i l iCircular Motion
Speed/Velocity in a Circlep yConsider an object moving in a circle around a specific origin. The DISTANCE thearound a specific origin. The DISTANCE the object covers in ONE REVOLUTION is called the CIRCUMFERENCE. The TIME that it takes to cover this distance is called the PERIOD.
Tr
Tdscircle
π2==
Speed is the MAGNITUDE of the velocity. And while the speed may be constant, the VELOCITY is NOT. Since velocity is a vector with BOTH magnitude AND direction, we see that the direction o the velocity is ALWAYS changing.
We call this velocity, TANGENTIAL velocity as its direction is draw TANGENT to the circledirection is draw TANGENT to the circle.
Centripetal AccelerationpSuppose we had a circle with angle, θ, between 2 radii You may recall:
=rsθ vs ∆
==θ
radii. You may recall:
metersin length arc=s
vtsvr
∆=
vv
rvt ∆=
∆v∆v
vr
v
θvov
atv
rv
c=∆∆
=2
voonacceleratilcentripetaa
trc =
∆
Centripetal means “center seeking” so that means that the acceleration points towards the CENTER of the circleacceleration points towards the CENTER of the circle
Drawing the Directions correctlyg y
So for an object traveling in aSo for an object traveling in a counter-clockwise path. The velocity would be drawn TANGENT to the circle and theTANGENT to the circle and the acceleration would be drawn TOWARDS the CENTER.
To find the MAGNITUDES of each we have:
22rva
Trv cc
22==
π
Drawing the Directions correctlyg ycircumference of the circle is C = 2π r. If the period for one rotation is T, the angular rate of rotation also known asangular rate of rotation, also known as angular velocity, ω is:ω = dθ / dt.
Circular Motion and N.S.LvamaF
2
Recall that according to Newton’s Second Law, th l ti i mv
ramaF cNET ==
2the acceleration is directly proportional to the Force If this is true:
rmvFF cNET ==
the Force. If this is true: ForcelCentripetaFc =
Since the acceleration and the force are directlySince the acceleration and the force are directly related, the force must ALSO point towards the center. This is called CENTRIPETAL FORCE.
NOTE: The centripetal force is a NET FORCE. It could be represented by one or more forces. So NEVER draw it in an F.B.D.
Examples The blade of a windshield wiper movesp The blade of a windshield wiper moves through an angle of 90 degrees in 0.28seconds. The tip of the blade moves on the arc of a circle that has a radius of 0.76m. What is the magnitude of the centripetal acceleration of the tip of the blade?
2T
rvcπ2
= smvc /26.4)4*28(.)76(.2==
π
222
/9223)26.4( smva === /92.2376.0
smr
ac
ExamplespWhat is the minimum coefficient of static friction necessary to allow a penny to rotate along a 33 1/3 rpm record (diameter= 0.300 m), whenth i l d t th t d f th
Top view
FF =
the penny is placed at the outer edge of the record?
min1rev
mvF
FF
N
cf
2
=
=
µFN sec801sec1sec555.0
sec60min1*
min3.33 =
T
revrev
rmvmg
r2
=µ
NFf
/5240)15.0(22
sec80.1555.0
===
==
smrv
Trevrevππ
rgv
r2
=µ
mg
Side view 187.0)524.0(
/524.080.1
22
===
===
v
smT
vc
µrg 187.0)8.9)(15.0(rg
µ
ExamplespVenus rotates slowly about its
axis, the period being 243 d Th f V i
Fgdays. The mass of Venus is 4.87 x 1024 kg. Determine the radius for a synchronous satellite in orbit around
g
satellite in orbit around Venus. (assume circular orbit) 2mvMmGFF
==
==
2
2
2πT
rvvGMrr
GFF
c
cg
=→=→= 32
2
2
23
2
22
444
πππ GMTrGMTrT
rr
GMTr c
==−
32
272411
4)101.2)(1087.4)(1067.6(
44
π
ππ
xxxr
Tr
1.54x109 m4π
ExamplespThe maximum tension that a 0.50
m string can tolerate is 14 N. A mgTg0.25-kg ball attached to this string is being whirled in a vertical circle. What is the maximum speed the ball can have (a) the top of the circle, (b)at the bottom of the circle?
rmvmaFF ccNET
2
2
===
mgTr
mvmgTrr
mvmgT
))89)(250(14(50)(
)( 22
++
=+→=+
smvm
mgTrv
/74.525.0
))8.9)(25.0(14(5.0)(
=
+=
+=
ExamplespAt the bottom? r
mvmaFF ccNET
2
2
===
mgTr
mvmgTrr
mvmgT
))8.9)(25.0(14(5.0)(
)( 22
−−
=−→=−
smvm
gv
/81.425.0
)))((()(
=
==
TT
mg
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