Section 3

5 min read

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GAMSAT almost always has some tricky Section 3 questions on Circular Motion. What is Circular Motion and how does it all work? You all experience Circular Motion ā for example moving around in circles, essentially whenever youāre sitting in a car and itās gone over a speed bump, which has a bit of a curved bit, or semi-circle. Or when you are at a roundabout and are taking a turn, and it kind of feels like a force on your body as you are moving away.

When you are going on a roundabout, you are not really changing your speed. You are maintaining a constant speed as you are turning. What you are changing is your velocity because velocity is a vector, and vectors not only have a magnitude, they also have a direction. Because you are potentially going from a straight path and towards your right, since you are changing direction, you are technically changing velocity. And when you are changing velocity, you are going to have an acceleration because acceleration equals to a change in velocity.

Acceleration is a change in velocity over time:

(a = change in V/t)

As Newtonās second law tells use, Force is the product of Mass and Acceleration.

F = m x a

Hence you end up feeling a force because of that acceleration which again is a result of that change in velocity. This process is really what Circular Motion is all about. The idea that even though the magnitude isnāt changing, since your direction is changing, you still experience a Force and then an Acceleration. Examples of common Circular Motion include a seat going around a round-about or a speed bumps. A rollercoaster loop is a vertical one which has a few different properties because it is also changing heights, so there is gravitational potential energy associated with it. There is also the moon and the earth, and gravitational satellites or the earth around the sun.

So now, letās look at the equations of Circular Motion. Letās start off with looking at the simple formula of how it was derived in the first place. You donāt need to understand the math behind this. The formula is: acceleration is equal to the square of velocity divided by the radius of the loop:

a = V2 /r

As Newtonās second law told us, Force is equal to mass multiply by acceleration, thus:

Force = mass xĀ V2 /r

Force = mv2 /r

Now letās consider how velocity is calculated. Now, if you are going round in a circle, velocity is going to be the displacement you covered, or in this case, since we are only focusing on the velocity at a particular point, we are just looking for the speed because we are going to input the magnitude into this formula. So, velocity is going to be the circumference of the circle divided by the time taken:

V = 2š r/TĀ

Now if we look at v2, we end up with:

Ā V2 = 4š 2Ā r2Ā / T2Ā

If we forward this into the formula of acceleration, we end up with

aĀ = 4š 2Ā rĀ / T

Ā And hence we end up with:

FĀ = 4š 2Ā r m Ā / T2

As you can see, we have two formulas for Force, and these would be considered the regular formulas. You should be aware of them although, in the GAMSAT Physics, they are often given to you. However, it is not expected that they have to be given to you. So a knowledge of what the formula is and remembering it would be required (hint ā we have a great physics cheat sheet!!)

Now, letās look at satellites specifically. Satellites include artificial satellites; they are just one of those you see, going around the earth or natural satellites such as a planetās moon, or planets moving around the sun. If you want to calculate the force around them, the formula used is:

F = GMm/r2

Where āGā is the gravitational constant, āMā is the mass of the central body, āmā is the mass of the orbiting body and ārā is the radius.

This is a great formula to test on the GAMSAT PhysicsĀ because it often uses really large number for the mass of the planet and the mass of the sun or moon and it requires a lot of approximations to work with which is very good at testing the mathematical ability of students. So thatās how you calculate force. If you wanted the acceleration, it would simply be:

a = GM/r2

The āmā is added when calculating Force because of Newtonās second law where F = ma

Now letās look at Keplerās law. Kepler was a man who came with an equation which specifically was for satellites. You donāt need to remember it because it is completely derivable and we will now take you through how to do it. However, if you quickly remember it or if you at least understand the concept behind it, then itās quite easy to work with. Before we start, just remember that because satellites are also moving in Circular Motion, satellites also can use the same equations as the regular ones mentioned before and that is exactly what Kepler relied on when deriving his law.

4š 2Ā r m Ā / T2 = GMm/r2

In the equation above, we can cancel out the āmā on both sides. We might want to take all the ārā on one side and the āTā on the same side and every other thing on the other side. This gives:

r2 x 4š 2Ā r / T2 = GM/r2 xĀ r2Ā

4š 2Ā r3 / T2 = GM

4š 2Ā Ć· 4š 2r3 / T2 = GMĀ Ć· 4š 2

Ā r3 / T2 = GM/4š 2

The key to understanding the formula is that āGā is the gravitational constant which is the same anywhere in the universe, ā4ā is just a number which is a constant. The same with āĻ,ā itās just a number which is constant, and āMā is the mass of the central body. The mass of the central body is the mass of whatever the satellite is orbiting around. For that particular central body, the ratio of r3 and T2 for all of its satellites would be the same. So if you know the value of ārā value and āTā value for one satellite, and then you know the ārā for another satellite, you should be able to figure it āTā because r3 and T2 should be the same for each satellite.

So far, we have talked about gravitational motion; we have talked about horizontal motion and round-about and satellites. What about if we look at speed bumps or rollercoaster tracks in a vertical motion. The key to doing well in this is remembering about the energy transformations. As you go around a loop in a rollercoaster, you are faster at the bottom and slower at the top because as you go up, the kinetic energy you had converts to gravitational potential energy. As you come down, that potential energy is converted back into kinetic energy. Remembering about this conservation is important because you are not going to have a constant magnitude in your velocity anymore because you have to incorporate this:

At the top of the loop, low K.E, high GPE

At the bottom of the loop, high K.E, low GPE

The good thing about this is that often in GAMSAT Physics, if you get a vertical circular motion question, then they would actually usually ask you to find the velocity unless it is right at the top or right at the bottom because it is too variable as it changes it magnitude and direction during it travels. So usually, just the top or the bottom but you should be able to find the velocity.

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