HSC Physics Question Thread

[jamonwindeyer]

Hey Meckenza!

Very nice question there! I completely agree with your reasoning and you should also be satisfied that you understand the main concern of this question, well done!

Like you have correctly stated, as speed increases, back emf increases and this opposes the supply current, hence causing the current to decrease. Now let's analyse what happens physically. Initially, at speed = 0, the DC motor is first started. At this time, the coil is stationary so the back emf is zero. Therefore the current flowing through the coil is high and equal to supply current. Once the armature rotates faster, the back EMF increases (This explains the relatively sharp drop in current at the start) and the difference between constant supply EMF and back EMF becomes smaller. However, because back EMF can never exceed the value of supply emf, there has to be a point where supply current pretty much stops decreasing because the impact of back emf upon it becomes smaller and smaller. This explains that the current is decreasing at a decreasing rate. Hence you have that curve that looks like as if its reaching a horizontal asymptote (you can think about it this way if you are maths-orientated).

Hope that helped! If you have further questions please dont hesitate to ask!

Best Regards
Happy Physics Land

Teamwork man 

[Happy Physics Land]

Teamwork man 

You and your engineering swag have illuminated my understanding of physics Jamon! :L
Never realised you can use hyperbola here, maybe this graph A is a conic section? 

[Meckenza]

Hey Meckenza! Ah, Back EMF, this concept drove me nuts in the HSC aha 

So you've correctly said that A makes the most sense. I'll try to explain the curved shape as best I can, though it actually isn't something I know about for sure!

Back EMF is caused by the change in magnetic flux experienced by the coil in a motor as it spins. As it rotates, the coil moves through different parts of the field, and this changing flux induces a current. By Lenz's Law, this current acts against the supply. It sounds like you understand this pretty well, but let me know if you need a little more!

To answer your question on the shape of the curve, it probably makes sense to first consider the alternative. What if we had a straight line relationship? That would imply that we could push the speed of a motor high enough that no current flows. As speed increased, eventually the current would drop to zero. That can't be possible, since that would mean the motor would not be spinning! So a straight line relationship definitely doesn't work.

In actuality, Back EMF only increases until it reaches a point very similar to the supply EMF. This is the horizontal part of the graph you are seeing!

What we have here is an inversely proportional relationship. As speed increases, current decreases at a proportional rate, since Back EMF is proportional to speed. This creates a hyperbolic graph, roughly represented by:



If you graph a hyperbola, it has a shape similar to what you are showing in that question.

Now I am sure it is more complex than this, but this explanation gives you the idea of why the line is curved. I hope it helps! If you wanted a more concrete, detailed explanation I can definitely do some research for you, I am sure I have some more detailed explanations in my Electrical Engineering textbooks 

Wait, so how did you derive that current is proportional to 1/speed?

[Happy Physics Land]

Wait, so how did you derive that current is proportional to 1/speed?

Hey Meckenze!

I will answer this question for jamon (If he doesnt mind). Essentially that expression "supply current is proportional to 1/speed of motor" is the same as "supply current is inversely proportional to the speed of motor". Like what you have correctly stated, supply current is inversely proportional to back emf. But because speed is proportional to back emf, then it follows that current is also inversely proportional to speed.

[Meckenza]

Ahhhk, ok. Thanks Happy Physics Land and Jamonwindeyer!!! 

[jamonwindeyer]

Ahhhk, ok. Thanks Happy Physics Land and Jamonwindeyer!!! 

You are most welcome! Yeah HPL is on the answer, it is basically just the fact that as Speed goes up, current goes down, an inversely proportional relationship (if you take it to the simplest level). Sorry for being a little ambiguous! 

Hey Meckenze!

I will answer this question for jamon (If he doesnt mind). Essentially that expression "supply current is proportional to 1/speed of motor" is the same as "supply current is inversely proportional to the speed of motor". Like what you have correctly stated, supply current is inversely proportional to back emf. But because speed is proportional to back emf, then it follows that current is also inversely proportional to speed.

You can answer any question at any time HPL! 

And with regard to your comment, it most definitely could be, it is most definitely more complex than the relationship in the previous message. I might even do some reading and see if I can find a proper equation! I've tried a couple of derivations but can't come up with anything that would suggest the curved shape mathematically just yet, I'll ask around the brainiac electrical engineers at uni 

[Happy Physics Land]

You are most welcome! Yeah HPL is on the answer, it is basically just the fact that as Speed goes up, current goes down, an inversely proportional relationship (if you take it to the simplest level). Sorry for being a little ambiguous! 

You can answer any question at any time HPL! 

And with regard to your comment, it most definitely could be, it is most definitely more complex than the relationship in the previous message. I might even do some reading and see if I can find a proper equation! I've tried a couple of derivations but can't come up with anything that would suggest the curved shape mathematically just yet, I'll ask around the brainiac electrical engineers at uni 

Engineering god can I please have a hand with a mechanics question???

