Ch5.1 - Centripetal Force Top View Centripetal Force
INERTIA (velocity) Top View Centripetal Force INERTIA (velocity)
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. Ex 1) What agent exerts the centripetal
force in each?
m earth Fg Fnet = Fg Ex 1) What agent exerts the centripetal force
in each? m
earth Ex 1) What agent exerts the centripetal force in each?
m Fg Fnet = Fg m.ac = Fg earth Ex 1) What agent exerts the
centripetal force in each? Ex 1) What agent exerts the centripetal
force in each?
FT Fnet = FT FT Ex 1) what agent exerts the centripetal force in
each? Fnet = FT
m.ac = FT Ex 1) What agent exerts the centripetal force in
each?
(Top view of record player) (coin) . Ex 1) What agent exerts the
centripetal force in each?
(Top view of record player) Fnet = Ff,s (coin) . Ff,s Ex 1) What
agent exerts the centripetal force in each?
(Top view of record player) Fnet = Ff,s m.ac = Ff,s (coin) . Ff,s
Centripetal Acceleration Centripetal Acceleration
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? Centripetal Acceleration
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? CENTRIPETAL FORCE CENTRIPETAL FORCE
CENTRIPETAL FORCE 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? Fnet = FN Fn CENTRIPETAL
FORCE
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? Fnet = FN Fn Fnet = FN m.ac = FN Fn CENTRIPETAL
FORCE
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? Fnet = FN m.ac = FN Fn CENTRIPETAL FORCE 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?
Fnet = FN m.ac = FN 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? Fnet = Ff,s 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? A foreign-made space station only has a radius of 3
meters
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? At head:At feet: A foreign-made
space station only has a radius of 3 meters
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? At head:At feet: 4.The Talladega
Speedway 1500kg cars complete turns of radius 100m
at a speed of 80m/s. What is net force on car? What agent exerts
it? Fnet = Fn m.ac = Fn kg flat track car rounds a corner of radius
30m at a speed of 20m/s, without slipping. What is the minimum
coefficient of static friction that can accomplish this? Fnet =
Ff,s m.ac = .Fn Ch5.2 - Gravity All matter is attracted to all
other matter. - More matter = larger force - Distance apart is
important (Inverse square law) Newtons Law of Universal Gravitation
Ch5.2 - Gravity All matter is attracted to all other matter. - More
matter = larger force - Distance apart is important (Inverse square
law) Newtons 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 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 Earths mass to
be 5.98x10 24 kg, what force keeps the moon in her orbit? Ch5.2 -
Gravity All matter is attracted to all other matter. - More matter
= larger force - Distance apart is important (Inverse square law)
Newtons Law of Universal Gravitation - 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
Earths mass to be 5.98x10 24 kg, what force keeps the moon in her
orbit? 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? M 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. Ch5 HW#2 1.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.
I want you to calculate my weight 2 ways:
(My mass = 75 kg; Earths mass = 6x1024 kg; Earths radius = 6x106 m)
1) Fg = m.g 2) (Roughly the same with rounding error.) What is the
force of gravity on me (75 kg) on the surface of Jupiter? MJ =
2x1027 kgRJ = 7x107 m What is the gravitational attraction between
the earth and the Sun? Msun = 2x1030 kgDES = 1.5x1011 m I want you
to calculate my weight 2 ways:
(My mass = 75 kg; Earths mass = 6x1024 kg; Earths radius = 6x106 m)
1) Fg = m.g= (75kg)(9.8m/s2) = 748.5N 2) (Roughly the same with
rounding error.) What is the force of gravity on me (75 kg) on the
surface of Jupiter? MJ = 2x1027 kgRJ = 7x107 m What is the
gravitational attraction between the earth and the Sun? Msun =
2x1030 kgDES = 1.5x1011 m I want you to calculate my weight 2
ways:
(My mass = 75 kg; Earths mass = 6x1024 kg; Earths radius = 6x106 m)
1) Fg = m.g= (75kg)(9.8m/s2) = 748.5N 2) (Roughly the same with
rounding error.) What is the force of gravity on me (75 kg) on the
surface of Jupiter? MJ = 2x1027 kgRJ = 7x107 m What is the
