Thread: Maths question lul

Originally Posted by ThePieMan

(ii) If the same plane is given a roughened surface, with a coefficient of friction 0.5, find the distance travelled before the particle comes to rest.

Not sure if this approach is correct at all, but I'll just have a go at it.
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F=mu x R as Xei had previously stated.

If you consider the particle the second it comes to rest, the reaction force = the component of gravity perpendicular to the slope upwards.
In other words R=gcosθ.

Now consider all the forces acting on the particle during its motion: -mgsinθ - mF = ma. (since the decelerating component is -gsinθ parallel to the slope downwards, the accelerating component upwards is +gsinθ. F is friction, which is equal to mu x R)

m terms cancel, leaving: a = gsinθ - (mu x R)
substituting θ=pi/4: a = gsin(pi/4) - (0.5 x gcos(pi/4)) = (5root2)/2

Now use the equation of motion to find distance travelled.

That's looking pretty good, there's just a couple of points to be made.

Firstly, be clear about what's a force and what's an acceleration. Especially in the context of weight. The weight force on the particle is mg, not just g, and so the weight acting down into the plane is mgcos@. As the particle doesn't accelerate above or below the plane, there must be a reaction force balancing it, giving R = mgcos@ away from the plane. The m divides out later. You shouldn't find yourself having to multiply a force by m.

Second, make sure you're clear about what the positive directions are. I suggest you chuck out the term 'deceleration' when actually doing problems, and just think in terms of accelerations in different directions. You can do it either way but it makes more intuitive sense to say the positive x direction is up the plane (and the positive y direction away from the plane). Both forces, the weight component and the friction (which opposes motion) will be pointing down the plane as the particles moves up it, in other words they are negative, so the the total force = ma = -mgsin@ -mu*mgcos@.

I think Wayfaerer was right, you could actually do this with energy if you want, but I imagine that that's not what the intention of the question is. It would be good practice though. There would be three things to take into account: the kinetic energy, the gravitational potential energy, and the energy lost from the system due to friction doing work on the particle.

Thanks, that does clear up some issues I've been having with inclined plane questions.
I'm quite bad at mechanics actually because I find it rather difficult to express information in the form of diagrams etc. But I guess this is something I have to learn by rote memorisation?

Not really by rote, but yes, definitely practice makes perfect. But you seem to be getting there, it's essentially just vectors. Once you get used to the idea that forces, accelerations, velocities and displacements are essentially arrows in space and behave like you would expect arrows to, the Mechanics modules should be a bit of a gift. In any case they only ever give you the same kinds of question over and over, so any stuff that initially seems tricky like separating a weight force into parallel and perpendicular components will eventually come naturally. I never learned the actual specifics of how to do the stuff, I would always just redraw the diagram with the different angles and arrows and go from there.

Tells us what the answer was when you find out!

Use equation of motion now that you know acceleration? So s=ut+1/2at^2 = V*2Vcos(pi/4)/g + 1/2*(5root2/2) + (2Vcos(pi/4))^2. I think you'll end up with an expression for distance in terms of V.

Right, lets leave the mechanics aside now .
New post - I'll put up some complex number questions I'm been doing recently, as well as ones I'm having problems with.

(i) By considering the ninth root of unity, show that: cos(2pi/9) + cos(4pi/9) + cos(6pi/9) + cos (8pi/9) = -1/2

(ii) Show that (1+i)^51 = 2^25(-1+i)

(iii) (This one is hard).

Write the equation (z-1)^5 = z^5 in the form ((z-1)/z)^5 = 1, and show that its roots are 1/2(1+icot*1/5kpi)

where k = 1,2,3,4

Originally Posted by Wayfaerer

I'm guessing you'd have to find the mass, right? Then maybe use the work-energy theorem?

Yeah, I apologise. This question can be answered by considering initial KE, and final PE = mgh = work done.

No, like I said, you need to take into account the work done by friction...

As for the complex number questions, i) is just a standard A Level technique which will be in your textbook, ii) will be easy if you convert to modulus-argument (Re^i@) form, iii) you really need to show how far you've gotten by yourself because the initial steps should be straightforward.

(i) Completed.

(ii) (1+i)^51 into mod-arg form: (root2, pi/4)^51 = ((root2)^51, 51pi/4) = (2^51/2, 3pi/4). Got 3pi/4 because 51pi/4 = 3pi/4 + 48pi/4 (which is just 6 rotations of the circle)

Not sure how to get from (2^51, 3pi/4) to 2^25(-1+i). Must be something blindingly obvious because arg(-1+i) = 3pi/4...

(iii) I started by finding the fifth roots of unity:
z^5 = 1
cos(2pi*k)+isin(2pi*k)=1

for fifth roots: cos((2pi*k)/5+isin((2pi*k)/5) for k = 0,1,2,3,4
roots are:
k=0: 1
k=1: cis 2pi/5
k=2: cis 4pi/5
k=3: cis 6pi/5
k=4: cis 8pi/5

(cis = cosθ+isinθ)

We had this equation: ((z-1)/z)^5 = 1
So I tried equating the fifth roots of unity with ((z-1)/z).

That's good work so far... with ii) if you split 51/2 into an integer and a fraction like you did for the argument, you should see how it's equal to the other thing. With iii) you just need to keep going, solve (z-1)/z = a for z, where a is a fifth root of unity.

Originally Posted by ThePieMan

Yeah, I apologise. This question can be answered by considering initial KE, and final PE = mgh = work done.

For a fun but rewarding experience, try it using the Lagrangian or Hamiltonian.