find diameter of small pulley if given both speeds and size of big pulley?
968006
The D=8" pulley is rotating at 120 rpm, so you can use its circumference (C) and its rotation speed (W) to find the length of rope (or whatever) that passes over it every minute:\n\nL = W*C = W*πD = 120*π*8 = 3015.9"\n \nNow, this must equal the length of rope passing over the smaller pulley (because its all 1 rope). So use the same equation as before, but with the smaller pulley's circumference (c) and rotation speed (w):  \n\nL = w*c = w*π*d\n\nSolve this for the diameter of the smaller pulley (d):\n\nd = L / (w*π) = (3015.9) / (150*π) = 6.4"\n\nThat's the long way.  It boils down to the (much simpler) expression:\n\nd*w = D*W\n\nFrom which you can get the smaller diameter:\n\nd = (D*W) / w = (8*120) / 150 = 6.4"\n\nBut you have to understand the first way in order to understand the easier way...
580938
All you have to do is find one that is either\na) larger than .5 or\nb) smaller than .45\n.501 > .500\nHence, your answer is .501
470650
diameter = 9 inches\n\nRadius = 4.5 inches\n\npi * r^2 = 3.14159 * 4.5^2 = 63.61725...
850826
From what I can see you forgot the moment of inertia of the pulley. \nRemember Torque = I*alpha. TensionxR = I*a/r. There will be another acceleration term from this torque.\n\nEDIT: No, wait sorry. I jut didn't see it. Ha it was right there in part i) The photo's kinda hard to read. Yeah, It's good.
69843
work = force times distance\nW=Fs
858739
Use Bernouillis equation and you will have your answer. You can use the equivalent lengths method to find the friction.
367853
F = m*a \n\nm = 2 kg (two masses; two strings)\na = gravitational constant which happens to be 9.81 m/s^2 on the surface of the earth (or 32.2 ft/s^2 in EE units)\n\nsum the forces in the y-direction:\n\nT = F = 2 * m * a = 2 N\n\n(assuming massless strings)
129926
Either to extract by pull or collision.
909202
a= r * w^2 = .12 * 2^2 = .12 *4 = .48 m/s^2
160871
Divide 34 by PI...that will give you the diameter.\n\nDiameter = Circumference / PI
506291
For the slower machine, it would take 3 times the length of time, 8*3=24min\n\nFor the faster machine, it would take 1.5 times, 8*1.5=12min
870849
Depending on the size 3-5 mph
861607
The frequency, f, of a vibrating string of m kg/m and subject to a tension of T Newtons is given by the relation\n\nf=sqrt(T/m)/2/L\n\nHence the frequency is inversely proportional to the length of the string, assuming the tension remains unchanged.\n\nThus the required length of the string is\n40*474/624=30.385 cm.
769928
T = 2pisqrt(l/g)\nl = g(T/(2pi))^2 = 9.80(1.0/(2pi))^2 = 0.25 m\n\n(The object's weight is irrelevant for small oscillations.)
970313
r=radius\nd=diameter\n\n1 C=2r(pi)\n2 C=d(pi)\n\n(Same thing because 2r=d)
970512
According to energy conservation law: m2*g*h=m1*g*h+m1*v^2/2+m2*v^2/2+I*w^2/2.\nm2*g*h is initial potential energy of the system;\nm1*g*h is final potential energy of m1;\nm1*v^2/2 is kinetic energy of m1;\nm2*v^2/2 is kinetic energy of m2;\nI*w^2/2 is kinetic energy of pulley with inertia moment of I & final rotary speed w rad/s^2.\nNow m1=0.46kg, m2=0.5kg, h=0.75m, r=0.05m, t=5s, g=9.8m/s^2.\nAs h=a*t^2/2, then a=2*h/t^2 – linear acceleration of m1 and m2. \nAs v=a*t, then v=2*h/t – linear speed of m1 and m2.\nAngular acceleration of pulley w’=a/r; angular speed w=v/r.\nThus 2*(m2-m1)*g*h=(m1+m2)*v^2+I*(v/r)^2, so I=(extract it please and plug values in).\nT1=m1*(g+a), T2=m2*(g-a). Don’t worry, be happy!
964672
The answer is  - (0.635 - 0.324)* r * g = 2.9 Nm.\n\n\nYour equation for torque and other things are correct.\n\nFor the system to be in equilibrium the NET torque must be zero.\n\nSuppose that 0.635 kg mass rotates the pulley in clockwise direction.\n\nThen the other mass 0.324kg will rotate the pulley in the anti-clockwise direction.\n\nTorque produced by 0.635 kg about the axis is F x r = 0.635*r*g. (Let us take clockwise rotation  as positive)\n\nTorque produced by 0.324 kg about the axis is F x r = - 0.324*r*g. (anticlockwise direction)\n\nThe net torque about the axis is (0.635 - 0.324)* r * g = 0.311*r*g. \n\nIf there is no friction, the pulley will rotate in the clockwise direction by the action of this torque\n\nIf the frictional torque acts opposite to this torque, then the net torque will be zero and the system will be in equilibrium.\n\nTherefore the frictional torque is - 0.311*r*g. (anticlockwise direction)\n\n\nSubstitute the value for r = 0.95 m and g = 9.8m/s^2\n\nThe frictional torque is 2.9 Nm.