gravitational attraction between the earth and the Sun? Msun =
2x1030 kgDES = 1.5x1011 m Ch5.3 - Gravity Applications
Acceleration 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/s a = 10 m/s2 Satellite
Orbits Orbital Speed: Satellite Orbits Orbital Speed: Ex1) A
satellite in geostationary orbit must be at a height of 3
Ex1) A satellite in geostationary orbit must be at a height of
3.6x107m above the earths surface. What speed must it travel at to
stay directly above good ol AV? Keplers 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 Problem
Hard 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 planets 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 planets diameter is 6.8x106 m, determine the mass of the
planet and compare it to Earth. Fnet = FG 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. E m
Fnet = FER FRM 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 RER = 3.46x108m (346,000km)
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 RER = 3.46x108m(346,000km) 46. Three
very small spheres of mass 2. 50 kg, 5. 00 kg, and 6
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 6 2 46. Three very small spheres of
mass 2. 50 kg, 5. 00 kg, and 6
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 6 2 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.DONT
compute the surface acceleration due to gravity for a
one-solar-mass neutron star, just trip out on this unfathomable
fact! Ch5.5 Roller Coasters Ex1) 60kg person sits on bathroom scale
on roller coaster that does a loop of radius 10m.What does the
scale read at 1-4? v = 11 m/s 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 4 v = 14 m/s v = 14 m/s v = 20 m/s Ch5.4 Roller
Coasters Ex1) 60kg person sits on bathroom scale on roller coaster
that does a loop of radius 10m.What does the scale read at 1-4? v =
11 m/s 2 4 v = 14 m/s v = 14 m/s v = 20 m/s Fnet = FN 1. Fnet = FN
3. Fnet = FN + Fg 2. Fnet = FN Fg 4.
v = 14 m/s Fnet = FN 1. Fnet = FN 3. v = 14 m/s v = 11 m/s Fnet =
FN + Fg 2. Fnet = FN Fg 4. v = 20 m/s 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 = 14 m/s v = 20 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? m.ac = FN mg FN FN = mv2/r + mg = 71N + 44N = 115N Fg
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 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 Side: Fnet = FN Bottom: Fnet = FN Fg
2. While whirling around in a vertical circle with a radius of
1
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= N 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 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? Fnet = FN + Fg
v = 40 m/s Ch5 HW#5 p ,27,30 25. 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. Th =
____________N Tv = _____________N
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 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 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. 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.
Ch5 HW#6 RC FRQ Ch5 HW#6 Roller Coaster FRQ
A 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 FRQ
A 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 = mv2 ii. Calculate the value vmax of this
maximum speed v = 42 m/s 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. a) i. indicate on the
figure the point P at which PE KE
the maximum speed of the car is attained mgh = mv2 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 = mv2 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. b) Calculate the speed vB of the car at point
B.
mgh = mv2 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 More PE needed PE = KE + Wf
b) Calculate the speed vB of the car at point B. mgh = mv2 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 Ch 8.1
Rotational Motion Angular Displacement- distance around in circles
r Ch 8.1 Rotational Motion Angular Displacement- distance around in
circles 1 radian when s=r or r or 6.28 rad around circle Ex1) A
pendulum of length .40m sweeps out an angle of 30 to cover half its
period. What is displacement of the pendulum bob? Angular speed -
how fast you go in circles
Linear speed (v) relates to angular speed () radians omega sec
Angular speed - how fast you go in circles
Linear speed (v) relates to angular speed () rad omega sec 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?
