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Wheel size and speed help

  1. Jun 16, 2003 #1
    Wheel size and speed help!!!

    Okay, my father and I have been an argument for days on this. We were arguing about wheel sizes and their effected distance traveled. A larger wheel has a greater circumference; therefore one revolution will travel a farther distance. Here's the argument. If you have two wheels of different sizes, one small and one large, and rotate them with the exact same force on an axle- Will the larger wheel travel a farther distance in a given time because it can cover more distance, or because it actually rotates faster at the point of contact on the ground. My argument is the smaller wheel and the larger wheel can actually be "spinning" (rotational speed) at the exact same speed, but because a larger wheel will cover a greater distance on one complete turn, will move faster along a plain. I understand that if they spin the same speed, the smaller wheel will have made a complete turn before the larger wheel. It is my belief that even though this is true, the larger wheel has still made enough distance (even though it has not yet made a complete revolution) to eventually cover more ground; thus, creating more distance; thus having a faster linear speed. So I guess after all that mumbo jumbo, I'm trying to ask do two wheels of different sizes travel at different linear speeds with the same rotational speed at the axle because of size, or because of the speed of the two wheels (at the point of contact on the ground) actually changes (rule out all variables: surface friction, wind, etc.)? I hope someone can make sense of all of this and put a stop to our bickering :).

    P.S. Also, If I took two wheels, again of different sizes- one large, one small, and started them on top of a slope. If I let go of both wheels at the same time (these wheels are the same weight) would the larger wheel reach the end before the smaller wheel? Or would they reach at the same time? Also, let's say that the slope was long enough for each wheel to reach maximum speed. Please explain the answer to this.
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  3. Jun 16, 2003 #2


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    I'll take a shot. I believe that as you increase the diameter of the wheel you increase the inertia (Inertia = 1/2 mass * radius ^2). Because you stated that both wheels recieve the same amount of force I believe that the smaller wheel will have a greater acceleration than the larger wheel since the smaller wheel will have the least amount of inertia ('laziness') and thus require the least amount of force to get it going. The larger wheel will travel the greater distance and that is the trade off when you increase or decrease size you trade distance for force. I know for sure that the large wheel does travel a longer distance per revolution than the smaller wheel but the 2 wheels are not travelling at an equal revolution/unit of time.

    To your 2nd question I believe that the smaller wheel will beat the larger b/c you stated that the wheels were the same weight (mass when gravity is applied) and again I = mr^2, if the mass remains constant and the diameter is increased the inertia and amount of force needed to move the larger wheel increases.

    I'd wait for somebody else to confirm or correct me before you do anything I'm just Microbiologist studying physics for the MCAT and reading these boards is one of my favorite ways of studying. I think the bigger question is why you're trying to show up your dad so close to father's day. :wink:
  4. Jun 16, 2003 #3
    It's the greatest gift I could have gotten him. He loves to argue :).
  5. Jun 16, 2003 #4


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    Honestly I'm not sure I completely understand what you're asking, even after having read the question several times. Perhaps you should spend a little more time formulating your questions so they are easily understandable.

    What I understood is that you want to put two wheels on either end of an axle; one wheel is smaller than the other. You want to know what will happen. The answer is that given the same angular velocity, the larger wheel will cover more linear distance on the ground. You can build an example out of Legos quite easily. You could also go let some air out of one of the front tires on your car (I don't recommend this, of course, it's dangerous) and discover that your car pulls in the direction in the deflated tire.

    I should also mention that the point of contact of a wheel on the ground is NOT MOVING. The wheel rim and ground have zero relative velocity in normal rolling motion. The only time the point of contact is not stationary is when the wheel is sliding.

    The second question was answered properly by E8. The wheel with the smaller rotational inertia will reach the bottom first.

