PHYS140 FAQ

How does changing gears make a bicycle easier or harder to ride?

2006-02-03T20:00:05Z

Q. Changing gears on a bicycle makes it easier or harder to pedal. Changing the front gears to bigger gears will make the bicycle harder to ride, while changing the back gears to bigger gears makes it easier to ride. Why are they different?

]]> A. If we first look at the front gears, we notice that they all turn one full rotation when we push the pedals through one rotation. The difference comes in their size. The big gear has a larger circumference than the smaller gears and so pulls more length of chain for every full rotation than the smaller ones. Since the chain is, to a good approximation, under uniform tension, the length of chain pulled corresponds to an amount of work done. Thus the larger gear must be doing more work per rotation of the pedals, making it harder to turn.

While the front gears get harder with increasing size, the back gears perform the opposite, making it easier to turn the pedals with the larger gears. The way to understand this is that the front gears are pulling the chain, while the chain is pulling the back gears. We can understand this in just the way we understood the front gears:

The back gears turn one full rotation when the wheel turns one full rotation, not when the pedals turn one rotation. If we imagine that the wheel takes a set amount of work to turn it one full rotation, then the smaller gears are doing the same amount of work as the larger gears, but are doing so with a lesser amount of chain. This means that the chain must be pulling with a greater force on the smaller gears, meaning that we have to push harder on the pedals.

If this still seems confusing, try to separate the front and back gears into two separate systems. The front gears interact with the pedals and the chain, while the back gears interact with the chain and the wheel.

Drinking Brid

2005-11-01T20:16:22Z

Q. There are many many explanation of the drinking bird on the web - just google "Drinking Bird" and you'll find more information than you could ever want! Here I'll write only a brief description of what causes the bird to drink.

]]> A. First, the bird, with his wet beak, sways in the air. As the water evaporates it cools the air in the bird's neck. Cooler air has less thermal energy and so the air molecules move around less. As we saw when we dipped a balloon in liquid nitrogen, cooled air exerts less pressure (the cool molecules move slower and are no longer able to push out on the balloon and keep it inflated). So, the air in the neck exerts less pressure on the alcohol (the red liquid inside the bird), allowing the alcohol to rise. The rising alcohol moves the center of mass of the bird up. Eventually the center of mass rises above the pivot point and the bird is no longer statically stable. The swaying of the bird moves the center of mass away from the pivot (the base of support of the bird) and the bird falls over. This allows the bird to tip back into the water, which allows the process to continue, and also "resets" the bird, equalizing the pressure and returning the liquid to the bulb at the bottom.

But isn't this perpetual motion?

No! Perpetual motion would mean that this could go on forever without ever adding anything to the system. It may not look like we're adding anything to the system, but in fact by having the bird in an open room we are constantly supplying dry air to evaporate the water from the bird's beak. If we enclosed the bird it would eventually stop when the air couldn't dry the bird's beak anymore.

Lamp and spinning fan

2005-10-25T21:52:23Z

Q. In DR we saw a little fan in a glass bulb. When we turned on a light shining on the fan it began to spin. Why?

]]> A. Because the light from the lamp is not in direct contact with the fan (there is no conduction) and there is no air that moves from the light to the fan since the fan is enclosed (there is no convection) the only method of heat transfer from the lamp to the fan is radiation. The radiation from the lamp creates some radiation pressure on the blades of the fan, but this effect is very small and negligible. More notably, the radiation heats up the blades of the fan. The black side of the blades is a better absorber than the white side. Because the black side is absorbing more heat it is also radiating more heat. Thus, the air near the black side of the fan is heated up more than the air near the white side of the fan. Hot air means that the molecules have more energy and bounce around more. This random motion means that the hot air has higher pressure; it wants to expand. The high pressure region near the black side of the fan blade pushes on the blade. This causes the fan to spin.

Heated metal plate

2005-10-20T05:42:27Z

Q. What happens to a metal plate when it's heated? What about if the plate has a hole in the center?

]]> A. When we heat up a metal plate it expands. But what happens to the size of the hole in the center?
Let's start by thinking about a metal plate with no hole. We heat it up and the entire plate expands. The volume of the plate has increased and so the density of the plate has decreased since we have not changed the mass. Everywhere in the plate is uniformly a new, smaller density.

Now imagine we have a cool plate and we cut a circle out of the center. We then place both the plate with the hole and the circle that we cut out in the oven. Both the plate and the circle heat up and expand, lowering their densities. Since the plate and the circle are the same material they are both at the same new, smaller density. Now what happens if we try to put the hot circle back into the hot plate? They should fit together perfectly! We haven't done anything different than when we left the circle in the plate, and in that case nothing funny happened in the center.

