Pedal Angular Velocity: Pi/2*f or 2pi*f?

AI Thread Summary
The discussion centers on determining the angular velocity of a pedal that sweeps through pi/2 in one complete cycle, with a known frequency. The pedal's movement is linked to a mechanism, specifically a four-bar linkage, where the upper flywheel rotates through 2pi while the pedal does not. The confusion arises from whether the angular velocity should be calculated as (pi/2)*f or 2pi*f, with some participants asserting the latter despite the pedal's limited sweep. Clarification indicates that the angular velocity should be based on the angular distance traveled (pi/2) divided by the time taken for that distance, considering the pedal's unique motion. Ultimately, the correct interpretation hinges on understanding the relationship between frequency and the specific angular displacement of the pedal.
max3
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So I've got a pedal which sweeps through pi/2 in 1 complete cycle (frequency is known). Is its angular velocity (pi/2)*f or 2pi*f just the same? (considering that it doesn't sweep through 2pi but pi/2)
 
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max3 said:
So I've got a pedal which sweeps through pi/2 in 1 complete cycle (frequency is known). Is its angular velocity (pi/2)*f or 2pi*f just the same? (considering that it doesn't sweep through 2pi but pi/2)

Welcome to the PF.

How can one complete cycle be PI/2? What is the geometry involved?
 
well the mechanism itself is the lower half of a pedal operated sewing machine. you may think of it as a four-bar linkage or a 'double crank'. while the upper flywheel rotates through 2pi, the pedal (connected by a rod) does not. I am given the frequency of operation of the pedal and need to know the pedal's angular velocity at 1 instant (assuming angular acceleration = 0).
 
max3 said:
well the mechanism itself is the lower half of a pedal operated sewing machine. you may think of it as a four-bar linkage or a 'double crank'. while the upper flywheel rotates through 2pi, the pedal (connected by a rod) does not. I am given the frequency of operation of the pedal and need to know the pedal's angular velocity at 1 instant (assuming angular acceleration = 0).

Ah, so the pedal rotates through PI/2 as the flywheel rotates through PI. So the angular velocity of the pedal is angular distance traveled divided by the time it takes to travel through that angular distance, right? What does that give you for an answer?
 
That's what I figured but many have contradicted me in saying that the angular velocity is 2PI*f irrespective of the pedal NOT sweeping through 2PI. Many thanks for your feedback.
 
max3 said:
That's what I figured but many have contradicted me in saying that the angular velocity is 2PI*f irrespective of the pedal NOT sweeping through 2PI. Many thanks for your feedback.

Be a little careful with the "f" part, though.

"f" refers to a frequency, or number of somethings per second. For circular motion, it would normally refer to the number of rotations per second. If one cycle is the pedal going down and coming back up, then there are two half periods per cycle. The angular velocity is the angular distance divided by the time taken, so the angular velocity of the pedal is the PI/2 traveled in one half of the period of the system (not the whole period).

Does that make sense?
 
berkeman said:
The angular velocity is the angular distance divided by the time taken, so the angular velocity of the pedal is the PI/2 traveled in one half of the period of the system (not the whole period).

Does that make sense?

Yes, only that a half period would correspond to PI/4 and not PI/2 as you state since in my case PI/2 is swept in 1 full cycle. Many thanks.
 
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