how to caluate motor belt speed?
607030
INTRODUCTION TO MOTOR, PULLEY AND BELT\nA motor is a prime mover means it provides power to mechanical system or machines. A motor converts electrical energy into mechanical energy which is transmitted by belt drive to perform some mechanical work. The armature (extension of the rotating part) of a motor is coupled with a shaft by a coupling (coupling is a device which connects two shafts so that if one is rotating, the other also rotates). From this shaft power is transmitted to a distant shaft by belt drive. Separate pulleys are present and keyed (means fixed with shaft so that both shaft and pulley rotate as a whole) to each of the driver and driven shaft and joined by a belt in such a way that if one pulley rotates the other also rotates.\n\nDRIVER AND DRIVEN PULLEY SPEEDS\nLet the driver pulley rotate at 'N1' rpm (revolution per minute) and the driven pulley rotate at 'N2' rpm. If the diameters of the respective pulleys be 'D1' metre and 'D2' metre respectively then we have the relation N1/N2 = D2/D1.\n\nBELT SPEED CALCULATION\nNow, we are in a position to start the belt speed calculation. The surface speed of the driver pulley is given by V1 = pi*D1*N1 metre/minute = pi*D1*N1/60 metre/second. In ideal condition i.e. if no slippage was present then the belt speed would have been equal to the surface speed of the driver pulley i.e. equal to 'V1'. But actually there is aways some slippage between the belt and pulley surface so the speed of the belt is always less than 'V1'. If this slip factor is 's' then the belt speed is (1-s)*V1 \n= pi*D1*N1*(1-s)/60 metre/second.
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Hand-held VSR drills use a geared motor to vary the speed. A drill press with a belt system uses a single speed AC motor and a system of different sized pulleys to vary the speed. If you want to vary the speed without moving the belt, you'll have to do it electronically.\n\nThere are speed controllers for AC motors on the market. Here's a link to a footswitch type.
263422
Distance=rate*time\n\nJust plug in what you know and solve for what you're missing.
677244
One really needs the manufacturers part number to get that information. If the dataplate is gone, you can ohm out the windings, and through the process of elimination determine the  proper connections. You might want to check out the stepper motor seciton at CNC zone.\nhttp://www.cnczone.com/forums/forumdisplay.php?f=193
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Take a rope ( that doesn't elongate/stretch easily), wind round both pulleys (just the way belt goes)and mark one end where it touches the other. Measure this and the size in inches is the size of the belt. As to whether 'A' or 'B' or 'C', it is based on cross-section. The smallest is 'A'. I hope I am understood.
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1st link \n\nb) F = 40 N\n\nc) F = 60 N\n\n2nd link\n\nb) linear velocity = 0.59 m/s\n\nc) Taking g = 9.8 m/s^2\n\nv = 0.83 m/s\n\nd) It would not effect the answer as the expression for v which is sqrt(mu * r * g) is independent of the mass. \n\n3rd link\n\na) Rotate the setup, note the reading on the balance. Find the time required for 1 circuit.\n\nb) No work is done\n\nc) T = 5.48 N\n\nd) 5.9 %\n\ne) If the cord is exactly horizontal, it won't touch the rod and hence cannot be rotated.
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1.\nConvert 80 mph to feet per second:  80 * 5280 / 3600 = 117.333\n\na = (Vf - Vo) / t  = (117.333 - 0) / 6 = 19.556 ft /sec^2\n\n2.  \n\nConvert 70 mph to feet per second: 70 * 5280 / 3600 = 102.667\n\nVf^2 = V0^2 - 2*a*d\n\n0 = 102.667^2 - 2*a*157\n\n10540.44 = 314 a\n\na = 33.57 ft / sec^2\n\n\n3.  a = (Vf - Vo) / t = 80 / 5 = 16 ft/sec^2\n\nV = 80 - 16t\n\nIntegrate that equation from 0 to 5\n\n80t - 8t^2 = 80 * 5 - 8*25 = 200 ft
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Top Speed 330.0 kph / 205.1 mph \n0 - 62 mph 3.8 seconds
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False.  m/s (meters per second?) is not a measure of rotary speed.  Until you get that straightened out, no answer can be calculated for a) or b)\n\nBefore posting a ?, do a search here.  This same ? came up yesterday.........!
