PHYS140 FAQ

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.

Posted by Rebecca Lamb at 03:52 PM | Permalink

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.

Posted by Rebecca Lamb at 11:42 PM | Permalink

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!

Posted by Rebecca Lamb at 12:22 AM | Permalink

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.

Posted by at 01:10 PM | Permalink

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.

Posted by Rebecca Lamb at 11:29 PM | Permalink