Trigonometry/ProOneSoftware-Trigonometry-Win31.iso/trig/chapter0.6r

Text File | 1995-04-09 | 7KB | 253 lines

 251
à 0.6ïLinear and Angular Velocity
äïPlease answer the following questions about arc length,
êêlinear velocity, and angular velocity.
âèA wagon wheel of radius 2 feet is turning at an angular ve-
locity of 100 radians per minute.ïFind the linear speed of the wagon
in miles per hour.ïAlso find 1/4 of the circumference of the wheel.
êêêè 100ë60 minë1 mi
# vï=ïr∙wï=è(2 ft)∙ ────── ∙ ────── ∙ ────────è≈è2.27 mi/hr
êêêè1 minë1 hrè 5,280 ft
êêèsï=ïr∙Θï=ï(2 ft)∙(π/2)ï≈ï3.14 ft
éS
êêêèImagine an angle with a stationary initial
@fig0601.bmp,15,25
êêê side, but with a terminal side that is opening up
êêê at a uniform rate of speed.ïAs this angle opens
êêê up, the point, P(x, y), on the circle moves a-
êêê round the rim of the circle at a uniform linear
êêê velocity.ïThe arc length cut off by the angle
êêê increases at a uniform rate, and the radius re-
mains constant.ïLet "w" represent the angular velocity, "v" represent
the linear velocity of the point, "s" represent the arc length, and "Θ"
represent the angle in radian measure.ïWe know from the definition of
radian measure that Θ = s/r.ïOn solving for "s", we have s = r∙Θ.ïThe
linear speed of the point on the rim is the distance the point moves
divided by the time that has ellapsed.
êêêèsêr∙ΘêèΘ
#êêïvè=è─è=è───è=èr∙──è=èr ∙ w
êêêètê têè t
Notice that "s" was replaced with "r∙Θ", and "Θ/t" was replaced with
"w".ïThis gives us the relationship between the linear velocity of the
point on the rim and the angular velocity, "w".ïThus, vï=ïr ∙ w.ïIt
is important to note that when using either this formula for linear ve-
locity or the formula for arc length, that Θ and w must be in radian
measure.ïIf the given information is in degrees or revolutions, then
you should change it to radian measure first.ïLater, we will graph
expressions such as y = sin(wt) where "w" is the angular velocity.
è In the example, the wagon wheel of radius 2 feet is turning at an
angular velocity of 100 radians per minute.
êêêêêê 100ê 200 ft
#êë vè=èr ∙ wè=è(2 ft) ∙ ─────è=è──────
êêêêêê1 minê 1 min
This linear speed in feet per minute is converted to miles per hour by
multiplying by the appropriate factors.
êêë 200 ftè60 minë1 mi
#êëvè=è────── ∙ ────── ∙ ────────è≈è2.27 mi/hr
êêê1 minè 1 hrè 5,280 ft
Also, the length of 1/4 of the circumference can be found by using the
formula for arc length.ïThe radius is known to be 2 feet, and the an-
gle that subtends 1/4 of the circle is π/2.ïThus, s = (2 ft)∙(π/2) ≈
3.14 ft.
 1ïIf Θ = 3 radians is a central angle on a circle of radius
3 feet, what is the arc length, "s", cut off or subtended by this angle?

êêëA)ï1.5 feetêè B)ï5.5 feet

êêëC)ï7.5 feetêè D)ïå

üêêUse the formula for arc length.

êêêêèsï=ïr ∙ Θ

êêêïsï=ï(2.5 ft) ∙ (3 radians)

êêêêïsï=ï7.5 ft
Ç C
 2ïIf Θ is a central angle on a circle of radius 6 inches and
Θ subtends an arc of length 14 inches, find "Θ".

êêèA)ï2.33 radiansêB)ï4π radians

êêèC)ïπ/4 radiansê D)ïå

üêêUse the formula for arc length.

êêêêèsï=ïr ∙ Θ

êêêë14 inï=ï(6 in) ∙ Θ

êêêïΘï=ï14/6ï≈ï2.33 radians
Ç A
 3ïIf Θ = π/6 is a central angle on a circle and Θ subtends an
arc of length 23 centimeters, find the radius of the circle.

êêèA)ï12.1 cmêëB)ï29.28 cm

êêèC)ï18.6 cmêëD)ïå

üêêUse the formula for arc length.

êêêêèsï=ïr ∙ Θ

êêêë23 cmï=ïr ∙ (π/4)

êêêïrï=ï(23 cm) ∙ (4/π)ï≈ï29.28 cm
Ç B
# 4ïIf Θ = 30ò is a central angle on a circle of radius 14 yards,
find the lenth of the arc subtended by this angle.

