Question

What is the rpm of the last driven pulley, if a motor runs at 1440 revolutions /min is
connected by means of a multiple belt drive and the individual transmission ratios are
3:2, 4:1?
​

Answer 1

Answer:

3:2

Explanation:

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Answer 2

Answer: 8640 rpm

Explanation:

The motor runs at 1440 revolutions/min

We know, that the transmission ratio is the ratio of the angular velocities or angular frequencies of the driving component to that of the driven component.

So, if we consider that the revolution per minute of the motor to be “f1”, “f2” & “f3” of pulley 1 & pulley 2 respectively.

Then, we can say

f2/f1 = 3/2 …… (i)

&  

f3/f2 = 4/1 ….. (ii)

Taking eq. (i), we get

f2/f1 = 3/2

⇒ f2/1440 = 3/2

⇒ f2 = 3 * 1440 / 2 = 2160 rpm

Now, taking eq. (ii), we get

f3/f2 = 4/1

⇒ f3/2160 = 4/1

⇒ f3 = 2160 * 4 = 8640 rpm

Thus, the rpm of the last driven pulley is 8640 rpm.