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# Tutor profile: Mykolas S.

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Mykolas S.
Oxford University Physics Student, over 2 years of tutoring experience
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## Questions

### Subject:Physics (Newtonian Mechanics)

TutorMe
Question:

A circular disk of uniform density and radius R has a hole in it. The hole is also circular, its radius is R/2. The centre of of the hole is R/2 away from the centre of the disk. Where is the disk's centre of mass? (If the setup is unclear, a simple drawing will make things much easier to understand)

Inactive
Mykolas S.

If you draw out the shape of this disk with a hole, it is at first entirely unclear where the centre of mass might be. One way to start is to draw a line that goes through the centre of the hole and the centre of the disk. You can then see that our disk is symmetric with respect to the line! Therefore, the centre of mass is somewhere along that line! To proceed further, you need a bit of inspiration. You might think, "I wish it was two different circles, rather than one circle with a circular hole. Then I would at least know where the centres of mass of the individual circles were - at their geometrical centres." But indeed, you CAN think of this as two different circles! If you say the density of the disk is $$\rho$$, then what we have in front of us is a circle or radius R with density $$\rho$$ ON TOP OF a circle of radius R/2 with density $$-\rho$$; the negative mass cancels with the positive mass and you get a hole! Alright, so we actually have to find the centre of mass of two circles. Let's say the 'negative-mass circle' has mass $$-m$$, then the 'positive-mass circle has mass $$4m$$, because its area is 4 times larger. To find the centre of mass of this arrangement, let's take a step back. Suppose we have any composite object, for example, a knife with a wooden handle and metal blade. We can find the centres of mass of the blade and handle (you don't need to know how). With that knowledge, the centre of mass of the whole knife can be calculated as follows: 1) Imagine a point mass where the blade's centre of mass is, with a mass equal to that of the blade. 2) Do the same thing with the handle and its centre of mass. 3) The centre of mass of the ACTUAL KNIFE is precisely the same as the centre of mass of the TWO POINT MASSES. Why do we bother doing this? Because its much easier to calculate the centre of mass of two point masses than that of a continuous object. You can simply imagine connecting the point masses with a weightless rod and trying to balance the rod on a pivot point. When the rod is balanced, the pivot point is the centre of mass! So we turned our problem into a torques problem! Going back to the disk with a hole, remember we showed it is equivalent to a composite object (two circles with masses $$4m$$ and $$-m$$) , so we can employ this new trick of ours! Going through our steps, outlined above: 1) Put $$4m$$ at the centre of the disk 2)Put $$-m$$ at the centre of the hole, R/2 away 3)The CM of the disk-with-hole coincides with the CM of these two point masses If we label the distance from the centre of the disk to the CM as $$L$$, then the distance from the centre of the hole to the CM is $$R/2 - L$$. To find the balance condition, we equate the torques acting on the CM clockwise to the torques acting on the CM anti-clockwise: $$4mgL = -mg(\frac{1}{2} R - L)$$ $$L = -\frac{1}{6}R$$ We get that L is negative. Did we make an error somewhere?? No, the negative sign simply tells us that the CM is to the opposite side of the disk's centre compared to our guess! In fact, we should have expected that a hole on the right 'pushes' the CM to the left, and vice versea.

### Subject:Physics (Electricity and Magnetism)

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Question:

A +1 Coulomb point charge is placed 2 metres above an infinite grounded conducting plate (let the electric potential of the plate be 0). Imagine a line connecting the charge to the closest point in the plate. Find the magnitude of the electric field in the middle of that line.

Inactive
Mykolas S.