êêèA)ï7.33 ydêëB)ï12.4 yd

êêèC)ï27.1 ydêëD)ïå

üèUse the formula for arc length, but change the angle from de-
êïgree measure to radian measure.
êêêêèsï=ïr ∙ Θ
êêêêêêë π
#êêêïsï=ï(14 yd) ∙ (30°) ∙ ────
êêêêêêë180°
êêêë rï≈ï7.33 yards
Ç A
 5ïFind the length of the arc on the earths surface if it is
#subtended by an angle of 7ò and the radius of the earth is 3,960 mi.

êêëA)ï423.6 milesê B)ï86.7 miles

êêëC)ï483.8 milesê D)ïå

üïUse the formula for arc length, but change the angle from de-
gree measure to radian measure.
êêêê sè=èr ∙ Θ
êêêêêêêèπ
#êêêïsè=è(3,960 mi) ∙ (7ò) ∙ ────
#êêêêêêêï180ò
êêêë sè≈è483.8 miles
Ç C
 6ïIf a shaft is turning at 300 revolutions per minute, what
is the angular velocity in radians per minute?

êêë A)ï200/minêëB)ï600π/min

êêë C)ï150π/minêè D)ïå

üïJust convert the angular velocity from revolutions per minute
ê to radians per minute.
êêê300 revè 2π
#êë wè=è─────── ∙ ─────è=è600π radians/min
êêê 1 minè 1 rev
Ç B
 7ïA pulley of diameter 12 inches is turning at 40 revolutions
per minute, find the speed of a point on the rim in feet per second.

êêè A)ï186 ft/secêèB)ï23.6 ft/sec

êêè C)ï2.094 ft/secê D)ïå

üïUse the formula for linear velocity, but change the angular
ê velocity to radians per second.ïThe radius is .5 ft.

êêêêïvï=ïr ∙ w

êêêè 40 revè 2πë 1 min
#êïvï=ï(.5 ft) ∙ ────── ∙ ───── ∙ ──────ï=ï2.094 ft/sec
êêêë1 minè1 revè60 sec
Ç C
 8ïA cart has wheels that are 4 feet in diameter and is tra-
veling at 5 miles per hour.ïHow many revolutions per minute are the
wheels turning?
êêèA)ï7.5 rev/minêïB)ï35 rev/min

êêèC)ï26.2 rev/minê D)ïå

üïUse the formula for linear velocity, but change the units for
linear velocity from miles per hour to feet per minute.ïSolve for "w"
in radians per min and then change ç units to revolutions per
minute.êêê vï=ïr ∙ w

êêè 1 hrè 5,280 ftè5 mi
#êêè────── ∙ ──────── ∙ ────ï=ï(2 ft) ∙ w
êêè60 minë1 mië1 hr

êêê 440 ft/minï=ï(2 ft) ∙ w
êêêë220/minï=ïw
êêêï1 revè 220
#êêwè=è ───── ∙ ─────è≈è35 rev/min
êêêè 2πè 1 min
Ç B
 9ïA satellite circles the earth four times a day at a radius of
6000 miles.ïHow many miles per hour is it moving?

êêèA)ï6,283.2 mi/hrë B)ï7,800 mi/hr

êêèC)ï4,836 mi/hrê D)ïå

ü
êêêë vï=ïr ∙ w

êêêë4 revè 2πë1 day
#ê vï=ï(6000 mi) ∙ ───── ∙ ───── ∙ ──────ï≈ï6,283.2 mi/hr
êêêë1 dayè1 revè 24 hr

Ç A
 10ïA pulley with radius 1 inch is turning at 500 revolutions
per minute.ïIt is attached by belt to a pulley with radius 6 inches.
What is the angular velocity of the second pulley in rev/min?
êêè A)ï27.9 rev/minê B)ï264 rev/min

êêè C)ï83.3 rev/minê D)ïå

üèFirst find the linear velocity of a point on the rim of the
small pulley.êêè 500 revè 2π
#ë vï=ïr ∙ wï=ï(1 in) ∙ ─────── ∙ ─────è=è1000π in/min
êêêêè 1 minè 1 rev
Assuming that the belt is not slipping, a point on the rim of the
second pulley is also traveling this fast.ïThe angular velocity of
of this pulley can now be found.
êêêê vï=ïr ∙ w

êë1 revè1000π in
#êë───── ∙ ────────è=è(6 in) ∙ w
êë 2πê1 min

êêê wè=è83.3 rev/min
Ç C