Tangential acceleration:
Angular 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: Ch8 HW#1 p278
7,13,14,19,25,27,35,42 Ch8 HW#1 p278 7,13,14,19,25,27,35,42 7.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 = 65mC = 2r r = .15m= 2(.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. Ch8 HW#1
p278 7,13,14,19,25,27,35,42 7.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. 5.3 rev2 rad= 33.3 rad 1 rev 13.If a ball 30 cm
in diam rolls 65 m without slipping, how many revs did it make in
the process? s = 65mC = 2r r = .15m= 2(.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. 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 s, what is the angular speed of the wheel? =
? = 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. i = 0 f = 500 rev/s = 100 rev 2 rad =
rad/s t = 30 sec s rev = ? = = 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? f = 0 i = 200 rpm
= 21 rad/s = - 5 rad/s2 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 s, what is the
angular speed of the wheel? = ? = 100 rev 2 rad = x104 rad/s 0.02 s
rev 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. i = 0 f = 500 rev/s = 100 rev 2 rad =
rad/s t = 30 sec s rev = ? = = 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? f = 0 i = 200 rpm
= 21 rad/s = - 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 citys latitude (the angle
measured above the equator). vNYC = 353 m/sradius to NYC t = 86,400
s =7.27x10-5 rad/s r = 6.37x106 m = 40 1. vf = vi + at 1. Ch8.2
Angular Equations
Linear EquationAngular Equivalent 1. vf = vi + at1. 1. vf = vi + at
1. f = i + t 2. s = (vi+ vf)t 2
Ch8.2 Angular Equations Linear EquationAngular Equivalent 1. vf =
vi + at1. f = i + t 2. s = (vi+ vf)t2 2. s = (vi+ vf)t 2. = ( i+
f)t 3. s = vit + at2 3.
Ch8.2 Angular Equations Linear EquationAngular Equivalent 1. vf =
vi + at1. f = i + t 2. s = (vi+ vf)t2. = ( i+ f)t 3. s = vit + at
2. s = (vi+ vf)t 2. = (i+ f)t
Ch8.2 Angular Equations Linear EquationAngular Equivalent 1. vf =
vi + at1. f = i + t 2. s = (vi+ vf)t2. = (i+ f)t 3. s = vit + at =
it + t2 4. vf2 = vi2 + 2as4. 2. s = (vi+ vf)t 2. = (i+ f)t
Ch8.2 Angular Equations Linear EquationAngular Equivalent 1. vf =
vi + at1. f = i + t 2. s = (vi+ vf)t2. = (i+ f)t 3. s = vit + at =
it + t2 4. vf2 = vi2 + 2as4. f2 = i2 + 2 Ex1) Mounted on a bus
engine is a 2.0m diameter flywheel, a massive disk 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
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 43,44,49,53,55 43. A videodisc revolves at 1800rpm beneath a
laser read-out head. If the beam is 12cm from the center of the
disc, how many meters of data pass beneath it in 0.10sec? Ch8 HW#2
p280 43,44,49,53,55 43. A videodisc revolves at 1800rpm beneath a
laser read-out head. If the beam is 12cm from the center of the
disc, how many meters of data pass beneath it in 0.10sec? = 1800
rpm r = 0.12m = .t t = 0.10s = (188.5 rad/s)(0.10s) s = ? = rad s =
r . = (0.12m)(18.85 rad) = 2.26m 44. A variable speed electric
drill motor turning at 100 rev/s
is 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/s
is uniformly accl at 50 rev/s2 up to 200 rev/s. How many turns does
it make in the process? i = 100 rev/s f2 = i2 + 2 f = 200 rev/s =
(50) = 50 rev/s2 = ? = 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 rev. How much time elapsed
and what was the angular deceleration? i = 20 rad/s f = 0 rad/s =
rev 100 rad = ? t = ? 53. A chimp sitting on a unicycle with a
wheel diameter of 20in (
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 = 100 rad/s 10.5 rad/s vi = ? 53. A chimp sitting on a
unicycle with a wheel diameter of 20in (
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 = 100 rad/s 10.5 rad/s vi = ? v = r . = (0.254m)(10.5 rad/s)
= 2.6 m/s 55. An electric circular saw reaches an operating speed
of 1500rpm
in the process of revolving thru 200 turns. Find the angular accl
and time elapsed. i = 0 rpm 0 rad/s f2 = i2 + 2 f = 1500 rpm 157
rad/s = 200 revs 1257 rads =? t = ? = (i+ f)t Ch8.2B More Rotation
Equations
Ex1) 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) Abicycle 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 bikes brakes can decelerate it at 0.5 rad/s2,
thru what angle does it spin? c. What linear distance does it move?