So, since we know that the circle expanded to a larger radius then the hole in the center of the plate should also have expanded.

Gears sizes and force

2005-10-12T06:22:31Z

Q. If you want to minimize the force you exert on the pedals what sizes gears should you have at the crank sprocket (at the petals) and the freewheel sprocket (at the wheel)?

]]> A. At the crank sprocket (at the petals), you want to be on the smallest gear. On the smallest gear you are moving the smallest amount of chain. Since the amount of chain you move changes how much the wheel goes around, moving less chain means moving the wheel less, which means the bike is doing less work on the road (less distance), so you must be doing less work on the bike. Since you're feet are going the same distance around the pedals, you must not be pushing as hard (less force).

At the freewheel sprocket, it's the opposite; you'll have to apply the smallest force on the largest gear. There are two ways to think about this. First, on a large gear you are further from the center of the wheel when you apply the force. This means you have a longer level arm, so it takes less force to spin the wheel. If it's easier to spin the wheel then it's easier to move the chain and thus easier to pedal. The other way to approach this is that on the largest gear it takes more chain to spin the wheel. That means when you go once around on the pedals and move some amount of chain this amount of chain only makes the wheel go around once or twice (as opposed to five or six times on a smaller gear). Moving the wheel less means the bike is again doing less work on the road (less distance) and thus you must be doing less work on the pedals (less force).

When the pedals are easiest to push (least force) you are doing less work each time you push the pedals once around, but you are also moving less. To go as far as on other gears you're going to have to turn the pedals around more times (a greater distance). Regardless, you have to do the same amount of work to get where you're going!

Bike gears and work

2005-10-10T19:10:26Z

Q. How do the concepts of work and energy relate to the gears on a bicycle?

]]> A. Remember the question in your homework (#2) concerning “jacking” up the corner of a car? Well, the physics involved is the similar to what we use to understand pedals on a bike.

The work you do on the handle of the jack is equal to the work that the jack does on the car (neglecting the effects of friction). Work equals the force applied times the distance over which that force is exerted. Pedals are a method for transferring work from you to the bike (just like the jack transfers work from you to the car). The product of the force (F) you exert on the pedals and the distance (D) the pedals move equals the force exerted on the rim of the wheel times the distance it moves. That is, F_pedal * D_pedal = F_tire * D_tire.

For the freestanding bike that we used in DR, one can vary the force applied to the tire over a large range (by changing gears) while keeping the force one applies to the pedals constant.

The spring constant

2005-10-04T05:29:34Z

Q. What is the spring constant? How do you calculate it? Does it change if we use a different mass to calculate it?

]]> A. The spring constant is a measure of how stiff the spring is. A spring that is very hard to stretch out has a large spring constant. A spring that is easy to stretch has a small spring constant. As the term spring constant implies, the spring constant is always the same for a given spring, assuming you don't put so much force on it that you break it.

In DR we had some springs, a meter stick, and some known masses. You can calculate the spring constant using the equation F = -k*x.
In this equation, the minus sign tells you that if you displace the spring (x) in one direction, the force is in the opposite direction. This is because both x and F are vectors and so the minus sign simply reverses the direction.

So, to find k we first need to exert a force on our spring. We do this by placing a mass on the spring. This mass exerts a force F=mg on the spring, its weight. Since after a few seconds we see that the spring and the mass stop moving we know that the net force is 0. The spring must be exerting a force on the mass to hold it up. This force is exactly the same amount of force that the mass is exerting on the spring. Thus, we can conclude that the F in our equation is just the weight that we place on the spring.

For a 500g mass, F = m*g = (0.500 kg)*(10 m/s^2) = 5 N

Notice that I converted the mass from grams to kilograms so that the force would then be in Newtons.

Now we just need to find x. x is the displacement of the spring. If the end of the spring was originally at 30cm and was then at 45cm after the mass was placed on it the displacement is then:

x= 45cm - 30cm = 15 cm = 0.15 m

Again notice I converted, this time form centimeters to meters. It's not absolutely necessary, but it's a good habit to always put things in their standard units.

So now we can calculate k. Since we have F = -kx we can rearrange to get:

k = -F / x

Again, the minus sign is just telling us about the direction. The force the spring is exerting is up while the displacement is down. Plugging in our numbers we have:

k = 5 N / 0.15 m = 33 N/m

If we were to use the same spring and put a 200g mass on it, by how much would it be displaced?

