- #1

Amaroq Zev

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Problem: old-fashioned trucks used a chain to transmit power from the engine to the wheels. Suppose that the drive sprocket had a diameter of 6 in and the wheel procket had a diameter of 2o in. If the drive sprocket goes 300 rpm:

A. Find the angular velocity of the drive sprocket in radians per minute(I have concluded that the answer is..uh..100 rpm but I am not certain. I divided the rpm by 3 inches, the driving sprocket's radius.)

B. find the linear velocity of the 20-in wheel sprocket in inches per minute.(18849.566in/m 300 rev/m*2(pie) rad/1 rev*10in/1 rad

C.)Find the angular velocity of the wheel in radians per minute(30 rpm) (got that by dividing the 300 rpm by the wheel sprocket's radius, since the wheel and sprocket share the same axel, the angular velocity is the same for the two of them, so 300 rpm divided by 10 in)

D.(this is my question) Find the speed of the truck to the nearest mile per hour.(I started setting this up but I don't know if I did it right so far, and I also forgotten how many ft ar in a mile so I can't continue really...anyway here is what I got so far. 30 revolutions/1 minute*60minutes/1 hr*2(pie)rad/1 revolution*19inches/1 radian*1mile/? from there I would have put the feet in there then 1 ft over 12 inches...or...I would have figured out how many inches are in a mile seperately and put the answer in the corresponding position in the equation...have I done this right so far? The picture has a small circle(drive sprocket)connected to another circle(wheel sprocket) by a chain, along the outside of the wheel sprocket is a axel sharing circle(wheel) of 38" diameter, if anyone knows if I am doing this right please tell meh...pwease?

alright I got it, now to continue the equation I still need to know if I am setting this up right or not.

30 revolutions/1 minute*60minutes/1 hr*2(pie)rad/1 revolution*19inches/1 radian*1mile/5280 feet*1 ft/1in

okay so then that would make...40.69790483 which rounds off to 40.698 mph...is this correct?