Mr. Kovall’s Neighborhood

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Circular Motion Notes

(revised 07-24-02)

Refer to the following pages in your text:

            Chapter 3-page 62, section 3-5; Chapter 6-page 125, sections 6-1, 6-2, 6-3; Chapter 9-page 208, sections     9-1, 9-2, 9-3.

 


                                                                     

                                                                                                                                         T = 1/f

                                                                                                                                         T = seconds/orbit

                                                                                                                                         f =  orbits/sec, rev/sec

 

                                                                    

                                                                  

                                                                       

 

In uniform circular motion, the direction of the body’s velocity vector is NOT constant, as it was in the Horizontal and Vertical Motion chapter studied earlier. This constantly changing orientation of the body’s velocity vector with time, produces an acceleration.

 

a = ΔV / ΔT  =  change in velocity vector’s magnitude / time  or  change in vector’s direction / time

 

      a is due to the motion of the body on a circular path and is directed toward the center of the circle. This

      “centripetal”  or center-seeking acceleration is illustrated below:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


                       

 

                       

 

      A body in uniform, non-linear accelerating, circular motion, has a velocity that is constant in magnitude but not in direction.  This leads to a “center-seeking” or centripetal acceleration that is constant in magnitude but technically, not in direction.  Even though the acceleration vector always points to the center of the circular path, its absolute orientation relative to you changes:

 

 

 

 

 

 

 

 

 

 

 

 

 


                               

            Figure 6 and Figure 7 represent similar triangles, with Figure 6 rotated below to clarify this point:

 

 

 

 

 

 

 

 

 

 

 

 


                        ΔV / V  =  S / R

 

                         ΔV  =  V·(S / R)  =  (S·V) / R

 

Dividing both sides by ΔT:

            ΔV / ΔT  =  (S·V) / (R·ΔT)  =  (S / ΔT) · (V / R)

                         

V  =  S/ΔT, and a  = ΔV / ΔT,  so:

 

                       a  = (V) · (V / R)

                       a  =  V2 / R

The acceleration vector, a, in this case is a centripetal acceleration, so:

 

                aC  =  V2 / R

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


            An object not tethered to “C”, will follow the tangential path “AD”, obeying Newton’s First Law of Motion.  The object “falls” from point “D” to point “B” each moment to remain on the circular path – for small angle θ, DàB points toward the center of the circle.

 

 

 

Comparison of aC between curves of differing “R”:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


            aC1/aC2  =  (V12/R1) / (V22/R2)  =  R2/R1

 

aC1/aC2  =  100/200 

 

aC1  =  1/2aC2 

 

Centripetal Force

 

            According to Newton’s Second Law of Motion:  F  =  m · a

“F” is the net or unbalanced force applied against a body having mass “m”, which produces an acceleration “a” in the body’s motion.  This increase (or decrease) in the body’s velocity over time is usually very apparent to the observer.

When the same body is placed in uniform circular motion, motion at a specified and unchanging velocity, the force applied against the body, and the resulting acceleration is much less apparent to the observer.  Newton’s Second Law of Motion still applies, and the equation is written as: FC  =  m · aC

“FC” is the “centripetal” force directed against the body’s mass, and points toward the center of the circle.  This force may be found within the string tethering a revolving body to the center, may be a lateral thruster on a space probe, or may be the friction holding a car’s tires to the pavement of bend in the road.  As discussed in the previous section, acceleration, aC, is directed toward the center of the circle as well.  This acceleration is due to a change in direction of the velocity vector, and not to a change in its magnitude.

 

FC  =  m · aC

 

aC  =  V2/R

 

FC  =  m · V2/R

 

 

 

 

 

The force of friction (static) between the pavement and tires of a car provides the centripetal force necessary to keep the car moving along the curve (unbanked).

The example below compares the centripetal force necessary to keep a body moving along a curve at 20 m/sec (V1), then at 40 m/sec (V2):

 

 

 

 

 

 

 

 

 

 

 

 

 


            V2  =  2V1

 

 

FC1 / FC2  = ( m·V12/R) / ( m·V22/R) = V12/V22

 

FC1 / FC2  =  V12/(2V1)2  =  V12/4V12  =  1/4

 

FC2  =  4FC1 

 

When the body’s velocity is doubled from the first trial to the second, the associated centripetal force in the second trial is four times that required during the first trial.

