Ch 7.1 Projectile Motion

Updated 5/26/03
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Catapult Design: 
Design Project             #1 http://academic.engr.arizona.edu/instructor/projects/catapult.htm
Compliant Catapult     #2 http://www-personal.engin.umich.edu/~vaughand/design.html
Mouse Trap Catapult  #3 http://ceeps.uscolo.edu/designcontest/mt_ctplt.html
Design Competition    #4 http://www.bae.uky.edu/~shearer/EGR_199/CatapultDesignComp.htm

1 mi/h = 1.47 ft/s = 0.447 m/s     MACH 1 = 750 mi/h
  1 lb = 4.445 N                     1 mi = 5,280 ft
USE PROTRACTOR & WASHER TO DRAW DIAGRAMS.

1     a] DEFINE PROJECTILE:  An object moving through air or space under the influence 
        of gravity only;  No drag, lift, or thrust
     b] GIVE 3 EXAMPLES OF PROJECTILES. i] A thrown ball, rock, bullet, or arrow. 
     [ii] a planet, moon or satellite, [iii] A rocket after the fuel has burned out.

2A    A STONE IS THROWN HORIZONTALLY AT 25 ft/s FROM A 50 ft HIGH CLIFF.
        a] HOW LONG DOES IT TAKE THE STONE TO REACH THE BOTTOM? 
        b] HOW FAR FROM THE CLIFF DOES THE STONE STRIKE THE GROUND?                        .
        c] DERIVE THE EQUATION FOR THE DISTANCE FROM THE CLIFF. Vh √(Dv/2g)

2B    A STONE THROWN HORIZONTALLY AT 15 mi/h HITS THE GROUND 47 ft FROM THE CLIFF.  
        a] HOW LONG DOES IT TAKE TO HIT THE GROUND;
        b] HOW HIGH IS THE CLIFF?  Dv = g(Dh/Vh)²/2

2C    FIND A CAR'S HORIZONTAL VELOCITY IN mph IF IT TRAVELS 25 ft HORIZONTALLY WHEN 
         PROJECTED FROM A HEIGHT OF 6.0 ft.
        Vh = Dh/√(Dv/g/2)/1.47 ft/s/mi/h

4     A BALL IS HIT AT A VELOCITY OF 30 mi/h & AN ELEVATION OF 40^o.
        a] DRAW & LABEL THE DIAGRAM (Use protractor between 50^o & 130^o to draw trajectory.)
           (Protractor angles = (90-40) & 180 - (90-40) = 50^o & 130^o)
        b] THE INITIAL VERTICAL VELOCITY                    [=Vm SIN(q)]
        c] THE TIME TO REACH THE PINNACLE OR TOP OF THE ARC; [Vm SIN(q)/g]
        d] THE MAXIMUM HEIGHT IT REACHES.                 gt²/2 =  g/2(Vm SIN(q)/g)² = (Vm SIN(q))²/2g
        e] HORIZONTAL VELOCITY.                             [Vm COS(q)]
        f] THE TIME IT IS IN THE AIR.                       2*Time to Top
        g] THE HORIZONTAL DISTANCE IT TRAVELS.              Vm²SIN(2q)/g

5     IF YOUR CANNON HAS A MUZZLE VELOCITY OF 400 ft/s, FIND THE:
a] RANGE:  Vm²/g = (400 ft/s)²/32 ft/s² = 5,000 ft
b] DRAW & LABEL THE DIAGRAM  
c] FIND TWO ANGLES NEEDED TO HIT A TARGET 4,000 ft AWAY. 
2q = Sin^-1(Target Distance/Range) = Sin^-1(4000 ft/5000ft) = 53.1º
Be sure to have your calculator in degree mode.
2q = 53.1^o --> q = 53.1/2 = 26.6^o
The second angle is a compliment of the first:  90^o - 26.6^o = 64.4^o
d] DIAGRAM SIN^-1(15/35)             
              /|
      35 ft  / | 15 ft
           /q   |   |
DESCRIBE IN WORDS:  SIN^-1(15/35) is the angle whose sin = 15/35

