Pulley Question
in [Math]
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From: John Albert on 2 Jul 2010 07:00 Hello, Could you please help me in solving this problem: "Two pulleys of diameters 6 m and 3 m are connected by a belt. The larger pulley rotates 44 times per min. Find the angular speed of the smaller pulley." I know that I'm supposed to use s = radius*theta. Could you please help me solve this step by step? Thanks so much. I appreciate your response.
From: cwldoc on 2 Jul 2010 07:47 > Hello, > > Could you please help me in solving this problem: > > "Two pulleys of diameters 6 m and 3 m are connected > by a belt. The larger pulley rotates 44 times per > min. Find > the angular speed of the smaller pulley." > > I know that I'm supposed to use s = radius*theta. > Could you please help me solve this step by step? > > Thanks so much. I appreciate your response. For a fixed speed of the belt, the angular velocity is inversely proportional to the diameter of the pulley; so the smaller pulley must rotate twice as fast as the larger one. Or if you want to more of a "step by step" explanation: Note that s = radius*theta implies v = radius*w, where v is speed of the belt and w is the angular velocity of the pulley For the larger pulley, v = (r2)(w2) For the smaller pulley, v = (r1)(w1) Solving for w1 gives w1 = (r2/r1)(w2) = [(3m)/(1.5m)](44 rotations/min) = 88 rotations/min = (88 rotations/min)(2 pi radians/rotation) = (176 pi) radians/min
From: Ray Vickson on 2 Jul 2010 16:00 On Jul 2, 8:00 am, John Albert <bflo_p...(a)yahoo.com> wrote: > Hello, > > Could you please help me in solving this problem: > > "Two pulleys of diameters 6 m and 3 m are connected by a belt. The larger pulley rotates 44 times per min. Find > the angular speed of the smaller pulley." > > I know that I'm supposed to use s = radius*theta. Could you please help me solve this step by step? > > Thanks so much. I appreciate your response. How many meters of belt move around the large pulley in 1/44 min? For that amount of belt, what is the angle (radians) through which the small pulley rotates? So, you know the angle in radians per 1/44 min, hence can find the angular speed in radians/min. R.G. Vickson
From: George Jefferson on 2 Jul 2010 18:24 "John Albert" <bflo_prof(a)yahoo.com> wrote in message news:1590469722.60663.1278082839649.JavaMail.root(a)gallium.mathforum.org... > Hello, > > Could you please help me in solving this problem: > > "Two pulleys of diameters 6 m and 3 m are connected by a belt. The larger > pulley rotates 44 times per min. Find > the angular speed of the smaller pulley." > > I know that I'm supposed to use s = radius*theta. Could you please help > me solve this step by step? > > Thanks so much. I appreciate your response. Visualize a very very large pully and a very very small one. When the large pully rotates such that the belt has moved a distance of L then the small pull will have also had to move a distance of L but will have to rotate faster to keep up(think of an adult walking and a kid walking trying to keep up). If this did not happen the belt would break or slip. You can visualize this as putting a mark on the belt and starting it at some point then rotating the large pully and measuring the distance the mark traveled(on the path of the belt) from the start point. This obviously has to be the same for any other mark because you can have one part of the belt going faster than another part(again, because the belt would break). now, suppose we go one full revolution in a min. The large belt then traveled 2*pi*r1 in that time. If instead it is traveling at S1 RPM's then the it has traveled 2*pi*r1*S1. For example, if the radius is 1 and the RPM's is 1 then it it travels 2*pi. If we double the speed then it's 2*pi*1*2 = 2*2*pi because we doubled the speed we went twice as far in the same amount of time. Now the other belt has a speed too and has to travel the same distance(because of the same argument above) so 2*pi*r1*S1 = 2*pi*r2*S2. If S2 is unknown then S2 = r1/r2*S1 But this is really saying that the total distance the large pully travels in a certain among of time has to be EXACTLY the same distance the small pully travels in the same time. If they are not the same then we know the belt must have broken(since they are connected). If r1 = r2 then S2 = S1 which should be obvious(the speeds are the same if the radii are the same). BUT again, the thing you should understand is that we are simply getting the total distance that the "mark" on the belt travels and realize that it has to be the same for both pullies. When the large pully rotates ONE time the belt must travel 2*pi*6m = 37.7 m In one minute it will travel 37.7m * 44 = 1659m BUT the small one will travel 2*pi*3m = 18.9m in one revolution BUT how many revolutions does it have to go in 1 min to keep up with big pull? We have to have 18.9m * X = 1659 X = 1659/18.9 = 88 rpm. Note because the large pully is twice as big it goes half the speed of the small one. i.e., S = 6/3*44 = 2*44 = 88
From: Gc on 4 Jul 2010 07:03
On 2 heinä, 18:00, John Albert <bflo_p...(a)yahoo.com> wrote: > Hello, > > Could you please help me in solving this problem: > > "Two pulleys of diameters 6 m and 3 m are connected by a belt. The larger pulley rotates 44 times per min. Find > the angular speed of the smaller pulley." > > I know that I'm supposed to use s = radius*theta. Could you please help me solve this step by step? > > Thanks so much. I appreciate your response. Notice that the speed of the belt is radius*angular velocity. Then make an equation. |