Unformatted text preview: -n:o Imwpmms; ._ .. . ._,__._...-..ﬁ..-.__ K 4. :mnal' r._..-_c...v ‘ 44 Chapter 3 Motion in Two and Three Dimensions Find the orbital period (the time to complete one orbit) of a space shut-
tle. in circular orbit at an altitude of .250 km, where the acceleration of
gravity ’13-‘51th of its surface value. [dreamer This: is a problem about uniform c-imular motion. altitude, not the orbital radius, and We want the period, not the speed.
' so outplan is to write'the‘ speed in terms ofthe'pcriod and use the re- sult'in Equation 3.:1_6. The. crb‘tal altitude-is the distance from Earth’s
surface, so we’ll need-to add Earth’s radius to get the orbital radius r.
swim-rs The speed i; is-‘tha orbital circumference, 27m divided by
the period "I; Using this in Equation 3.16 givas ‘ I (Iltrrri‘T)2 b 4.,117. r T2 v2 '
7' _ Uniform Circular Motion: Engineering a'Road
a. engineer is designingaﬂat, horizontal road for an so kin/h speed
limit (that’s 22.2 mls): If the mind urn acceleration of a vehicle on- this road is 1.5 mfsﬁ what‘s the minimum safe radius for curves in thermal? "ETERPRET Even though acurve is only a portion of a circle, we can " st'dl. interpret this: problem as involving'nniforin circular motion. -: until—93 Equation 3 .16. c: ﬁr; deft-amines the acceleration - given
' the speed and radius: Here website the acceleration and speed, so our
- ‘plén: is, to solve for the farms Uniform Circular Motion: Calculating a Spacé'Shuttlé.Orbit. Appendix 5 lists Earth’s radiusasRE = 6.37 Mm giving an orhital ra-
dius r e. R3 _+ 259 km : 6.62 Mm. Solving our acoelc'rat'ion expression
for the period then giVes T i V 47am: 5355 3': 89 min, where We
used a}: mag; ' , “ . ' assess“ Make sense?-You’ve_ probably heard thatzastronauts orbit Earth in about an hour and a half,_expcriﬂncing multiple sunrises and sunsets
in a 24-hour day. Our answer of 89 ruin is certainly consistent with
that. There’s no choice here; for a given orbital radius,.Earth’s size and.
mass dete’miine the period. Because astronauts’ orbits are hunted to a
few hundred Idiomctcrs, a distance small compared 'with' R5, variaw
tionsin g and T are minimal. Any such “low Earth. orbit” has a period
of approximately 90. ruin At higher altitudes. the decline in the
strength-of gravity‘becomes more noticeable and periodslengthen; the
Moon-Tor example, orbits in?! days._ We‘ll discuss orbits more in
Chapter 8. - _ - - . ' l warrants Using the given numbers, we - have r=vi2ln :
(mam/simjmsi = 329' m. - mess-s Make sense? Aspee'd of 80' km/h is pretty fast,_ so we need a. wide curve to keep the required acceleration below its design value, if _' the curve is sharper,.'yehiclcs may slide off the read. We‘ll see'mOre
clearly: in subsequent. chapters how vehicles manage tonegotiatc higha - speedcnrvca . Nonuniform Circular Motion The car kw, so
its tangentia acceleration E, is oppositeits velocity. nitride is still vzlr, 516133.22 Acceleration of a car that
slows as it rounds a curve. What if an. object moves in a circular path but its speed changes? Then it has components
of acceleration both perpendicular and parallel to its velocity. The former, the radial ac-
celeration an is what changes the direction to keep the object in circular motion. Its mag-
With 1) now the instantaneous speed. The parallel component of
acceleration; also called tangential acceleration 0, because it’s tangent to the circle,
changes the speed but not the direction. Its magnitude is therefore die rate of change of The radial acceleration a, changes only the Speed, or dv/dt. Figure 3.22 shows these two acceleration components for a car rounding
direction of motion. a curve Finally, what if the radius of a curved path changes? At any point on a curve we can de— ﬁne a radius of curvature. Then the radial acceleration is still v2/r, and it can vary if either
v or r changes along the curve. The tangential acceleration is still tangent to The curve. and
it still describes the rate of change of speed. So it’s straightforward to generalize the ideas
of uniform circular motion to cases where the motion is nonuniform either because the speed changes, or because the radius changes, or both. GOT IT? 3.4 The ﬁgure shows velocity vectors for four points on a noncircular
path. Rank order the centripetal accelerations at these points given v. = v4 and v2 = V}. V3 _ ...
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- Spring '07