Ch8 HW#3 p ,46,47,50 Ch8 HW#3 p ,46,47,50 45. A wheel is released
from rest on an incline and rolls for 30.0s until it reaches a
speed of 10.0 rad/s. If accl is const, thru what angle did it
rotate? 46. A large motor-drivengrindstone is spinning at 4 rad/s,
when
the 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 rpms? Thru what
angle does it spin in this process? i = 100 rpm 10.5 rad/s f = i +
t f = 0 rpm 0 rad/s = 2 rad/s2 t = ? = ? = (i+ f)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. = ? 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 isnt
moving? 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)
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? Ex4) What does the scale read?
mmeterstick = 100g, m1 = 1000g, m2 = 500g m2 m1 Ch8 HW#4 p282
63,64,67,68 Ch8 HW#4 p282 63,64,67,68 63.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? = 35 Nm = 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
F r r Ch8 HW#4 p282 63,64,67,68 63.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? = 35 Nm = F.r F = ? r = .25m F = =35Nm= 140N
r m 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
F r r Ch8 HW#4 p282 63,64,67,68 63.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? = 35 Nm = F.r F = ? r = .25m F = =35Nm= 140N
r m 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
= (320N)(r) (200N)(5 r) r = 1.9m from Harry F r r 67. Two campers
carry their gear (90
67.Two campers carry their gear (90.72 kg) on a light, rigid
horizontal pole whose ends they support on their shoulders m
apart.If Selma experiences a compressive force of 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 r 68. The bridge has a uniform
weight of 20,000N. Both cars weigh 8000N.
Calculate the forces on the bridge supports (a) and (b). 15m m m20m
a b Fg8000 Fg8000 FA FB 1 2 Fnet = FA + FB Fg1 Fg2 FgBr
15m m m20m 1 2 a b Fg Fg8000 Fnet = FA + FB Fg1 Fg2 FgBr 0= FA + FB
8000N 8000N 20,000N Unfortunately there are 2 unknowns We need a
different way to solve this Fg20000 FA FB 15m m m20m 1 2 a b Fg
Fg8000 Fg20000Pick a support to be a fulcrum, I picked A The net
torque around A is: Tnet = Fg1.r FgB.r (-FgBr.r) + (-Fg2.r) 0=
(8000N)(15m) + FgB.(70m) (20000)(35) (8000N)(50m) FgB = 14,000N Now
you can set up a 2nd net Torque eqn around support B or solve the
original Fnet eqn for A: FgA = 36,000N 14,000N = 22,000N Ch8.3b Net
Torque Equations Ex1) What does the scale read?
(mass of meterstick= 75g) 2kg Ex2) What do both scales read? (mass
of meterstick 100g) Ex3) A symmetric bridge weighs 30,000N is 200M
long
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 p Bonus
problems Ch8 HW#5 p282 69 +2 Bonus Questions
69.The beam is of negligible mass, what value does the scale read?
1 2 1.Find the reading on scale 2 and the unknown mass m2.
m1 = 15N, mass of meterstick = 2N, Scale 1 reads 20N Fs1 = 20N FS2
= ? m1 m2 2. What do the scales read? m1 = 10N, m2 = 5N,
massmeterstick = 2N
Tnet = F S1.r + FS2.r + (-Fg1.r) + (-Fg2.r) + (-Fms.r) Fs1 = ? FS2
= ? 10N 5N Lab8.1 Mass of a Meterstick
- Ch8 HW#5 p Bonus Questions due at beginning of period. Go over it
before lab. - Ch8 HW#6 p278 12,48,66 due tomorrow - Lab8.1 end of
period 48. Record player at 78rpm. Brake brings to stop in 1
sec.
Ch8 HW#6 p278 12,48,66 12. An ant positioned on the very edge of a
Beatles record that is 26cm in 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. 1 2 1kg M 2kg 48.
Record player at 78rpm. Brake brings to stop in 1 sec.
Ch8 HW#6 p278 12,48,66 12. An ant positioned on the very edge of a
Beatles record that is 26cm in diam revolves around 100 as the disk
turns.What distance traveled? = 100 rads = r 180 = 1.75 rad =
(.13m)(1.75rad) = .23m 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. 1 2 1kg M 2kg Ch8 HW#6 p278 12,48,66 12.
An ant positioned on the very edge of a Beatles record that is 26cm
in diam revolves around 100 as the disk turns.What distance
traveled? = 100 rads = r 180 = 1.75 rad = (.13m)(1.75rad) = .23m
48. Record player at 78rpm. Brake brings to stop in 1 sec. How many
radians does it turn? i = 78rpm = 8.2rad/s= (i+f)t f = = 4.1rad t =
1s=? Pivot at mass M: netT = T2 + T1 Ts1 Pivot at scale 1:
66.Weightless ruler, find mass M and scale 1. scale2 = 4.17kg 1 2
1kg M 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 __________. Ch8.4 Rotational
Motion 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 moving, there must be no net _Force_.