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first one is easy\n\nfirst find the normal force\n\nFf (friction) = Fn(normal force ) (.35)\n0(motionless ) = Fn .35\nFn = -.35\n\nfrom there its easy\n\nFn = mgsin(theta)\n\nsolve for theta\n\n-.35 = (200)(9.8) sin(theta)\n\nand solve it ur self\n\nsecond one, tell us wat u want us to find and we can tell u, but here are the things u can find, ill give u the formulas\n\nacceleration = 0 (constant speed)\ntheta = 15 degrees\nvelocity = constant\n\nthats pretty much it, from what u gave me
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The answer is 147.12385cm . Assumption PI = 3.14159\nUnless a picture is drawn it could be difficult to explain.\nThe belt would be like tangents to the circle, so the Free Length(length of belt not circumscibe the circle) should be 50cm. And only half of one circle would be circumscribe by the belt, so the length can be deduced by\ntotal length of belt= 2* Free Length + 2* Circumscribe Length\nFree Length=50cm\nCircumscribe Length = PI*Dia/2 = 3.14159 * 7.5 = 23.561925cm\nSo the total length would be 2*50+2*23.561925=147.12385cm
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acceleration=rate of change of velocity\n=80*5280/3600*6ft/sec^2\n=19.6 ft/sec^2\n\n2.0=(70*5280/3600)^2-2f(157)\n314s=67.1ft/s^2
828583
We make small display rotators if you check out our web site you can get an idea how we build them. You probably just need a slower gear motor. You can change the pulleys for different speeds but then you need to change the belt size. It is always easier to get a motor close to the final speed you want.\n I don't know what the cabinet looks like or what kind of space you have for the motor but it may help to see how we build them. www.semcomotion.com
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Ok.  From the hint start by figuring out the ratio of wheel speeds\nCircumference of A =2*pi*10\nCircumference of C = 2*pi*25\nso C/A=2.5  ie A must go around 2.5 times to make C go around 1 time.\nThere are 2*pi radians to 1 revolution so 100 rev/min =200*pi rad/min.  A must go 2.5 times as fast as this or 2.5*200*pi=500 rad/min\nAccelleration is the derivative of velocity which means that we must integrate the angular acceleration to obtain the speed at any given point of time.\nIntegrating 1.6rad/s^2 with respect to time and setting velocity equal to zero at time equals zero we obtain angular velocity= 1.6*time elapsed (rad/sec).  Note that this speed is in "per seconds" and we need it to be in "per minutes" so we must multiply by 60 for the units conversion.\nThis gives us the angular speed in rad/min for any chosen passage of time.  We shall set this equal to the goal speed of 500*pi rad/min as mentioned above. So\n1.6*time elapsed*60 =500*pi\nSolving for time elapsed we obtain a time of 1.63 minutes.
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If initial speed is V0 and the speed after a time t second is V,\n\nAcceleration is change in speed per unit time.\n\nAcceleration a = (V - VO) / t.\n\nThe average speed during this time t is \n\nV (average) = (V + V0) / 2.\n\nNote; if any quantity is changing uniformly, the average is found by adding the initial and final value and dividing the sum by 2.\n\nThe average speed can also be determined by dividing the total distance by total time of travel.\n \nV (average) = S / t.\n\nTherefore S/t = (V + V0) / 2.\n\nS = (V + V0)* t / 2.\n\nFrom a = (V - V0) / t we have V = V0 +at.\n \nTherefore S = {V0 +at + VO} * t /2 \n--------------- = {2 V0 t + at t } /2\n\n-------------- S = V0t + 0.5 a tt
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Well the first thing to do in a problem like this is extract information from it.\n1.) Weight of Block = 10kg\n2.) Velocity of block(Vi) = 6m/s\n3.) Final velocity of block(Vf) = 0m/s\n4.) Friction acting on block = 20N The friction is always opposite of the direction of the block.\n\nSecond:\nWhat are we calculating? Distance traveled by block! DISTANCE.  We need a formula that will give us distance as an answer! So...we have final velocity, initial velocity, and distance....lets see...how about this one...\n vf^(2) = vi^(2) + 2*a*d \n\nThis equation has Vf, Vi, and D.  \nVf = 0 m/s\nVi = 6 m/s\nD = What we are trying to find.\nThe only thing missing in this equation that we have to find is (a) or acceleration. Hmm.\n\nThere is actually acceleration on  the block.  Friction is acting on the block in the opposite direction; therefore the block is decelerating.  We can use F=ma to find the force of deceleration.\nFriction Force = 20N\nMass = 10kg\nAcceleration = ?\n\nUse the following equation:\nF=ma\n1.) -20N = 10kg * a\n2.) a = -2 m/s (Yes it's negative b/c we are decelerating!)\n\nNow we go back to the original problem!\n vf^(2) = vi^(2) + 2*a*d \n\nThus\n(0^2) = (6^2) + (2)*(-2)*(d)\n0 = 36 - 4d\n-36 = -4d.\nDistance = 9meters! (Make sure to get the units correct!)\nI think I got this right...if not tell me what I did wrong pls\n\n-----------------------------------------------------------------------------------\nThe guy who did it using Energy equations forgot to add one thing....\nmass x (velocity)^2 = Force x distance\nso 10kg * (6m/s)^2 = 20N * x\nso (10 * 36)/20 = 18 meters\n\nThis is another way to do it.  He is using the energy equations to calculate energy in the block. When the block is in motion there is Kinetic Energy in it.  When the block has no motion, there is zero Kinetic Energy in it.  However, the individual that did this messed up the equation as follows:\nKe = (1/2) * Mass * Velocity^2\n\nThus the answer is not 18 but 9 meters!!!!!\nHere is the correction:\n1.) .5*mass x (velocity)^2 = Force x distance\n2.) .5*10kg*(6m/s)^2 = 20N * x distance\n3.) (5 * 36)/20 = distance&#92;\n4.) Distance =9 meters!
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Read Newton's 3rd law of motion.\nAlso, use F=ma to figure out your acceleration.