Abeka Physics Section 10.1 Flashcards
Define periodic motion
> periodic motion - any motion that occurs repeatedly after the same lapse of time
Describe uniform circular motion
> an object that follows a circular path at constant speed is in uniform circular motion
examples: valve of bicycle tire rotating at constant speed and blade of flying helicopter
although speed of object is constant, velocity is changing because direction of motion is changing
Describe velocity and acceleration at 2 points in uniform circular motion
> any change in velocity is acceleration
imagine a circle with 2 points along its circular boundary: both point branch out to create perpendicular vectors
bottom vector line goes toward the circle’s center and signifies acceleration
the top vector line acts perpendicular to acceleration and signifies velocity
because the 2 vectors are perpendicular, this indicates that acceleration has no effect on speed of object but causes object to change direction
as object approaches point 2, velocity vector turns inward toward center as acceleration changes direction so that vectors of velocity and acceleration remain perpendicular
as object continues beyond point 2, effect is another change in direction without change in speed
object maintains same speed as it goes round and round circular path, and vector of acceleration in uniform circular motion is always directed toward center of circular path
Describe and define centripetal acceleration
> centripetal means “center-seeking”
acceleration vectors point toward center
centripetal acceleration - an orbiting (or rotating) object’s acceleration toward center of circle due to force causing constant change in direction
Describe how to find rate of centripetal acceleration
> to find change in velocity (delta v), perform head-to-tail addition of vectors:
delta v = v2 - v1 = v2 + (-v1)
where v1 is same length as v1, but in opposite direction
first, draw v2, and then draw -v1 at head of v2
change in velocity is resultant drawn from tail of v2 to head of -v1 (making vector triangle)
to find arc s (curved path followed) use formula: s = vt
delta v/t = v^2/r
change in velocity per unit time is same as acceleration (because this acceleration is always directed toward center of circle it is called centripetal acceleration which is symbolized as a(sub c)
a(sub c) = v^2/r
Solve this:
You are riding in a gondola on the edge of a 30ft. tall Ferris wheel at a constant speed of 2.0ft/s. What is your centripetal acceleration?
Substitute given values into equation 4; v is 2.0 ft/s and r is 15ft., half of the height.
a(sub c) = v^2/r
so,
a(sub c) = (2.0ft/s)^2
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15 ft.
a(sub c) = 0.27 ft/s^2 directed toward center
Describe and define centripetal force
> acceleration is always caused by net force acting in direction of acceleration
when object has uniform circular motion, centripetal acceleration is caused by force pulling or pushing object toward center of circle
centripetal force - any force producing centripetal acceleration
relationship between acceleration and force producing it is given by Newton’s second law, F = ma. For circular motion, this same relationship relates centripetal force and centripetal acceleration:
F(sub c) = ma(sub c)
since a(sub c) = v^2/r, you can substitute this making
F(sub c) = mv^2/r
this equation may be used to calculate centripetal force whether caused by gravity, friction, or electromagnetic attraction
like centripetal acceleration, centripetal force is vector directed toward center of motion
Solve this:
The moon has a mass of 7.349 x 10^22 kg. Its mean distance from the earth is 3.91 x 10^8 km from the center of the earth. It completes one revolution in 27 days, 7 hr., and 43 min. What is the gravitational force necessary to keep it in a circular orbit around the earth?
> to use equation F(sub c) = mv^2/r, replace m with 7.349 x 10^22 kg and r with the distance of the moon from earth’s center 3.91 x 10^8 km. Also find the moon’s speed; to do so, determine the distance the moon travels in one orbit about the earth and find the time required to complete that orbit
To find the distance of the orbit, find the circumference of the circle traced by the moon:
C=2|~|r=2|~|(3.91 x 10^8 m) = 2.4567..x10^8m Since forces are measure in newtons, find the speed in meters per second. Thus, the time of one orbit in seconds is
t(sub orbit) = 27days + 7hr. + 43min. = 2.360580… x 10^6s
Thus, the speed of the moon is
v= 2.4567… x 10^8m/2.360580… x 10^6s = 1040.72…m/s
substitute into equation 6 to obtain
F(sub c) = (7.349 x 10^22 kg)(1040.72…m)^2
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3.91 x 10^8 m
F(sub c) = 2.04 x 10^20 N
Describe centrifugal force
> since forces occur only in action-reaction pairs, any centripetal force must be balanced by an equal force in opposite direction
this reactive force directed away from center of circular motion is known as centrifugal force
while centripetal (“center-seeking”) force pushes or pulls toward center, centrifugal (“center fleeing”) force pushes or pulls away from center
Distinguish centripetal force from centrifugal force
> centripetal force may always be regarded as action
centrifugal force may always be regarded as reaction
as earth follows its orbit about the sun, earth’s centrifugal force on sun may be seen as reaction to sun’s centripetal force on earth
since these forces are action-reaction pairs, centrifugal force does not exist without centripetal force