Since we know k is constant, we can find x. F = mg = (0.200kg)(10m/s^2) = 2N
F = -kx
x = - F / k
x = 2N / (33 N/m) = 0.06m = 6cm

As you should have expected this smaller mass stretched the spring less than our original 500g mass.

Blocks and balls

2005-09-19T17:07:11Z

Q. In lecture and in DR, we saw two balls crash into two wooden blocks. One ball was like a superball and the other was made out of clay. The block that the superball hit fell over, while the block that the clay ball hit did not. I don't get it -- what happened?

]]> A. Check out this movie that I made:

Blocks and balls movie, quicktime format

Blocks and balls movie, avi format

Play around with the movie, and watch carefully what happens when the balls crash into the blocks. I personally notice two things:

The superball rebounds farther after it hits the block. It's hard to see this, since you have to account for the perspective.
The collision with the clay ball lasts a lot longer than the collision with the superball.

We can use the concept of impulse to understand what happens here. The impulse that each ball transfers to the block is equal to the change in momentum of the ball; let's call that change in momentum Dp. That impulse must be equal to the force that the ball exerts on the ball (call that force F) multiplied by the time of the collision (call that dt). So, we have Dp=F x dt, or F=Dp / dt. For the collision with the superball, dt is small and Dp is big (since it rebounds farther we know that the velocity is higher after the collision), so the force is higher compared with the collision with the clay ball. The superball exerts a strong enough force to knock over the block, while the clay ball does not!

If you look even more carefully at the movie, you'll notice something quite neat. Each block really rotates around the edge in contact with the table. In the collision with the clay ball, the torque associated with the force from the ball is not large enough to cause the block to rotate far enough to tip over.

Spinning Wheel and Rotating Stool

2005-09-17T19:37:00Z

Q: In DR I sat on a stool and held the spinning bicycle wheel. When I flipped over the wheel I started to spin. What was going on??

]]> A: The key point here is that angular momentum (just like linear momentum) is conserved. That means that, if there are no external torques, the angular momentum is constant. No matter when we measure it we will get the same value. We have to be careful to always take all of the angular momentum into account to find that the total angular momentum is conserved.

So, first we had the wheel spinning with some angular momentum. Let's call that L1. With the right hand rule we find that the angular momentum of the wheel is up. You are sitting on the stool, but since you're not spinning you have no angular velocity and thus no angular momentum. So, at the beginning the total angular momentum is just that angular momentum of the wheel, L1 (up).

Then you flipped over the wheel. Since the wheel still has the same angular speed and the same moment of inertia the magnitude of the angular momentum is the same as before, L1. However, since you flipped it over the right hand rule will tell you that the angular momentum points down. Now you are spinning as well, so we will say that your angular momentum is L2. With the right hand rule we find that your angular momentum is up. So, the total angular momentum at the end is L1 (down) + L2 (up)

The conservation of angular momentum says that the total angular momentum can not change (since there are no torques from an external source acting on you or the wheel). So, that means that the angular momentum before and after you flipped the wheel should be the same. That is:

L1(up) = L1(down) + L2(up)

Since down and up are opposite directions we can switch the direction just by adding a negative sign. We rewrite L1 (down) as -L1(up) to get:

L1(up) = -L1(up) + L2(up)

For this equation to be true, it must be the case that:

L2 (up)=2 * L1(up)

Well isn't that interesting! We've just discovered that our angular momentum due to the spinning stool is twice the angular momentum of the wheel.

To explain a bit more conceptually, when you flipped over the wheel you changed its angular momentum. Since the total angular momentum of the system has to be conserved this caused you to spin in the opposite direction with an angular momentum twice as large as the wheel’s (since you have to both zero out the downward angular momentum of the wheel and add more upward angular momentum to equal the original upward angular momentum)

Two falling balls of different masses

2005-09-17T19:18:04Z

Q: In DR we dropped 2 balls of different masses from the same height and saw that they hit the floor at the same time. This is confusing to me - shouldn't the lighter ball hit first since it has less inertia and the same gravity as the heavier ball?

]]> A: I believe the confusion here is due to a loose use of the word gravity. When we say gravity we may either mean the acceleration due to gravity, which refers to the g=9.8 m/s^2 acceleration experienced by all objects on earth, or we may be referring to the force due to gravity, by which we mean F=mg=weight. When we drop the two balls they both experience the same acceleration due to gravity, 9.8m/s^2, which is why they hit the ground at the same time. However, the force of gravity is different for each ball since F=mg and the masses are different. The less massive ball has a smaller inertia and is easier to accelerate, but it also has a smaller force due to gravity on it. The result is that both balls have the same acceleration due to gravity and thus hit the floor at the same time.