What is the formula for calculating pulley diameters and rates of revolution?

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The discussion focuses on calculating pulley diameters and rates of revolution using specific formulas. A 220 mm diameter driving pulley rotating at 160 r/min drives a driven pulley rotating at 450 r/min. The diameter of the driven pulley is calculated using the formula \(D_2 = \frac{R_1}{R_2}D_1\), resulting in a diameter of approximately 78 mm. Additionally, the mass distribution of a bronze casting composed of copper, tin, and antimony is analyzed, confirming the mass of each component based on their ratios.

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Eabzolid
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I got a couple of questions to ask. And if you can just tell me how you would solve these like how would you setup the equation:

1. A bronze casting is made up of six parts copper, five parts tin, and one part antimony. The casting has a mass of 60KG. Find the mass of each component in Kg.

2. A 220 mm diameter pulley turning at 160 r/min drives another pulley at 450 r/min. What is the diameter of the driven pulley (to the nearest millimetre)?

Thank you
 
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Eabzolid said:
I got a couple of questions to ask. And if you can just tell me how you would solve these like how would you setup the equation:

1. A bronze casting is made up of six parts copper, five parts tin, and one part antimony. The casting has a mass of 60KG. Find the mass of each component in Kg.

Thank you

If there are six parts copper, five parts tin and one part antimony, then how many parts are there in total? What proportion of the total is each ingredient?
 
Eabzolid said:
I got a couple of questions to ask. And if you can just tell me how you would solve these like how would you setup the equation:

1. A bronze casting is made up of six parts copper, five parts tin, and one part antimony. The casting has a mass of 60KG. Find the mass of each component in Kg.

If we add all the parts given in the ratio, we see there are 6 + 5 + 1 = 12 parts. We find 60/12 = 5, so we can multiply the given ratio 6:5:1 by 5 to equivalently state the ratio of copper to tin to antimony as what?

Eabzolid said:
2. A 220 mm diameter pulley turning at 160 r/min drives another pulley at 450 r/min. What is the diameter of the driven pulley (to the nearest millimetre)?

Thank you

In one minute, the driving pulley rotates 160 times, and so (assuming no slipping between this pulley and the belt) a point on the belt will move 160 times the circumference of the driving pulley. Given that the driven pulley rotates 450 times during this same minute, and the linear relationship between diameter and circumference, then we know 160 times the diameter of the driving pulley has to be the same as 450 times the diameter $D$ of the driven pulley:

$$160\cdot220=450D$$

So, solve for $D$, what do you find?
 
I put 160x220/450=78.22. Basically 78mm

Thanks a lot.

How would you approach a question or any questionn like that ??
 
Eabzolid said:
I put 160x220/450=78.22. Basically 78mm

Thanks a lot.

How would you approach a question or any questionn like that ??

I would approach it just like I posted, but another approach would be to use the arc-length formula:

$$s=r\theta$$

In terms of the diameter $D$, this would be:

$$s=\frac{D}{2}\theta$$

Since there are $2\pi$ radians per revolution, we could then state:

$$s=\frac{220}{2}\cdot2\pi\cdot160=\frac{D}{2}\cdot2\pi\cdot450$$

This reduces to:

$$160\cdot220=450D$$

which matches my previous approach. Solving for $D$, we find:

$$D=\frac{160\cdot220}{450}\approx78$$
 
Oh wow. That formula is a bit complex, isn't there a different formula that can be used?
 
Eabzolid said:
Oh wow. That formula is a bit complex, isn't there a different formula that can be used?

Well, you could reason the the linear velocity $v$ of two given points, one on the edge of the driving pulley, and one on the edge of the driven pulley, must be the same:

$$v=r_1\omega_1=r_2\omega_2\implies r_2=\frac{\omega_1}{\omega_2}r_1$$

where $r_i$ are the radii and $\omega_i$ are the angular velocities. Since the angular velocity varies directly as the rate of revolution $R_i$, and the constant of proportionality must be the same for similar geometries, we can then state:

$$r_2=\frac{R_1}{R_2}r_1$$

And, since the diameter varies directly as the radius (again with identical constants of proportionalities), we have:

$$D_2=\frac{R_1}{R_2}D_1$$

The basic idea here is that the rate of revolution of a pulley varies inversely as its diameter (for some fixed belt speed):

$$R=\frac{k}{D}$$

Since the constant of proportionality $k$ will be the same for both pulleys (similar shapes), then we may state:

$$k=R_1D_1=R_2D_2\implies D_2=\frac{R_1}{R_2}D_1$$
 

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