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Circular Motion - Definition Path of Motion – Circle. Uniform Circular Motion (UCM) – constant speed v. v v v v r r: radius of circle Period T (s): time for a complete rotation If n rotations happen in time t: T=t/n. Frequency f (s-1, Hertz- Hz): number of rotations per unit of time - f=n/t. T=1/f, f=1/T, Tf=1
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Chapter 4
Motion in two and three dimensions
4.7: Uniform Circular Motion
Circular Motion - DefinitionPath of Motion – Circle.
Uniform Circular Motion (UCM) – constant speed v.
v
v
v
v
r
r: radius of circle Period T (s): time for a complete rotationIf n rotations happen in time t: T=t/n.Frequency f (s-1, Hertz-Hz): number of rotations per unit of time - f=n/t.
T=1/f, f=1/T, Tf=1
Tangential (Linear) Velocity
rfv 2
C = 2πr = vT
Trv 2
or
Exercises Set 11. A car’s tire rotates at 1200 RPM.
a. What is the frequency? f = 1200/60s = 20 Hz
b. What is the period? T= 1/f = 1/20 Hz = 0.05s
c. What is the speed (R= 0.15 m)
v = 2πRf = 2 x 3.14 x 0.15m x 20 Hz = 18.85 m/s
Ex2• Problem: The earth is 1.50 x1011 m (93 million miles) from the
sun. What is its speed in m/s (neglecting the motion of the sun through the galaxy)?
sec
Angular Measure
The position of an object can be described using polar coordinates—r and θ—rather than x and y. The figure at left gives the conversion between the two descriptions.
Uniform Circular Motion and Centripetal Acceleration
A careful look at the change in the velocity vector of an object moving in a circle at constant speed shows that the acceleration is toward the center of the circle.
4.7: Uniform Circular Motion
As the direction of the velocity of the particle changes, there is an acceleration!!!
CENTRIPETAL (center-seeking) ACCELERATION
Here v is the speed of the particle and r is the radius of the circle.
4.7.1. A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. By which one of the following means can the centripetal acceleration of the ball be increased by a factor of two?
a) Keep the radius fixed and increase the period by a factor of two.
b) Keep the radius fixed and decrease the period by a factor of two.
c) Keep the speed fixed and increase the radius by a factor of two.
d) Keep the speed fixed and decrease the radius by a factor of two.
e) Keep the radius fixed and increase the speed by a factor of two.
4.7.1. A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. By which one of the following means can the centripetal acceleration of the ball be increased by a factor of two?
a) Keep the radius fixed and increase the period by a factor of two.
b) Keep the radius fixed and decrease the period by a factor of two.
c) Keep the speed fixed and increase the radius by a factor of two.
d) Keep the speed fixed and decrease the radius by a factor of two.
e) Keep the radius fixed and increase the speed by a factor of two.
Exercises Set 4• 2) A car is traveling along a circular path of 50 m radius at 22 m/s.
222
/68.9)50()/22( sm
msm
rvacp
a. What is the car’s acceleration?
b. How much time to complete a circuit?
ssm
mxxv
rTT
rv 28.14/225014.3222
Example Set 5• 3) Satellite, radius 1.3 x 107 m, g =
2.5 m/s2.
smravrva cpcp /88.570010*3.1*5.2 7
2
a) What is the speed?
b) Period?
hssm
mxxxv
rTT
rv 98.386.327,14/88.5700
103.114.3222 7
Ex. 6
smravrva cpcp /8.28.*8.9
2
Hzm
smr
vfrfv
56.08.0*14.3*2
/8.22
2
n = 0.56 x 60 = 33.44 rev/min
4.7: Centripetal acceleration, proof of a = v2/r
cos,sin
,
vvvv
vdtdxv
dtdy
yx
xP
yP
Note
Sample problem, top gun pilots
We assume the turn is made with uniform circular motion.Then the pilot’s acceleration is centripetal and hasmagnitude a given by a =v2/R.Also, the time required to complete a full circleis the period given by T =2R/v
Because we do not know radius R, let’s solve for Rfrom the period equation for R and substitute into the acceleration eqn.
Speed v here is the (constant) magnitude of the velocity during the turning.
To find the period T of the motion, first note that the final velocity is the reverse of the initial velocity. This means the aircraft leaves on the opposite side of the circle from the initial point and must have completed half a circle in the given 24.0 s. Thus a full circle would have taken T 48.0 s.Substituting these values into our equation for a, we find