    - Warren
  6. Jun 19, 2003 #5


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    There's your problem. Since you specifically said that by "spin at the same speed" you meant the same angular velocity, both wheels will make a complete turn (2 [pi] radians) in exactly the same time.
    The larger wheel, with greater circumference, will cover the greater distance. If you have two wheels of different diameter attached by a rigid axle, they will move in a circle with the larger wheel on the outside of the circle.
  7. Jun 19, 2003 #6
    I don't see what's so hard to understand. He's asking that if you have say, 2 carts side by side, one having a larger set of wheels than the other, and they each have the same forward force applied, which one will travel farther. Now we have to also clarify if these carts are exactly the same weight. So the cart with the smaller wheels would need extra weight to compensate if they are to be an exact match. In that case, I believe the equal mass would cause them to travel the same distance. The smaller wheel would have more revolutions, and go faster, but the larger wheel, while moving slower would require less revolutions to make the same distance, and thus it balances out. If we're simply talking about 2 wheels side by side, Then yes, the small wheel with less mass will travel farther.
  8. Jun 19, 2003 #7
    Thanks for all of your replys guys. I think I need to simplify my question. All of the answers have been great but kind of not hitting my point (And that is totally my fault for not simplifying). My point I was arguing is larger wheels on an axel do not spin any faster but just cover more distance. Let me make an example: When you put larger tires/wheels on your car your speedometer will lose its accuracy. When your speedometer is calculated and configured originally, it is configured with a specific size tire. A sensor reads the rotation and your speed is displayed on the speedometer. So if you put a larger tire on, one complete turn of the axel is still going to cause the tire to make a complete revolution; and that complete turn is going to give you more distance covered along the ground. Thus, your speedo will read a lower speed than you are actually traveling. Okay, with that said, let me try to simplify our dispute. The argument was being made that the outer part of a wheel is actually spinner faster than the inner part. (I know this wasn't my initial post, but I guess he's changing his argument) Take any spinning object that from where the axel is attached, has a significant distance to the outer portion of the spinning object. Let's use a helicopter blade for this example. A huge blade is connected in the center to an axel. The axel spins the blade with a supplied force. The outer most edge of the blade, has to cover way most distance than the inner part of the blade, but yet they both make a trip around at the exact same time (of course, they are all one piece). So, does this mean the outer most edge of the blade is traveling faster than parts closer to the axel? If you were to take a radar gun and clock the speed of the outer part and than clock the speed closer to the axel, would the gun actually read faster near the outer edge?
  9. Jun 19, 2003 #8


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    Uh yes, of course points on the outside edge of a rotating body move faster than points near the center.

    The velocity is called linear or tangential velocity, and it is described by:

    v = w r

    where w is the angular velocity (in radians/sec) and r is the distance from the axle.

    - Warren
  10. Jun 19, 2003 #9


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    Yes. In fact, this is one of the limits to helicopter design discovered by Igor Sikorsky. If you design a rotor blade that is a standard length and spins above a certain number of RPM's, or one that rotates at standard RPM's but is above a certain length, the tips of the blades will actually break the sound barrier. Now, if you've ever seen film footage or heard stories of aircraft accelerating to supersonic speed, you have probably heard that just before reaching Mach 1, a great deal of turbulence is experienced. This is called "transonic turbulence". A helicopter blade never breaks through this turbulence because the tips may be moving at supersonic speed, but the inner part of the blade is going much slower and, somewhere in between, some part of the blade is stuck in the transonic envelope. Even if we had the materials can make a blade that could survive such a beating for an acceptable service-life, the insane amount of turbulence this causes makes the blade very inefficient at lifting.

    The way in which this relates to the tires are your car is as follows. First, it is important to remember that all motion is relative, so in deciding wether the tire is moving faster or not, you must first decide; "moving faster" relative to what?

    The center of your tire is stationary (relative to your car). The further from the center you get, the faster the tire is going relative to the car. However, relative to the road, the tire is stationary at the bottom. So at the same RPM's, the larger tire is moving faster at the outside edge. But moving faster only relative to the axle and, by extension, the car. This is why the seepometer reads false, because it is designed to count RPM's. Since the outside of the tire is moving faster relative to the car, and is stationary relative to the road, the road is going faster relative to the car.
  11. Jun 19, 2003 #10
    Wow, thanks guys. A great deal of great info! Keep it coming if you want to add your $.02.
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