Need to know if I set and got this problem right

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The radius of the wheel is 10 in, so the circumference of the wheel is 2*pi*10 in. This is how many inches the truck moves in one minute. So, you need to get miles per hour, not inches per minute.In summary, the old-fashioned trucks used a chain to transmit power from the engine to the wheels. By calculating the angular velocity of the drive sprocket (which was found to be 300 rpm), the linear velocity of the 20-inch wheel sprocket was determined to be 18849.566 inches per minute. The angular velocity of the wheel was also found to be 300 rpm, as the two are connected by the same axle. To find the speed of the truck in
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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?
 
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After receiving an affirmative or correction I have yet another question.
 
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is it that my question is so inferior to the other questions that this one gets viewed over? Or am I not placing it in the right spot? I am new and I would like to know...that's all...
 
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I think part of the reason is that, as a huge block of text, it isn't formatted attractively. The other part may be that it really belongs in the Introductory Physics forum.

A. Make sure you understand what rpm means. It means revolutions per minute. Angular velocity is basically how fast the object spins, so it has nothing to do with the size of the object (e.g. nothing to do with the radius of the drive sprocket): a large wheel turning once a minute is turning just as fast as a small wheel turning once a minute.

So we know the angular velocity is 300 revolutions per minute. However, The question asks you to find the answer in radians per minute. Radians is not the same as revolutions. One revolution equals 360 degrees = ? radians. 300 revolutions/min = ? radians/min.

B. I'm not sure how you got all of your numbers, so I'll run through it completely. Because the wheel and drive are on the same axle, they spin at the same speed. Hence, their angular velocity is the same. The linear velocity of the wheel is how far it moves every minute. This is connected to the angular velocity. For every time the wheel turns once, the entire circumference is traced over the ground, so the distance moved by the wheel (also the truck) is the circumference of the wheel.

The wheel has an angular velocity the same as you got in part A. It is in radians/min. How many inches does the wheel move per revolution (2pi radians)? How can you convert from radians/min to inches/min?

C. You had the right idea by saying their angular velocities are the same, but then you found different values for the drive and the wheel, because you divided by their radii. Remember that size has nothing to do with angular velocity. You found the answer in part A already.

D. There are 5280 feet in a mile. As far as I can tell, the truck moves only as fast as the wheel does, since all the wheels move at the same speed. You can find out how fast the wheel moves in the same method as part B.
 

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