Ch8.4 Rotational Motion 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
_torque_. (It can be spinning at const speed w no net torque.) If
you apple a force to an objects center of mass, it will accelerate
linearly. FFnet = m.a (Top View) . To keep an object from moving,
there must be no net _Force_.
Ch8.4 Rotational Motion 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
_torque_. (It can be spinning at const speed w no net torque.) If
you apple a force to an objects center of mass, it will accelerate
linearly. FFnet = m.a (Top views) If you apply a force to an object
outside its CoM, it will rotate (angular accl) F momentangular of
inertia accl . . 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) Ex1) A lab group sets up a torque experiment as
shown
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? 2.5N What is the linear accl of
the mass?
Ex2) A mass is attached to a string, that is attached to a bicycle
wheel. What is the linear accl of the mass? (Bike wheels are
essentially hoops:I = M.R2 ) 0.5N Ch8 HW#7 1 9 Ch8 HW#7 1 9
1.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 9 1.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 = mR2 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. = 68 N.m = 423 rad/s2 mr2. = 68 N.m (1.46kg)(.33m)2. = 68 N.m
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. 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 Freds 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 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 kgm2 . KEr = I.2 = 120. A spherical
space satellite having a moment-of-inertia of 250 kgm2is 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
kgm2 . KEr = I.2 = (1)(~30)2 = 900J 120. A spherical space
satellite having a moment-of-inertia of 250 kgm2is 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 = mv2 + I.2 (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 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.2 L mg(L) = (1/3ml2)(v/r)2 Ch8.5 - Angular
Momentum Ex1) Taking the Earth to be a uniform sphere of radius
6.37x106 m/s and a mass 5.98x1024 kg, compute its angular momentum
about its spin axis. Conservation of Angular Momentum
Ex2) The moons 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? Alternate formula:Can we derive?
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 moment of 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 p
,136,138, + 1 Bonus Question mvr 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. (m=1.9 x 1027kg, r = 7.8 x 1011 m, and vav = 13.1
x 103 m/s )
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) of Sun:2/t = 2/(2.2x106s)=
2.86x10-6 rad/s Sun: L = I. = 2/5mR2. 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. or Bonus #1) A rather unruly child hits a 2
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, p ,88 85cm 200g 90cm 100g 10 cm 50g
50g Lab8.3 Rotation Stations
- due tomorrow - Ch5,8 Rev p171 18,46, + 2 Bonus ?s, p ,88 Ch5,8
Rev p171 18,46, + 2 Bonus ?s, Ch8 p ,88 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? FN = (1000kg)(9.8m/s2)
(1000kg)(10m/s)2/(50m)
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
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 kg kg .1m.2m 3 1 2 46. Three very
small spheres of mass 2. 50 kg, 5. 00 kg, and 6
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 kg kg .1m.2m 3 1 2 On which planet
would you weigh the least? a. b. c. d.
Bonus #1) On which planet would you weigh the least? a.b.c.d. 4M 2R
1M 1R 1M 2R 2M 1R 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? v = 11 m/s 2 4 v = 14 m/s v = 14 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 = 0.5m f = 20 rpm 2.1 rad/s i = 0 rpm 0 rad/s = ? s =
? = 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 f2 = i2 + 2 f = 20 rpm 2.1 rad/s i = 0 rpm 0
rad/s f = .44 rad/s = ? s = ? s = r = 5 rad/s2 s = .22 m 88. An
essentially weightless 10
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. HW#30) The motor is mounted with a 20cm
diameter pulley and
revolves 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. = r. ((1/12)ml2). =
(1N)(.4m) + (1N)(0m) (2.5N)(.3m)
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? 2.5N 1N Tnet = T1 + Tms T2.5 I.
= F1r + Fmsr F2.5r ((1/12)ml2). = (1N)(.4m) + (1N)(0m) (2.5N)(.3m)
((1/12)(.1)(1)2. = -.35 = 42 rad/s2 (Not completely true, doesnt
rotate around total CoM)
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