what is a formula for a pendulum?
375701
Try those...\n\nhttp://www.superbase.com/sb_scientific_current/example_pendulum.htm\n\nhttp://www.8886.co.uk/pendex1.htm
202904
A regular pendulum moves back and forth while hanging from a point. A torsional pendulum twists clockwise and counterclockwise while hanging from a point. In each case, the heavier the lower end of the pendulum, the longer it will swing or rotate. A torsional pendulum also works better with a rubber or other "live" length of material. Using a rope will not give you much rotation after it unwinds this first time.
949347
The period of  simple pendulum is\n\nT = 2π√(l / g)      if on such planet g is 1/3 of the Earth then \n\ng = 1/3 (9.81) m/s² = 3.27 m/s²\nl = 1.53 m\nThen\n\nT = 2π√(1.53 / 3.27) = 4.3 s
491270
In contrast to a plane pendulum, which can just swing in one plane, a spherical pendulum can swing in all directions except that the distance to the suspension keeps constant, so the swinging mass is always on the surface a shpere around the suspension.
972284
T = 2.60    This should give the right answer in your formula, which appears correct.\n\nT, the period, refers to the time for one complete cycle of oscillation.
401476
Do practice .\n   do not learn formulae,\n   But see them and originate questions by yourself from it.\nlike \n   a formula for line is \n    y = mx + c;\n what happens if x = 0; what line it gives\n what happens when y = 0 :  "  "   "   "  \n if x = y                   "  "   "   "
1117824
Pendulums work like this...\n\nhttp://en.wikipedia.org/wiki/Pendulum\n\nNothing magical.  Just simple physics.
402767
In engineering terms pendulums are called mass dampers, usually either active mass dampers or passive/tuned mass dampers.  Check out the Earthquake Engineering Research Institute to start.
867152
The period equation for a simple pendulum is:\nT=2pi*(I/mgL)^0.5\nI =rotational inertia\nm=mass\ng= gravity\nL=length\n\nBecause the rock attached to the string is small then I=mL²\ninserting this into the period equation above you get:\nT=2pi*(L/g)\n\nRearranging to solve for L:\nL= (g*T²)/(4*pi²)\nNow plug in the given values:\n(9.8m/s²*9.22²s²)/(4*3.14²) = 21.1m\n\nThis will only be correct if the rock is just above the floor of the tower.
512039
I hope this is what you were looking for: \n\nLet A = your acceleration(RPS^2)\nLet D = your deceleration(RPS^2) (use positive value)\nLet RPS = your target velocity in RPS\nLet RPM = velocity in RPM\nLet N = the number of revolutions the motor needs to turn\nLet t (sec) = time needed\nSqrt = square root. Make sure you pay attention to the brackets.\n\nThen RPS = ( -t + Sqrt(t^2 + 2N(A+D)/AD))x AD/(A+D) \n\nAnd RPM = RPS/60\n\nI believe this formula is correct since I checked that both sides of the equation have the units of velocity.\nHowever, I have noted that many people on this site give out incorrect info. It is always a good idea to get a second source confirmation. Perhaps a servo motor manufacturer or distributor could provide the info and give helpful tips for a specific application. Also physics or math faculty at most colleges and universities are pretty co-operative when it comes to requests like this. If you fax the details they'll usually respond. Best wishes!
774149
There are quite a lot of variables involved in the equation. \n\nYou can calculate the force required from the "Gold standard" of decapitation devices the guillotine. A 30 kg weight dropping 226 cm is enough to sever the neck. Throw the earth's gravity into the equation and you can calculate the force the machine generated. \n\nTo calculate Kilo's per square inch that the guillotine blade delivered I suggest that you measure the average width of a neck. The Guillotine blade was pretty blunt so I would say you could calculate the cutting surface at about a mm. Jack's  knives were sharper than the guilotine blade so calculate using a twentieth or a fiftieth of a mm.\n\nThe second attached link shows some the physics of applying the strikes that Jack would have used. Your angles of motion calculation in pendulums and levers can tell you the best blows to deliver the necesary force. \n\nGood luck and I would like to see your thesis when you are done.
923444
Short Answer: \n\nvert comp = .161 meters\nhori comp = .936 meters                        \n\nLong Answer:\n\nMb=mass of bullet\nMp= mass of pendulum\nv=velocity of bullet\ng=9.81 m/s/s\nh=vertical displacement\n\n\nBy conversation of momentum:\n\nmomentum of pendulum with bullet=momentum of flying bullet\n\nsqrt(2*g*h)(Mb+Mp)=Mb*v= (28/1000)*230=6.44 \n\nsolving that you can find h=0.161 meters\n\n\nBy the distance formula:\n\nsqrt(2.8^2+0^2)=sqrt( (2.8-0.161)^2 + (horizontal position)^2)=2.8\n\nsolving that you can find horizontal position = 0.936 m
914856
OK as the ball is swung around in a horizontal circle the scale inside the tube is measuring the tension on the string. By knowing the time to complete one revolution  linear velocity can be found from v=r*omega(angular velocity). in the second part the v is given and needs to be converted into radians/second by using 3.7=.5*omega where omega=3.7/5=.74rad/s then using the mass radius and angular velocity you can find the tension in the string. With a= acceleration (into center) omega^2*r=.74^2*.5=.2738m/s^2and using f=ma the tension in the string will equal f which equals .2*.2738=.0548newtons Unless i have a silly math mistake somewhere.
792458
I'm assuming the amplitude is the horizontal distance from vertical.  In that case, the amplitude is the base and the pendulum length is the hypotenuse of a right triangle.  Figure that vertical side, subtract it from the length of the pendulum and that's the vertical distance it falls through.  Energy is the weight multiplied by this distance.
834149
The period T is 0.9s\nT = 1/f = 2pi/w so w = 2pi/T\n\nAs the movement is circular, the total of the forces is in this plan and to the centre and the value is mw²R.\n\nSo P + S = mw²R where P is the weight and S the tension of the string and R the radius and a the angle between the string and the vertical\nIn projection, we have :\nmw²R = S sin a\nmg = S cos a\n\nDividing the 2 equations :\ntg a = w²R/g\nand R = L sin a\n\nso w² = g/(Lcos a)\nso cos a = g/Lw² = g T²/(L 4 pi²) \n\nso cos a= 9.8 0.9²/ (0.3 4 3.14²) = 0.6702\n so a = 47.91°  or a =0.83 rad\n\nBut this angle is the angle between the string and the vertical so angle between the string and the horizontal is 90-a = 42.09°
795982
∫_0_2 g(t) dt = 5\n\n(a)\ny= ∫_0_4 g(t/2) dt\nlet t/2 =x ; so dt = 2dx; and the range becomes x:0 to 2\nso, y= ∫_0_2 g(x) 2 dx = 2  ∫_0_2 g(x) dx = 2 * 5 = 10\n\n(b)\ny= ∫_0_2 g(2-t) dt\nlet 2-t=x; so dt = -dx; range becomes x:2 to 0\ny= - - ∫_0_2 g(x) dx = 5
909370
Well, the equation for the time period of a pendulum is\n\nT = 2*pi*sqrt(l/g)\n\nWhere pi is the mathematical constant, 'l' is the length of the string, and 'g' is the acceleration of gravity.\n\nSo substitute your values, and you then have the equation\n\n0.80 = 2*pi*sqrt( l  /9.8 )\n\nYou could solve for the length and plug that back into the equation.\n\nT = 2*pi*sqrt( l / (.55*9.8) )\n\nSubstitute the length in for 'l' and solve for T