 

 

 

 

 

VMAX for a car traveling around an unbanked curve:

 

 

 

 

 

 

 

 

 


FWT  =  FN =  m·g

             

FC  =  m·V2/R

FS  =  μS·FN  = μS· m·g

 

FC  =  FS 

 

m·V2/R  =  μS· m·g

                 V2  =  μS·g·R

                 VMAX  =  (μS·g·R)1/2

 

      Note that the maximum velocity allowed for a car moving along an unbanked curve is independent of the car’s mass – light and heavy cars have the same VMAX for the same road surface conditions.

                 If the car’s wheels lock and slide as the car is negotiating the curve, the coefficient of sliding friction (μK) is less than the coefficient of static friction (μS), and the corresponding FS is reduced.  The FS is less than the FC required for the given speed, so the car slides outward, out of control.

 

 

 

 

 

 

Plane in a Banked Turn:

 

 

 

 

 

 

 

 

 

 

 

 

            FL represents the equilibrant lift vector, which supports FWT, allowing the plane to remain in the air.  FL is produced by the air pushing upward on the plane’s wings.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


FL Sin θ is the lift component actually responsible for providing the centripetal force necessary to hold the plane in its turn.

A greater speed (V) of the plane requires a greater banking angle (θ) to provide the necessary centripetal force (FC).

 

 

 

 

 

Car on a Banked Curve:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

            When a body is placed in circular motion and moves across an inclined surface as shown in Figure 14, the upward acting perpendicular contact force between the inclined surface and the body, FN, is different in magnitude and function than the FN illustrated in Figure 15.  FN (FWT Cos Ø) in Figure 15 is one of two vector components of the body's weight, and is responsible for supporting that component of the body's weight, Fn, which presses the body into the inclined surface.  The inclined surface here is not serving as a banked turn for a body following a circular path.  Also, the magnitude of FN in this case is less than the weight of the body. 

            However, the FN vector generated by the body moving in circular motion along the banked surface, in Figure 14, has two right-angle components.  The first component, FC (FN Sin Ø), is directed radially and holds the body into the turn as it moves over the banked surface.  The second component FNCos Ø, the equilibrant to the body's weight (FWT), is directed vertically and supports the weight of the body.  In this case, FN is larger than the weight of the body.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


Referring to Figure 14, and taking the ratio between the two vector components of FN,

 

FN Sin θ / FN Cos θ  =  (m·V2/R) / (m·g)

Tan θ  =  V2/R·g

 

            For a given speed (V), the centripetal force (FC) needed for a curve of radius ( R), is obtained by banking the turn at angle (θ).

            As a young civil engineer, what must you do to the angle θ when redesigning a banked turn whose posted speed limit is to be raised?  Why?

 

 

Satellites in Circular Orbit: Orbital Radius vs. Orbital Speed

 

            Newton’s Law of Gravitation states,

                        F  =  G · M1· M2 / R2  

 

                        F  =  gravitational force

                        G  = universal gravitational constant, 6.67 x 10-11 Nt m2/kg2

                        M1, M2 = masses of the interacting bodies

                        R = distance between the centers of mass

 

 

 

 

 

 

 

 

 

 

 

 

 

 


FGRAVITY = FC, and is the only force acting on the satellite in the radial direction., so:

 

FGRAVITY  =  G · M1· M2 / R2  

 

            and

 

FC  =  m·V2/R  =  MS·V2/R      

 

FGRAVITY  =  FCENTRIPETAL

 

         G · M1· M2 / R2   =   MS·V2/R

 

Solving for V = VORBITAL,

           

            VORBITAL  =  (G·ME/R)1/2

 

ME  =  5.98 x 1024 kg

 

            Low Earth Orbit (L.E.O.) satellites require larger orbital speeds than satellites occupying higher orbits (larger R).  For a given orbit, a satellite with a large mass requires the same orbital speed as a satellite with a small mass.

 

 

 

 

Synchronous Satellites (Geo-stationary)

            A synchronous satellite orbits the Earth is a circular path that is in the plane of the equator.. The period of the satellite is chosen to be one day, the same as the period of the Earth’s Rotation.

                        FIGURE 17

VORBIT  =  2πR/T  =  (G·ME/R)1/2

 

 

2πR2/2/T  =  (G·ME)1/2/(R)1/2

 

2πR3/2/(G·ME)1/2  =  T

Kepler’s third law of planetary motion above, states that the period of revolution is proportional to the three-halves power of the orbital radius.

