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Physics 101Lecture 6Circular Motion
Assist. Prof. Dr. Ali ÖVGÜNEMU Physics Department
www.aovgun.com
December 18, 2018
q What is the smallest value of the force F such that the 2.0-kg block will not slide down the wall? The coefficient of static friction between the block and the wall is 0.2. ?
Equilibrium, Example 1
F
mg
N
f
F
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Uniform circular motion
Constant speed, or,constant magnitude of velocity
Motion along a circle:Changing direction of velocity
Uniform Circular Motion: Definition
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Uniform Circular Motion: Observations
q Object moving along a curved path with constant speedn Magnitude of velocity: samen Direction of velocity: changingn Velocity : changingn Acceleration is NOT zero!n Net force acting on an
object is NOT zeron “Centripetal force” amFnet
!!=
v!
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Uniform Circular Motionq Velocity:
n Magnitude: constant vn The direction of the velocity is
tangent to the circleq Acceleration:
n Magnitude: n directed toward the center of
the circle of motionq Period:
n time interval required for one complete revolution of the particle
rvac2
=
rvac2
=
vrT p2
=
vac!!
^
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Centripetal Forceq Acceleration:
n Magnitude: n Direction: toward the center of
the circle of motionq Force:
n Start from Newton’s 2nd Law
n Magnitude:
n Direction: toward the center of the circle of motion
rvac2
=
rvac2
=
amFnet!!
=
vac!!
^
rmvmaF cnet
2
==
netF!
netF!
netF!
vFnet!!
^
netc Fa!! ||
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What provides Centripetal Force ?
q Centripetal force is not a new kind of forceq Centripetal force refers to any force that keeps
an object following a circular path
q Centripetal force is a combination of n Gravitational force mg: downward to the groundn Normal force N: perpendicular to the surfacen Tension force T: along the cord and away from
objectn Static friction force: fs
max = µsN
rmvmaF cc
2
==
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a
What provides Centripetal Force ?
rmvT
maTFnet2
=
==
rvmmgN
mamgNFnet2
+=
=-=
mg
N
v
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Problem Solving Strategyq Draw a free body diagram, showing and labeling all
the forces acting on the object(s)q Choose a coordinate system that has one axis
perpendicular to the circular path and the other axis tangent to the circular path
q Find the net force toward the center of the circular path (this is the force that causes the centripetal acceleration, FC)
q Use Newton’s second lawn The directions will be radial, normal, and tangentialn The acceleration in the radial direction will be the centripetal
accelerationq Solve for the unknown(s)
q In vertical circular motion the gravitational force must also be considered. An example of vertical circular motion is thevertical “loop-the-loop” motorcycle stunt. Normally, the motorcycle speed will varyaround the loop.
q The normal force, FN, and the weight of the cycle and rider, mg, are shown at fourlocations around the loop.
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Vertical Circular Motion
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q There is a minimum speed the rider must have at point 3 in order to stay on the loop.
q This speed may be found by setting in the centripetal force equation or point 3, i.e. in
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Problem :
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Ex2: The Conical Pendulumq A small ball of mass m = 5 kg is suspended from
a string of length L = 5 m. The ball revolves with constant speed v in a horizontal circle of radius r = 2 m. Find an expression for v and a.
mg
T θ
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The Conical Pendulum
44.0tan
4.0sin
sin
cos0cos
2 5 5
22
2
=-
=
==
==
=
=-=
===
å
å
rLr
Lr
rmvTF
mgTmgTF
mrmLkgm
x
y
q
q
q
q
q
22
2
2
m/s 3.4tan
m/s 9.2tansin
tan
tan
cos
sin
===
==
=
=
=
=
q
q
q
q
q
grva
Lgv
rgv
grvmgTrmvT
q Find v and a
December 18, 2018
Ex:3 Level Curvesq A 1500 kg car moving on a flat,
horizontal road negotiates a curve as shown. If the radius of the curve is 35.0 m and the coefficient of static friction between the tires and dry pavement is 0.523, find the maximum speed the car can have and still make the turn successfully.
rgv µ=
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Level Curvesq The force of static friction directed toward the
center of the curve keeps the car moving in a circular path.
rgv µ= smmsm
grmmgr
mNrv
mgNmgNF
rvmNf
sss
y
ss
/4.13)0.35)(/8.9)(523.0(
0
2
max
2max
max,
==
===
=
=-=
==
å
µµµ
µ
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Ex:4 Banked Curvesq A car moving at the designated
speed can negotiate the curve. Such a ramp is usually banked, which means that the roadway is tilted toward the inside of the curve. Suppose the designated speed for the ramp is to be 13.4 m/s and the radius of the curve is 35.0 m. At what angle should the curve be banked?
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Banked Curves
!6.27))m/s 8.9)(m 0.35(
m/s 4.13(tan
tan
cos0cos
sin
m 0.35 m/s 4.13
21
2
2
==
=
=
=-=
===
==
-
åå
q
q
q
q
q
rgvmgn
mgnFrmvmanF
rv
y
cr
Problem 1:
December 18, 2018ANS: For A : T=0.058 N ,
For C: T=0.258 N
Problem 3:
Problem 2:
R=25.5m
v=4 m/s
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Problem 4:
Problem 5:
R=39.24 m
v=4.1 m/s
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Problem 6:A string under a tension of 50.0 N is used to whirl a rock in a horizontal circle of radius 2.50 m at a speed of 20.4 m/s on a frictionless surface as shown in Figure. As the string is pulled in, the
speed of the rock increases. When the string on the table is 1.00 m long and the speed of the
rock is 51.0 m/s, the string breaks. What is the breaking strength, in newtons, of the string?
Problem 7:A puck of mass m1 is tied to a string and allowed to revolve in a circle of radius R on a frictionless, horizontal table. The other end of the string passes through a
small hole in the center of the table, and an object of mass m2 is tied to it Fig.
The suspended object remains in equilibrium while the puck on the tabletop
revolves. Find symbolic expressions for (a) the tension in the string, (b) the radial force acting on the puck, and (c) the speed of the puck.
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Problem 9:
Problem 8:
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Problem 10:
Problem 11:
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Problem 12:
