Upload
ishan
View
218
Download
0
Embed Size (px)
Citation preview
7/25/2019 physics project made by ishan gupta.ppt
1/22
Circular Motion
7/25/2019 physics project made by ishan gupta.ppt
2/22
1 Kinematics of Uniform Circular Motion
Uniform circular motion: motion in a circleof
constant radiusat constant speedInstantaneousvelocity is always tangentto
circle.
7/25/2019 physics project made by ishan gupta.ppt
3/22
1 Kinematics of Uniform Circular Motion
Looking at the change in velocity in the limit that the
time interval ecomes infinitesimally small! we see that
"#$1%
&his acceleration is called the centripetal! or radial!
acceleration! and it points towards the center of thecircle.
7/25/2019 physics project made by ishan gupta.ppt
4/22
Example 1
' 1#( kg all at the end of a string is revolving uniformly in a
hori)ontal circle of radius (.*(( m. &he all makes +.(( revolutionsin a second. ,hat is its centripetal acceleration-
aR =v2
r
v =2r
T=
2(3.14)(0.600 m)
(0.500 s)= 7.54 m/s
aR = v
2
r= (7.54 m/s)
2
(0.600 m)= 94.7 m/s2
7/25/2019 physics project made by ishan gupta.ppt
5/22
Example 2
&he moons nearly circular orit aout the earth has a radius of aout
/0!((( km and a period & of +2./ days. 3etermine the acceleration
of the moon toward the earth.
r = 3.84x108 m, T = (27.3 d)(24.0 h/d)(3600 s/h) = 2.36x106 s
v =2r
T a =v2
r =42r2
T2r =42r
T2 =42(3.84x108 m)
(2.36x106 s)2 = 0.00272 m/s2
7/25/2019 physics project made by ishan gupta.ppt
6/22
+ 3ynamics of Uniform Circular Motion
4or an o5ect to e in uniform circular motion! there
must e a net forceacting on it.
,e already know theacceleration! so can
immediately write the force:
"#$1%
,e can see that the force
must e inwardy
thinking aout a all on a
string:
7/25/2019 physics project made by ishan gupta.ppt
7/22
+ 3ynamics of Uniform Circular Motion
&here is no centrifugalforce pointing outward6
what happens is that the natural tendency of theo5ect to move in a straight line must e
overcome.
If the centripetal force vanishes! the o5ect flies
off tangentto the circle.
7/25/2019 physics project made by ishan gupta.ppt
8/22
Example 3
7stimate the force a person must e8ert on a string attached to a (.1#(
kg all to make the all revolve in a hori)ontal circle of radius (.*((m. &he all makes +.(( revolutions per second "&9(.#(( s%.
(F)R = maR = mv2
r
= m(2r/T)2
r
= (0.150 kg)42(0.600 m/0.500 s)2
(0.600 m)
=14 N
7/25/2019 physics project made by ishan gupta.ppt
9/22
/ ighway Curves! ;anked and Unanked
,hen a car goes around a curve! there must e
a net force towards the center of the circle ofwhich the curve is an arc. If the road is flat! that
force is supplied y friction.
7/25/2019 physics project made by ishan gupta.ppt
10/22
/ ighway Curves! ;anked and Unanked
If the frictional force is
insufficient! the car will
tend to move more
nearly in a straight line!
as the skid marks show.
7/25/2019 physics project made by ishan gupta.ppt
11/22
/ ighway Curves! ;anked and Unanked
's long as the tires do not slip! the friction is
static. If the tires do start to slip! the friction iskinetic! which is ad in two ways:
1. &he kinetic frictional force is smallerthan the
static.+. &he static frictional force can point towards
the center of the circle! ut the kinetic frictional
force opposesthe direction of motion! making
it very difficult to regain control of the car and
continue around the curve.
7/25/2019 physics project made by ishan gupta.ppt
12/22
/ ighway Curves! ;anked and Unanked
;ankingthe curve can help keep
cars from skidding. In fact! forevery anked curve! there is one
speed where the entire centripetal
force is supplied y the
hori)ontal component ofthe normalforce! and no
friction is re
7/25/2019 physics project made by ishan gupta.ppt
13/22
Example 6
A 1000. kg car rounds a curve o a flat road of radius
50. m at a speed of 50. km/h (14 m/s). Will the car
follow the curve or it will it skid! Assume (a) the
pavement is dr" and the coefficient of static friction is
s#0.$0% (&) the pavement is ic" and s#0.'5.
FN = mg = (1000. kg)(9.8 m/s2) = 9800 N
(F)R = maR = m v2
r= (1000 kg) (14 m/s)
2
(50 m)= 3900 N
(a) (Ffr )max =sFN = (0.60)(9800 N) = 5900 N
Since this is greater than 3900 N, the car will
follow the curve.
(b) (Ffr )max =sFN = (0.25)(9800 N) = 2500 N
Since this is less than 3900 N, the car will not
follow the curve, so it will skid.
7/25/2019 physics project made by ishan gupta.ppt
14/22
Example 7
(a) or a car traveling with speed v around a curve
of radius r determine a formula for the angle atwhich a road should &e &anked so that no friction is
reuired. (&) What is the angle for an e*presswa"
off+ramp curve of 50 m at a design speed of 50
km/h!
FR = maR FNsin=mv2
rFy = may = 0 FNcos- m g = 0 FN =
mg
cos
FNsin=mv2
r
mg
cossin=
mv2
rmgtan=
mv2
r
tan=
v2
rg
For r = 50 m and v =14 m/s, we have
tan=(14 m/s)2
(50 m)(9.8 m/s2)= 0.40= 22o
7/25/2019 physics project made by ishan gupta.ppt
15/22
=onuniform Circular Motion
If an o5ect is moving in a circular
path ut at varying speeds! it
must have a tangentialcomponent to its acceleration as
well as the radialone.
7/25/2019 physics project made by ishan gupta.ppt
16/22
=onuniform Circular Motion
&his concept can e used for an o5ect moving
along any curved path! as a small segment of thepath will e appro8imately circular.
7/25/2019 physics project made by ishan gupta.ppt
17/22
# Centrifugation
' centrifugeworks y
spinning very fast. &hismeans there must e a
very large centripetal
force. &he o5ect at '
would go in a straightline ut for this force6 as
it is! it winds up at ;.
7/25/2019 physics project made by ishan gupta.ppt
18/22
Example 9,he rotor of an ultracentrifuge rotates at 50000
rpm (revolutions per minute). ,he top of a 4.00
cm long test tu&e is $.00 cm from the rotation a*isand is perpendicular to it. ,he &ottom of the tu&e
is 10.00 cm from the a*is of rotation. -alculate
the centripetal accelerations in gs at the top
and the &ottom of the tu&e.
At top, 2r = (2)(0.0600 m) = 0.377 m per revolution
50,000 rpm = 833 rev/s so T =1
833 rev/s=1.20x10-3 s/rev
v =2r
T=
0.377 m/rev
1.20x10-3 s/rev= 3.14x102 m/s
aR =v2
r =(3.14x102 m/s)2
0.0600 m =1.64x106 m/s2 =1.67x105 g's
At the bottom, v =2r
T=
(2)(0.1000 m)
1.20x10-3 s/rev= 523.6 m/s
aR =v2
r=
(523.6 m/s)2
0.1000 m= 2.74x106 m/s2 = 2.80x105 g's
7/25/2019 physics project made by ishan gupta.ppt
19/22
7/25/2019 physics project made by ishan gupta.ppt
20/22
7/25/2019 physics project made by ishan gupta.ppt
21/22
7/25/2019 physics project made by ishan gupta.ppt
22/22