 

            The period of a satellite is equal to the time required for one complete revolution,

 

            If  T = 24 hours =  8.64 x 104 seconds,

 

 

R3/2  =  T(G·ME)1/2/2π

 

             R3/2  =  8.64 x 104 sec [(6.67 x 10-11 Ntm2/kg2)(5.98 x 1024 kg)]1/2/ 2π

 

 R3/2  = 2.746296245 x 1011 m3/2

 

 R  =  4.23 x 107 m, from the center of the satellite to the center of the Earth

 

However, the radius of the Earth is approximately 6.38 x 106 m, so the altitude of the geo-stationary satellite above the Earth’s Surface (SV) is,

 

 

SV  =  RORBITAL  -  REARTH  =  4.23 x 107 m – 6.38 x 106 m  =  3.59 x 107 m or 22,300 mi.

 

The orbit of a synchronous satellite can have only one radius.

 

The period T in the equation above assumes a 24-hour (8.64 x 104 sec) duration, known as a “solar day”.  This quantity is actually greater than one revolution relative to space (the stars), because the Sun moves slightly in its grand orbit about the galaxy as the planet is rotating on its axis.  Earth must spin a little past one space-relative revolution to place the Sun in the same position above you the next day.  The period of one space-relative revolution is known as a “sidereal day” and is 23 hours, 56 minutes in duration.

 

 

Artificial Gravity

 

           

 

 

 

 

 

 

 

 

 

 

 

 

 

                        M= mass of the astronaut

                        V = rotational speed of the space station

                        R = radius of the space station

 

                   FC  =  FART.GRAV.

                          FC = FWT

         M·V2 /R  =  M·g 

 

                   V  =  (R·g)1/2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


            Angle in radians = arc length / radius

 

                         θRAD  =  S/R   =   1/1 rad  =  1 rad

 

            1 revolution = 360º

            arc length = circumference = S

 

                        θ = 2πR / R  =  2π rad

                        360º =  2π rad  =  6.28 rad

 

                        1 rad = 360º / 2π  =  57.3º

 

 

 

Angular Velocity and Angular Acceleration

 

Average Angular Velocity (ω)  =  angular displacement (θ) / time elapsed (t)

                        ω  =  Δθ / Δt  in radian /second           ω - rpm (rev/min, rad/sec) directly proportional

                                                                                                + ω = ccw

- ω = cw

 

           

NOTE:  unit in revolutions / minute must be converted to radian / second:

            1rev/min = 1rpm = 2π rad / 60 sec  =  2π / 60 rad sec-1 = 0.1047 rad/sec

 

           

                       

                                   

Angular Acceleration  - whenever angular velocity changes (rev/min of device changes)

 

            linear acceleration, a = ΔV / ΔT     in meters/second/second (m/sec2)

            angular acceleration, α = Δω / ΔT    in radian/second/second (rad/sec2)

Equations for Rotational Kinematics

 

For constant angular acceleration, average angular velocity

            ώ = ½(ωF + ωO)                                

ώ = θ / T

 

θ = ½(ωF + ωO) ·T

 

Linear motion                                                                           Rotational motion

 

            VF  =  VO  + a·T                                                                                  ωF  =  ωO + α ·T

            S  =  VO·T + 1/2·a·T2                                                                                                    θ  =  ωO·T+ 1/2·α ·T2

            VF2  =  VO2  + 2·a·S                                                                             ωF2  =  ωO2 + 2·α ·θ

           

 

 

Tangential Velocity (VT) and Tangential Acceleration (a)

 

            Tangential acceleration (a) represents a change in speed, whereas angular acceleration (α) represents a change in direction.            

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


            θ = S / R  =  V·T / R

 

            θ / T  =  V·T / R·T

 

            ω =  V / R

 

            V  =  VT  =  R·ω  (tangential speed is directly proportional to the path’s radius)

(ω in rad/sec, while VT in m/sec.   magnitudes without + or -)

V1/V2  =  R1·ω/R2·ω  = 1ω/2ω

So,

V1  = ½V2 for the same ω

 

 

If rotational motion increases,

aT  =  ΔVT / T  =  (VF – VO) / T  =  (RωF – RωO) / T  =  R [(ωF –ωO) / T]

 

α =  (ωF –ωO) / T

 

aT = R·α        (remember radian measure)

 

 

 

Centripital Acceleration and Tangential Acceleration

 

            Remember, for uniform circular motion, VT remains constant.

 

 

 

 

 

 

 

 

 

 

 

 

aC  =  VT2 / R  =  (R·ω)2 / R  =  R·ω2

aC  =  R·ω2      (ω in rad/sec)

 

 

 

 

 

 

 

 

 

 

 

 

 

 


            If the tangential speed increases, the body undergoes non-uniform circular motion and experiences two acceleration components, aC and aT.

            aT  = R·α  or  aT  = FT / m  (from FT  =  m·aT )

 

            a  =  (aC2  +  aT2)1/2

 

            Φ =  Tan aT / aC