6     YOU HIT A BALL 200 ft. 
        a] HOW FAST (mi/h) DID IT LEAVE YOUR BAT?
        b] WHAT ARE YOUR ASSUMPTIONS?
Assume no air resistance and q = 45^o--->SIN(2*45^o) = 1
    Dh    = Vm²SIN(2q)/g
   200 ft = Vm²SIN(2*45^o)/32 ft/s²
   200 ft = Vm²/32 ft/s² ---> Vm² = 32 ft/s^{2*200 ft
       Vm = \/32 ft/s2*200 ft  = \/32*200ft2/s2 = 80 ft/s
       Vm = 80 ft/s x 1 mi/h/1.47 ft/s = 54.4 mi/h}  

4    a] Why are astronauts weightless?
     b] DRAW & LABEL THE DIAGRAM TO EXPLAIN WHY ASTRONAUTS ARE WEIGHTLESS.
         The earth curves 16 ft vertical for every 5 mi horizontal.
         An object will fall 16 ft every sec: 
         D = gt²/2 = 32 ft/s=2 (1 s)²/2 = 16 ft
The condition for orbit is that an object matches the earth's curvature by moving at 5 mi horizontally every second.  This allows it to fall 16 ft vertically without changing altitude.  The result is that the object falls around the earth.  [See diagram pg 108).]  Astronauts are weightless because they are in continuous free fall.   

     c] Use the diagram to explain how the space shuttle returns to earth.

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7.2 Periodic Motion
Updated 5/26/03

MERCURY [ 3.76 m/s², R =  2,439 km]     VENUS   [ 8.62 m/s², R =  6,052 km]
EARTH   [ 9.80 m/s², R =  6,380 km]     MOON    [ 1.62 m/s², R =  1,738 km]
MARS    [ 3.77 m/s², R =  3,398 km]     JUPITER [24.87 m/s², R = 71,350 km]
SATURN  [10.42 m/s², R = 60,268 km]     URANUS  [10.54 m/s², R = 25,559 km]
NEPTUNE [13.8  m/s², R = 22,300 km]     PLUTO   [ 0.60 m/s², R =  1,151 km]

1 mi/h = 1.47 ft/s = 0.447 m/s        MACH 1 = 750 mi/h    1 lb = 4.445 N     1 mi = 5280 ft

1   HAMMER THROW:  A 7 kg MASS IS ATTACHED TO A 2.1 m STRING. 
    THE MASS COMPLETES 4 HORIZONTAL REVOLUTIONS IN 2 s. 
    a] DRAW & LABEL THE DIAGRAM.
    b] DEFINE AND FIND THE PERIOD.        T = 2s/4 = 0.5 s
    c] FIND THE VELOCITY IN m/s.  V= 2piR/T= 2*3.14*2.1m/0.5s= 26.38m/s
    d] FIND THE ACCELERATION.     a= V²/R= (26.38m/s)²/2.1m = 332 m/s²
    e] FIND THE CENTRIPETAL FORCE IN N & lbs  
    F = ma = 7kg*332m/s² = 2321 N x 1 lb/4.445 N = 522 lb
    If my numbers look different from yours it is due to round off.
    I kept all numbers in my calculator an rounded off at the end.
Ex: V² = (2*3.14*2.1m/0.5s)²
    a = V²/R = (2*3.14*2.1m/0.5s)²/2.1
    F = ma = 7kg (2*3.14*2.1m/0.5s)²/2.1(1 lb/4.445N) = 521.7 lb

2     A CLOWN RIDES A MOTORCYCLE AROUND A "LOOP-THE-LOOP", WITH A 15 m RADIUS.  
       a] DIAGRAM, 
      b] DERIVE THE EQUATION & FIND THE SLOWEST SPEED, IN mi/h, HE MUST HAVE.  
       c] STATE THE ASSUMPTIONS. 

3A    a] DRAW & LABEL THE DIAGRAM FOR A SATELLITE.
      b] DERIVE THE EQUATION FOR ITS ORBITAL SPEED.
      c] SHOW THE UNIT CHECK. 
      d] STATE THE 3 ASSUMPTIONS.
FOR AN EARTH SATELLITE [Rp = 6,380 km, 9.8 m/s2] AT ALTITUDE = 160 km
      e] FIND ITS SPEED [mph & MACH#]
P: NEPTUNE [13.8 m/s2, 22,300 km]  HMWK: JUPITER [24.87m/s2, 71,350 km

3B*   Repeat 5A taking into account the change in g with altitude.