The school bus with a total mass of 10 tonnes travels at a constant velocity of 40km/h down a long hill with a slope of 15 degrees. Determine the braking force required to stop the bus within 50 metres from when the driver applies the brakes.

Thank you so much in advance! )

[Neutron]

Hey Neutron!

a)More reliable: DC, because commutator is in contact with carbon brushes, ensuring a more consistent flow of electricity
b) Suffers less wear on critical parts: AC, because slip rings are not directly in contact with carbon brushes, meaning that they dont suffer as much wear as commutators which will eventually wear out due to physical contact with brushes
c) Can be transmitted over long distances with less loss of energy: AC, because AC generators produce AC which are able to be stepped-up or stepped-down. Through stepping up AC to a very high voltage (550000V transmission voltage in Australia), the amount of currents being delivered has been reduced (Voltage and current are inversely proportional), hence less power loss because less heat dissipation.
d) Is easier to connect to houses: AC, because many appliances at homes require the standard 240V AC to operate
e) Is more likely to have shorting: AC (Im not too sure of the reason, perhaps when others come around they can provide you with the explanation?)
f) Can have cables with less insulation: DC, AC is a lot of dangerous than DC because it has the ability to cause fibrillation and can fatally propagate through human body. Hence theres more necessity to insulate AC than DC.
g) Produces less electrical interference: DC, because AC travels in a sinusoidal wave, hence its rising and falling propagation pattern makes it more likely to interfere with other waves. In contrast, DC currents only travel in straight lines, hence it is unlikely for it to come in contact and interfere with other waves.
h) Is likely to experience a higher impedance from coiled cables: DC, this is because DC travels only in a straight line, so if the cable is in the shape of a coil, DC would encounter a lot more resistance. Imagine yourself trying to run around in a circular maze, if you run in straight lines, you would frequently collide into walls and then you would turn and run straight in another direction, but again you would collide into another wall. Similarly, this is why DC would experience a lot more impedance as it travels through the coil. AC travels in a sine curve that is easier to propagate through the coil and because it has higher frequencies, it would experience less impedance

I hope these explanations all make sense to you, the technicality in my reasoning may not be very strong but Im sure the answers are correct.

Best Regards
Happy Physics Land

legend <3

[jamonwindeyer]

Engineering god can I please have a hand with a mechanics question???

The school bus with a total mass of 10 tonnes travels at a constant velocity of 40km/h down a long hill with a slope of 15 degrees. Determine the braking force required to stop the bus within 50 metres from when the driver applies the brakes.

Thank you so much in advance! )

Definitely not an Engineering God, but totally man! Let's have a look.

Let's start by converting everything to SI units.


Okay, so there is a few things to consider here. I think we should start by determining the acceleration required to slow the bus in that distance!

We have a formula linking velocity, acceleration, and distance:



Now this is directed along the slope the bus is travelling on.

Now, I am actually a little confused as to how the slope comes in to this question. At first I thought we'd have to account for the weight force of the bus, but that force is acting when the bus is travelling at the constant speed as well. Therefore, we have already accounted for it in the acceleration calculation. So, I have a funny feeling it is there as a trick, do you think I'd be right there?

Let's proceed assuming the simple case, that we now just need to calculate the force required to generate the above acceleration. We need to create the acceleration above, we have a mass, this is a job for Newton:



So this is my response for now, I hope it helps either as a solution or to get you started with one! I'm curious what you think about the slope thing (sorry if I explained it poorly)? My brain is telling me that we can ignore it!  the other option is we need to counteract weight force of the bus, which would be a matter of calculating the weight force, finding the component of that force which is directed down the slope, and then adding that value to the answer. I can go into more detail if you need, but yeah, super keen to hear your thoughts!

[jamonwindeyer]

Definitely not an Engineering God, but totally man! Let's have a look.

Let's start by converting everything to SI units.


Okay, so there is a few things to consider here. I think we should start by determining the acceleration required to slow the bus in that distance!

We have a formula linking velocity, acceleration, and distance:



Now this is directed along the slope the bus is travelling on.

Now, I am actually a little confused as to how the slope comes in to this question. At first I thought we'd have to account for the weight force of the bus, but that force is acting when the bus is travelling at the constant speed as well. Therefore, we have already accounted for it in the acceleration calculation. So, I have a funny feeling it is there as a trick, do you think I'd be right there?

Let's proceed assuming the simple case, that we now just need to calculate the force required to generate the above acceleration. We need to create the acceleration above, we have a mass, this is a job for Newton:



So this is my response for now, I hope it helps either as a solution or to get you started with one! I'm curious what you think about the slope thing (sorry if I explained it poorly)? My brain is telling me that we can ignore it!  the other option is we need to counteract weight force of the bus, which would be a matter of calculating the weight force, finding the component of that force which is directed down the slope, and then adding that value to the answer. I can go into more detail if you need, but yeah, super keen to hear your thoughts!

