Crossed belt connecting 20-cm and 40-cm pulleys

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A crossed belt connects a 20-cm pulley (10-cm radius) on an electric motor with a 40-cm pulley (20-cm radius) on a saw arbor. The electric motor runs at 1700 revolutions per minute.

http://s3.amazonaws.com/answer-board-image/09b77fee-f30b-4564-8c92-48e462481f83.png

(a) Determine the number of revolutions per minute of the saw.
(b) How does crossing the belt affect the saw in relation to the motor?
(c) Let L be the total length of the belt. Write L as a function of ?, where ? is measured in radians.

Homework Equations





The Attempt at a Solution



a) 3400 revs/min
b) it increases the angular velocity of the saw?
c)
I used the formula for circumference and arc length to find the length of the belt around both pulleys, but I don't know how to get the length of the belt that is not around the pulleys in terms of phi.

I don't know if I'm asking for too much but please, don't give the answer, just a hint.

Thank You
 
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Discussion

a) No I believe it would be half that, not double. Say the motor rotates once, with a radius r this means the pulley has moved a distance of [itex]2\pi r[/itex] since the circumference is just that. The saw has a circumference of [itex]2\pi (2r)=2(2\pi r)[/itex] so it will only rotate half a revolution since its circumference is double that of the motor.

b) I don't know very much about mechanics so I can't help you there sorry.

c) Let's say the length of the belt from one pulley to the other is x, and we cut this length up into two parts, x1 and x2. The cut is made where the belts cross.

[tex]x=x_1+x_2[/tex]

Now looking at the belt closest to the motor, let this pulley have a radius r. Now we already know (and it shows it on the diagram) that the belt creates a tangent to the pulley, which is perpendicular to the line running through the centre and through the point where the belt leaves the pulley. So we are dealing with a right triangle, where one angle is [itex]\phi[/itex].

Using trig, [tex]tan\phi=\frac{r}{x_1}[/tex] so [tex]x_1=r cot\phi[/tex]

Now do the same for the saw, and then add those lengths together to get the entire length of party of the belt from one pulley to the other. And because of symmetry, the other part of the belt crossing is the same length.
 
For part b, look at the directions of the motor and saw. Then, look at the directions if the belt were not crossed.
 
How would you find the length of the belt around the pulley?
 
cedar2 said:
How would you find the length of the belt around the pulley?

Notice that the point where the belt leaves the pulley, it create a right angle to the centre of the pulley (a line tangent to a circle is perpendicular to its radius at point of intersection) and since you know that one of the angles is [itex]\phi[/itex], you'll realize that you're dealing with a right-triangle. You can do this for every part of the belt not touching the pulleys.

As for the part of the belt touching the pulleys, you should be able to find the angle subtended by the two radii in each pulley, so then you should be able to find the length in terms of that.

Also, please make a new thread instead of bringing this 2 year old thread back to life.
 
The speed of the saw depends only upon the speed of the motor and the ratio of the radii of the two pulleys. The fact the belt is crossed changes the direction of the turn- if the motor pulley is turning clockwise the saw pulley is turning counter-clockwise and vice-versa.