3C     a] DERIVE THE EQUATION FOR A SATELLITE'S PERIOD OF ORBIT.
       b] SHOW THE UNIT CHECK.
          FOR AN EARTH SATELLITE [Rp =  6,380 km] AT AN ALTITUDE = 160 km
       c] FIND ITS PERIOD [hr,min,sec]
       P: NEPTUNE [13.8 m/s2, 22,300 km]  HMWK: JUPITER [24.87m/s2, 71,350 km

ANS: 3]  26.4 m/s; 522 lb (P) 25.1 m/s, 284 lb; (H) 12.6 m/s, 71.1 lbs
4]  27.2 mph; (H)  31.3 mph
5]  8,006 m/s = 17,909 mph = 23.9; (P) 39,385 mph = 52.5; (H) 125.8
6]  1 h 25 min 33 s; (P) 2 h 13 min 39 s; (H) 2 h 57 min 34 s

4   DEFINE: 
    a] PERIODIC MOTION:  A repeating motion that occurs in regular time intervals.
    b] RESONANCE:  A LARGE VIBRATION CAUSED BY A SMALL PERIODIC DISTURBANCE AT THE SYSTEMS NATURAL FREQUENCY.  EXAMPLES:  CHILD'MOTION WITH A SWING, A TUNED RADIO OR TV WITH THE RADIO/TV SIGNAL, IR LIGHT WITH THE ATOMS IN A MOLECULE.
    c] EQUATIONS FOR NATURAL FREQUENCY OF:
       A PENDULUM:           T = 2pi√L/g 
       WEIGHT ON A SPRING:   T = 2pi√m/k
       A SATELLITE IN ORBIT: T = 2pi√R/g    
    d] Simple Harmonic Motion:  PERIODIC MOTION OVER THE SAME PATH. 
THE RESTORING FORCE IS PROPORTIONAL TO THE DISPLACEMENT FROM THE EQUILIBRIUM POSITION.  OBJECTS THAT VIBRATE, OSCILLATE, ROTATE, OR ARE MOVED BY WAVE MOTION HAVE SIMPLE HARMONIC MOTION.

5    a] WAVES ARE CREATED BY OBJECTS THAT Vibrate.
     b] VIBRATIONS CAN BE DESCRIBED BY THEIR frequency, period, & amplitude.
     c] A VIBRATION CYCLE IS ONE COMPLETE back & forth MOTION.
     d] Amplitude IS THE MAXIMUM DISPLACEMENT FROM REST.
     e] DRAW & LABEL A PENDULUM (Heath pg 309)
     f] DRAW & LABEL A VIBRATING WEIGHT & SPRING AND (pg 147)

6    a] Frequency (f) = # EVENTS/UNIT TIME
     b] Hertz (Hz) = 1 EVENT/SECOND
     c] Period (T) = THE TIME/EVENT.
     d] f = 1/T, T = 1/f

7    On top of a mountain a pendulum 1.5 m long has a period of 2.5 s.  
     What is the acceleration of gravity at this location?  (Pg 149)

8    a] Find the frequency & period for a child on a swing that completes 20 cycles 
        in 25 s.
     b] A stroboscope is flashing so that the time interval between flashes is
        1/80 s. Find the strobe’s frequency.
     c] Calculate the frequency and period of a tuning fork that vibrates 24,000 
        times in 1.00 min.

9    a] DEFINE RESONANCE.  
     b] GIVE 3 EXAMPLES (MECHANICAL, ELECTRICAL, & CHEMICAL) OF 2 THINGS RESONATING 
        WITH EACH OTHER.  [Physics pg 126]

EXTRA QUESTIONS

1    Derive the equation for centripetal acceleration.

2     AN OBJECT ON A STRING IS SWUNG IN A HORIZONTAL CIRCLE.
      WHAT DIRECTION WILL IT MOVE IF THE STRING IS CUT?
      DIAGRAM AND STATE YOUR ANSWER. 
      In a straight line tangent to the circle at the point of release.

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