Power Transmission: Input Pulley Speed 2

[ElMicha]

Hey all, might be a dumb question but is the input pulley's angular speed different than the motor shaft driving it? It wouldn't be right, the radial velocity would be different but not the angular velocity right? In the Figure below the driving pulley is driven by a motor.

 

[desertfox]

Hi EIMicha

Yes the linear velocity at the radius end of each pulley would be different and the angular velocity would be the same.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[dvd]

What is the "input" pulley? The "driving" pulley as shown on left, or the "driven" pulley at right?

Motor and driving pulley have same angular velocity. Driven pulley has slower angular velocity. Linear speed of belt is constant, and larger diameter of driven pulley results in angular speed reduction.

 

[desertfox]

Hi Elmicha

Sorry I think I have mislead you and I am incorrect see this link

http://www.freestudy.co.uk/dynamics/pulleys.pdf

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[Strong]

Each pulley has a different diameter. The tangential speed (belt speed) is the same on each pulley. Rotational and angular speed are a direct ratio of the pulley diameters. I did not look at the "freestudy" reference.

Walt

 

[ElMicha]

Hey, finally have some time to spend on this project again. Thanks for the feedback. The freestudy helped a lot, I was struggling to understand how the driven pulley has the same power as the driving pulley. From what I am understanding now it's due to the tension on the driven pulley being greater, larger diameter and as Strong said the rotational and angular speed are a ratio of the diameter. The diameter is larger and will rotate slower but induce a greater tension in the belt on the driven pulley and in turn a larger torque, which will give an output power equivalent to the input power on the driving pulley. Is my understanding off?

 

[Strong]

Power (horsepower) at the driven shaft is the same as the input (drive) shaft minus belt slip and friction losses. Torque at the ourtput (driven) shaft is higher than the drive shaft if the sheave diameter is larger or torque at driven shaft is proportional to ratio of driven diameter to drive diameter (output/input).

Walt

 

[ElMicha]

Okay thanks Walt, this is making more sense now.

 

[Strong]

"The diameter is larger and will rotate slower [highlight #FCE94F]but induce a greater tension in the belt on the driven pulley[/highlight] and in turn a larger torque, which will give an output power equivalent to the input power on the driving pulley. Is my understanding off?"

Perhaps an understandikng of Tension would be helpful. Tension is an in-line Force in the belt that tries to stretch the belt and make it longer, but the belt strength resists most of the elongation. At rest the belt is placed under tension by moving the sheaves apart until the required tension is achieved. Both belt spans/strands have the same tension, assuming the sheaves/shafts are free to rotate. The required tension specified from machine or belt OEM can be measured by the elongation/stretch method, force spring-deflection method, or sonic/vibration natural frequency method. When the machine is operating one belt span has increased tension created by the drive sheave, and the other belt span has less tension. The tension Force in each span, while different, is the same at each end or at each sheave. Shaft Torque equals belt tension Force times radius of each sheave. A larger sheave diameter (therefore radius) will have greater Torque than a smaller sheave, irrespctive of which sheave is the driver or driven one. I am sure there are many references and youtube videos that can offer a better explanation.

Walt