[math]8.7 kg * 1864 m * 9.8067 m/s^2 = 160 kJ[/math] with significant figures
[math]8.7 kg * 1864 m * 9.8067 m/s^2 = 160 kJ[/math] with significant figures
[math]8.7 kg * 1864 m * 9.8067 m/s^2 = 160 kJ[/math] with significant figures
160000 J
Is the drawing weird, or is this an eccentric shaft? Assuming it's not an eccentric (though I don't think it will make a difference).Since there haven't been many wheel & axle problems:
What is the distance traveled by the green point for every rotation of the 12 spur gear? (Round to hundreths)
Okay so 12/20*4π=7.54 The wheel and axle don't matter. One rotation on the input of the axle is one rotation on the output of the axle
My reasoning was that the size of the circle made by the green point is equal to the size of the circle made by the center of the axle; since the axle is 0.65 meters offset from the center of the 20 spur gear, the circumference of that would be 1.3π, multiplied by 60% is 2.45... I think I just answered my own question. Anyway, I'll just give you the next turn.Can you explain your reasoning? The way I see it, the small gear makes the big gear rotate 60% of a rotation, which is directly connected to the wheel, and therefore must also go through 60% of a rotation. Thus it's just 60% of it's circumference of 4πm, which is 7.34m
If the non-spur wheel isn't off-center, then how is it eccentric? The green dot should have the same trajectory as the axle.Yeah, that is an eccentric path. Which was the other path, but that dot would probably describe some awful combination of the two rotations that I doubt either of us could imagine or derive in our head if it is the eccentric.
How hard do you want the question to be?
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