Let's start with analysing what happens to the conducting plate. It is said to be grounded (at V=0), so: 1) the potential is equal (to 0) everywhere along the infinite plate 2) the plate is conducting, so charges can move around until the plate's electric potential is indeed 0. Since the charge above the plate is zero (and it's implied there are no other charges nearby), it will create a positive electric potential around it, so negative charges will accumulate on the plate - they will each decrease the electric potential around themselves and "keep" the plate's potential at 0. In theory, we could now find how the negative charges are distributed along the plate. Then we could find the electric fields they produce at the midpoint of our imaginary line (see Question), and finally add them up. In reality, that is a lot of work! The distribution of charges is not obvious at all. We can employ a trick to make it much easier! To understand the trick, you need to understand something called the Uniqueness Theorem (fancy name, right?!). All it says is this: You DON'T need to know where ALL the charges of the world are to find the electric field (and potential) inside a volume that interests you. All you need for that is: 1) the potential along the BOUNDARY of that volume. 2) the charges (location & magnitude) INSIDE that volume. What that means is the charges OUTSIDE the volume can CHANGE in any way, as long the 2 things above stay the same! This allows us to solve our problem much more easily. We notice that the plate divides the world into two parts - let's call them the Top part (the part with our +1C charge) and the Bottom part (the one with no charge). We say that the Top part is the "Volume of interest". We can think of our volume as a half a sphere (i.e. hemisphere), with its flat part (equator) being the plate and the radius extending to infinity. Do we know the potential everywhere along the boundary of this volume? YES! Along the flat part, the potential is 0 (the question says so!). Out in infinity, the potential is also 0, because our charge is infinitely far away so it doesn't contribute. Do we know all the charges inside our "volume of interest"? YES! It's only the +1C charge mentioned in the question. So we need to remove the grounded plate and put something simpler, but: 1) keep the boundary potential the same, 2) not change the Top part charges. What does the trick? It's surprisingly simple - a -1C point charge 2 metres below our "equator" (almost like some kind of mirror image of our +1C point charge). You might not believe something so simple works, but if you manually calculate the potential created by these two charges, it is indeed zero on the "equator". And of course, infinitely far away, it's also zero. All that is left to do is calculate the electric field at the point we were asked about (see Question). The Uniqueness Theorem tells us that the field at that point in the original (charge & plate) configuration is exactly the same as the field in this new (2 charges) one. But it's much easier to calculate in the new configuration, so that's what we do! The point we care about is 1 metre below the +1 C charge, and 3 metres above the -1 C charge (if you don't believe me, draw a sketch!). They both create fields pointing down, so the total magnitude of the electric field will be $$E = \frac{ 1 C}{4 \pi \epsilon_0 {(1 m)}^2} + \frac{ 1 C}{4 \pi \epsilon_0 {(3 m)}^2} \approx 10^{10} \frac{V}{m}$$, with the field pointing down (towards the "equator"). And we're done! Take a minute to appreciate what we did here - we managed to solve a problem by pretending it's actually a much simpler problem, it's just that we knew the answers would be the same! This trick of adding charges and using the Uniqueness Theorem is quite well known and is called "Mirror Charges" - I encourage you to learn more about it, either with me or online!

### Subject:Physics

TutorMe
Question:

Imagine you are running a 42 kilometer marathon along a completely straight track at 10 kilometers/hour. As soon as you start, a little bee starts flying towards you from the finish line at 20 kilometers/hour. When the bee reaches you it turns around instantaneously and flies all the way back to finish line at the same speed, turns around again, flies until it encounters you, turns around, and so on until you finish. Meanwhile you just keep running at the same pace. Question: what distance does the bee travel?

Inactive
Mykolas S.

You might try and find where the two meet the first time, then the second time, then the third, on and on until you finish the marathon, and add up the distances in the end. That will give you the right answer! But it will be extremely tedious! We need a bit of inspiration here... The trick is this: find the time it takes you to finish the marathon. That’s easy – just divide the length of the track by your speed to get 4.2 hours. You know that the bee is doing its flying for exactly the same time! You know its speed, so the distance is just the bee's speed multiplied by the time - 82 kilometers! It does not matter where it turns around and how many times it does so!

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