Actually, yep, changed my mind, I think we need to account for weight force here. Sorry! 

So the acceleration bit is the same as above! Now, not only do we need to produce this acceleration against the bus, we also need to produce a force to counteract the weight force of the bus, which is pulling it down the hill! How do  we do this?

Well, the bus is on a slant. It's weight force acts directly downwards, but we only want the weight force directed parallel to the slope of the hill. We use vector components here and we get:



So instead, our braking force is:



I'm rolling with this now, sorry HPL! It has been a while since I have done mechanics, does this help you out at all?

[Happy Physics Land]

Actually, yep, changed my mind, I think we need to account for weight force here. Sorry! 

So the acceleration bit is the same as above! Now, not only do we need to produce this acceleration against the bus, we also need to produce a force to counteract the weight force of the bus, which is pulling it down the hill! How do  we do this?

Well, the bus is on a slant. It's weight force acts directly downwards, but we only want the weight force directed parallel to the slope of the hill. We use vector components here and we get:



So instead, our braking force is:



I'm rolling with this now, sorry HPL! It has been a while since I have done mechanics, does this help you out at all?

I understood everything you said Jamon, thank you so much!!!! It was one of those mechanics questions that was really confusing (came from a CSSA paper so no surprise there aye hehe). Initially l thought the angle didnt matter and I tried to do kinetic energy = work. Because the bus is coming down the slope with 40km/h, so we can calculate its kinetic energy Ek = 1/2 mv². And I saw that 50m, so I thought this probably has something to do with the amount of work (energy) to stop this bus (because work = Fs). Im not sure if you would approve of my method but yes I definitely benefitted from your solution! Thank you again Jamon you are my saviour!

[wils013]

http://imgur.com/YYMQBlW

How does it work that End X is negative?

[jamonwindeyer]

I understood everything you said Jamon, thank you so much!!!! It was one of those mechanics questions that was really confusing (came from a CSSA paper so no surprise there aye hehe). Initially l thought the angle didnt matter and I tried to do kinetic energy = work. Because the bus is coming down the slope with 40km/h, so we can calculate its kinetic energy Ek = 1/2 mv². And I saw that 50m, so I thought this probably has something to do with the amount of work (energy) to stop this bus (because work = Fs). Im not sure if you would approve of my method but yes I definitely benefitted from your solution! Thank you again Jamon you are my saviour!

I like that! Cool idea! Never even considered it that way. For anyone looking at this, HPL means:




Now this is a different solution than I got, but I realised I made a mistake calculating the component of weight force. Just the numbers were wrong. The actual answer I got was:



So, I would say we are in agreement! Both methods work very nicely! The difference is probably rounding errors, I rounded harshly when I calculated acceleration and such last night 

Great thinking HPL! I would never have thought to do it that way 

[jamonwindeyer]

http://imgur.com/YYMQBlW

How does it work that End X is negative?

Hey Wils013! This is actually a pretty obscure question concerning forces on charges! At least, that's the way I see it 

We know that moving charges inside a magnetic field experience a force, let's see what that force does.

So, let's consider individual charges inside that rod. The rod is moved up through the magnetic field. So the direction of motion is upwards. Now, the magnetic field lines move from North to South between the magnets. So they go to the right of the page, if you like.

Using these two directions, you should apply something like the Right Hand Slap Rule, but every school does something a little bit different. Whatever method you were taught, use it, and you'll find that the resultant force points towards End Y of the rod.

However! This represents the direction of the force on a positive particle. The force on a negative particle will be in the opposite direction! So, while the positive charges experience a force in the direction of End Y, the negative charges experience a force towards End X. Thus, all the negative charges accumulate at End X, and it becomes negatively charged!

I hope this helps! Let us know if you need a picture of that directional reasoning, it is a little hard to explain, but give it a go! 

[wils013]

Hey Wils013! This is actually a pretty obscure question concerning forces on charges! At least, that's the way I see it 

We know that moving charges inside a magnetic field experience a force, let's see what that force does.

So, let's consider individual charges inside that rod. The rod is moved up through the magnetic field. So the direction of motion is upwards. Now, the magnetic field lines move from North to South between the magnets. So they go to the right of the page, if you like.

Using these two directions, you should apply something like the Right Hand Slap Rule, but every school does something a little bit different. Whatever method you were taught, use it, and you'll find that the resultant force points towards End Y of the rod.

However! This represents the direction of the force on a positive particle. The force on a negative particle will be in the opposite direction! So, while the positive charges experience a force in the direction of End Y, the negative charges experience a force towards End X. Thus, all the negative charges accumulate at End X, and it becomes negatively charged!

I hope this helps! Let us know if you need a picture of that directional reasoning, it is a little hard to explain, but give it a go! 

Yes! Thank you so much I didn't recognise to use the hand rule, that makes so much sense thanks!