a horizontal axis. Around the cylinder is wound a number of turns of a fine gold thread. A 1.0kg monkey named Fred holds on to the loose end and descends on the unwinding thread as the cylinder turns. Compute Fred’s acceleration.

Ch8.4b - Rotational Motion and Energy

Ex1) A solid ball with a radius of 10cm and a mass of 1kg starts rolling from a height of 150 cm down a 30° incline. Find its velocity at the bottom.

h = 150cm 30°

221 IKER

Ex2) A 250g toy wind-up motorcycle is rolled backward, until the spring is completely wound. The bike is released, and after 3 sec is moving at 4 m/s. If the spring is attached to the rear wheel, with a diameter of 5cm, how much potential energy was stored in the spring initially?

Ch8 HW#8 p287 Bonus #1,119,120,125,126,129

Ch8 HW#8 p287 Bonus #1,119,120,125,126,129#1) A 0.5kg mass hangs on a rope wrapped around a freely rotating 2kg cylinder with a radius of .20m. What is the acceleration of the .5kg mass and the tension in the rope?

119. Determine the kinetic energy of a nontranslating disk that is spinning around its central symmetry axis at 300 rpm. The disk has a moment-of-inertia of 1.00 kg·m2 .

KEr = I.ω2 =

120. A spherical space satellite having a moment-of-inertia of 250 kg·m2 is to be spun up from rest to a speed of 12 rpm. How much energy must be imparted to the satellite?

∆KE = KErf – KEri

119. Determine the kinetic energy of a nontranslating disk that is spinning around its central symmetry axis at 300 rpm. The disk has a moment-of-inertia of 1.00 kg·m2 .

KEr = I.ω2 = (1)(~30)2 = 900J

120. A spherical space satellite having a moment-of-inertia of 250 kg·m2 is to be spun up from rest to a speed of 12 rpm. How much energy must be imparted to the satellite?

∆KE = KErf – KEri

125. A hollow cylinder, or hoop, of mass m rolls down an inclined plane from a height h. If it begins at rest, show that its final speed is given by

v=√gh PE = KEt + KEr mgh = ½mv2 + I.ω2 (mR2)

126. A solid cylinder of mass 2.0 kg rolls without slipping down a long curved track from a height of 10.0 m. Calculate the linear speed with which it leaves the track. PE = KEt + KEr mgh = ½mv2 + I.ω2

129. A long pencil is balanced straight up on its point on a horizontal surface. Without slipping, the pencil topples over. Show that the speed at which the eraser end strikes the surface is v=√3gL

PE = KEr mgh = I.ω2

L mg(½L) = (1/3ml2)(v/r)2

Ex1) Taking the Earth to be a uniform sphere of radius 6.37x106 m/sand a mass 5.98x1024 kg, compute its angular momentum about itsspin axis.

Ch8.5 - Angular Momentum

rvmL

Ex2) The moon’s velocity around the earth is 1000 m/s and is currently at a distance of 3.85x108 m away. 1 billion years from now the moon will be 10 million meters further. What will be the new velocity?

Conservation of Angular Momentum

ffii

Fi

rvmrvm

LL

Ex3) By pushing on the ground, a skater with arms extended manages to whirl around at a max speed of 1.0 rev/s. In that configuration, her momentof inertia about the spin axis is 3.5 kg.m2 . The point-masses making up her outstretched arms and leg are fairly far from the vertical axis, making her moment of inertia fairly large. What will happen to her spin rate as she draws her arms and leg in, making her moment of inertia only 1.0 kg.m2?

mr2. (v/r)Ch8 HW#9 p287+ 130,136,138, + 1 Bonus Question mvr

Alternate formula: Can we derive?

ffii

Fi

II

LL

Ch8 HW#9 p287+ 130,136,138, + 1 Bonus Question 130. A uniform disk of mass 800 kg and radius 0.5 m is rotating around its central symmetry axis at a rate of 60 rpm. Determine its angular momentum.

rrm

rvmL

136.Compute the orbital angular momentum of Jupiter

(m=1.9 x 1027kg, r = 7.8 x 1011 m, and vav = 13.1 x 103 m/s ) and then compare it to the spin angular momentum of the sun

(m = 1.99 x 1030 kg, r= 6.96 x 108 m). Assume the sun whose equator rotates once in about 26 days, is a rigid sphere of uniform density. It would seem that most of the angular momentum of the solar system is out there with the giant planets that also rotate quite rapidly.

Jupiter: L = (1.9x1027kg)(13.1x103 m/s)(7.8x1011 m)

ω of Sun: 2π/t = 2π/(2.2x106s) = 2.86x10-6 rad/s

Sun: L = I.ω = 2/5mR2.ω

138. A small mass m is tied to a string and swung in a horizontal plane. The string winds around a vertical rod as the mass revolves, like a length of jewelry chain wrapping around an outstretched finger. Given that the initial speed and length are vi and ri, compute vf when rf = ri/10.

orffii

Fi

rvmrvm

LL

ffii

Fi

II

LL

Bonus #1) A rather unruly child hits a 2.5 kg tether ball so that it swings around the center pole at its maximum radius of 1.2m with a speed of 10 m/s. As the ball winds its way around the pole it speeds up in order to conserve angular momentum, but of course you knew that. What is its speed when it is just about to hit the pole, with a radius of 0.2 m?

Lab8.2 – The Torque Experiment

- due @ end of period

- Ch8 HW#9 due at beginning of period (Go over before lab.)

- Ch5,8 Rev p171 18,46, + 2 Bonus ?’s, p278+ 51,88

10 cm

50g 50g

90cm

100g

85cm

200g

Lab8.3 – Rotation Stations

- due tomorrow

- Ch5,8 Rev p171 18,46, + 2 Bonus ?’s, p278+ 51,88

Ch5,8 Rev p171 18,46, + 2 Bonus ?’s, Ch8 p278+ 51,8818. A 1000 kg car traveling on a road that runs straight up a hill reaches the rounded crest at 10.0 m/s. If the hill at that point has a radius of curvature (in a vertical plane) of 50 m, what is the net downward force acting on the car at the instant it is horizontal at the very peak? What is the apparent weight of the car?

18. A 1000 kg car traveling on a road that runs straight up a hill reaches the rounded crest at 10.0 m/s. If the hill at that point has a radius of curvature (in a vertical plane) of 50 m, what is the net downward force acting on the car at the instant it is horizontal at the very peak? What is the apparent weight of the car?

Fnet = Fg – FN

mac = mg – FN

FN = mg – mv2/r

FN = (1000kg)(9.8m/s2) – (1000kg)(10m/s)2/(50m)

FN = 9800N – 2000N = 7800N

46. Three very small spheres of mass 2.50 kg, 5.00 kg, and 6.00 kg are located on a strait line in space away from everything else. The first one is a point between the other two, 10.0 cm to the right of the second and 20.0 cm to the left of the third. Compute the net gravitational force on it.

5kg 2.5kg 6kg .1m .2m

1 2 3

46. Three very small spheres of mass 2.50 kg, 5.00 kg, and 6.00 kg are located on a strait line in space away from everything else. The first one is a point between the other two, 10.0 cm to the right of the second and 20.0 cm to the left of the third. Compute the net gravitational force on it.

5kg 2.5kg 6kg .1m .2m

NxNxNxF

NxNx

GG

r

mGmF

r

mGmF

net888

88

22

223

32232

12

2112

108.5105.2- 103.8

105.2 103.8

)2(.

)6)(5.2(

)1(.

)5.2)(5(

1 2 3

Bonus #1)On which planet would you weigh the least?

a. b. c. d.1M1R

1M2R

2M1R

4M2R

Bonus #2) 60 kg person sits on bathroom scale on roller coaster that does a loop of radius 10m. What does the scale read at 1-4?

2

3 1

4

v = 14 m/s

v = 14 m/s

v = 11 m/s

v = 20 m/s

51. A 1.00 m diameter disk is made to accelerate from rest up to 20 rpm at a rate of 5 rad/s2. Through how many turns will it revolve in the process? How far will a point on its rim travel while all this is happening?

r = 0.5mωf = 20 rpm 2.1 rad/s ωi = 0 rpm 0 rad/sθ = ?s = ?α = 5 rad/s2

51. A 1.00 m diameter disk is made to accelerate from rest up to 20 rpm at a rate of 5 rad/s2. Through how many turns will it revolve in the process? How far will a point on its rim travel while all this is happening?

r = 0.5m ωf2 = ωi

2 + 2αθωf = 20 rpm 2.1 rad/s ωi = 0 rpm 0 rad/s ωf = .44 rad/sθ = ?s = ? s = r θα = 5 rad/s2

s = .22 m

88. An essentially weightless 10.0m long beam is supported at both ends. A 300N child stands 2.0m from the left end. A 6.0m long stack of newspapers weighing 100N per linear meter is uniformly distributed at the other end.Determine the 2 reaction forces supporting the beam.

ch5.1 - centripetal force inertia (velocity) top view centripetal force

Documents

coin f f

f n slide

f t slide

f g slide

centripetal force ex

view centripetal force

s f net

centripetal acceleration