HSC 2015 MX2 Marathon (archive) (1 Viewer)

InteGrand · Oct 4, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Still can you show what you did ...

$Set the speed of rotation to $\omega = 45 \text{ rpm} = 45 \times 2\pi \times \frac{1}{60}\text{ rad}/\text{s}=1.5\pi \text{ rad}/\text{s}$ (SI units) and solve for the tension in the string (answer will be in newtons if using all SI units) by resolving forces etc. (The forces on the particle are a tension force up the string, which has vertical and horizontal components, and a downward gravity force. The force in the horizontal direction is $mr\omega^2$, where $r$ is the radius of the circle (note that the string is given to be 3 m), and this is provided by the horizontal component of the tension force. The force in the vertical direction is 0 newtons, so the vertical component of the tension force perfectly cancels out the downwards gravitational force $mg$. Also label the acute angle between the string and the horizontal $\theta$.)$

Drsoccerball · Oct 4, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
$Set the speed of rotation to $\omega = 45 \text{ rpm} = 45 \times 2\pi \times \frac{1}{60}\text{ rad}/\text{s}=1.5\pi \text{ rad}/\text{s}$ (SI units) and solve for the tension in the string (answer will be in newtons if using all SI units) by resolving forces etc.$

umm... where does this formula come in $\frac{2 \pi}{w} =T$

Zlatman · Oct 4, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
$The four complex numbers $z_1, z_2, z_3, z_4$ are represented on the complex (Argand) plane by the points $A, B, C, D$ respectively. If $z_1 - z_2 + z_3 - z_4 = 0$ and $z_1 - iz_2 - z_3 + iz_4 = 0$, determine the possible shape(s) for the quadrilateral $ABCD$. (This question is from the 1970 HSC 4U paper.)$

$If $z_1 - z_2 + z_3 - z_4 = 0$, then $z_1 - z_2 = z_4 - z_3$ and $z_1 - z_4 = z_2 - z_3$.$

$Therefore, the quadrilateral has two opposite sets of equal sides.$

$If $z_1 - iz_2 - z_3 + iz_4 = 0$, then $z_1 - z_3 = i(z_2 - z_4)$, i.e. the two diagonals of the quadrilateral are equal and perpendicular.$

$Therefore, since the quadrilateral has two sets of equal sides opposite each other, and the two diagonals of the quadrilateral are equal and perpendicular, the quadrilateral must be a square.$

Is this right?

InteGrand · Oct 4, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
umm... where does this formula come in $\frac{2 \pi}{w} =T$

That formula is irrelevant for this one. That formula refers to the period T of particle's circular motion (time to complete one revolution) if its angular velocity is w, but we don't need that now.

For this one, we just need the following equations:

$T\cos \theta = m\times \underbrace{3 \cos \theta}_{r}\omega^2$ (i.e. horizontal component of tension (LHS) provides centripetal force (RHS))

$T\sin \theta = mg$ (weight force (RHS) equal in magnitude to vertical component of tension force (LHS) due to no acceleration in vertical direction).

where $T$ is the magnitude of the maximal tension force and m and ω are as given in my previous post. The only unknowns in these simultaneous equations are T and θ, and we just need to solve for T to get the answer.

Edit: and as we can see, only the first equation is needed to find T. Just cancel cos(θ)'s. This is the derivation of the formula $T = m\ell \omega^2$ (which I never memorised but which I think is in textbooks).

InteGrand · Oct 4, 2015

Re: HSC 2015 4U Marathon

Zlatman said:
$If $z_1 - z_2 + z_3 - z_4 = 0$, then $z_1 - z_2 = z_4 - z_3$ and $z_1 - z_4 = z_2 - z_3$.$

$Therefore, the quadrilateral has two opposite sets of equal sides.$

$If $z_1 - iz_2 - z_3 + iz_4 = 0$, then $z_1 - z_3 = i(z_2 - z_4)$, i.e. the two diagonals of the quadrilateral are equal and perpendicular.$

$Therefore, since the quadrilateral has two sets of equal sides opposite each other, and the two diagonals of the quadrilateral are equal and perpendicular, the quadrilateral must be a square.$

Is this right?

Yes, well done.

asdfghjklqwerty · Oct 4, 2015

Re: HSC 2015 4U Marathon

what i did was T=Mw^2r
and w= (45x 2pi)/60
and the rest is ezpz

InteGrand · Oct 4, 2015

Re: HSC 2015 4U Marathon

asdfghjklqwerty said:
what i did was T=Mw^2r
and w= (45x 2pi)/60
and the rest is ezpz

By 'r', you mean the string length (3 m), not the radius, right? Probably better to write $T=m\ell \omega^2$ , since 'r' normally refers to the radius.

Paradoxica · Oct 4, 2015

Re: HSC 2015 4U Marathon

$Prove that the right angled triangle with base 2 and height 11 has a base angle three times that of the right angled triangle with base 2 and height 1.$
There is a very easy way to do this.

InteGrand · Oct 4, 2015

Re: HSC 2015 4U Marathon

Paradoxica said:
$Prove that the right angled triangle with base 2 and height 11 has a base angle three times that of the right angled triangle with base 2 and height 1.$
There is a very easy way to do this.

Edit: nvm, misread

porcupinetree · Oct 4, 2015

Re: HSC 2015 4U Marathon

Here's a classic mechanics question:

$\textup{A car travelling at speed v metres per second along a banked}$
$\textup{curve of radius r metres experiences no lateral frictional force.}$
$\textup{Another car of mass m travels along the same curve at speed V}$
$\textup{metres per second, V} > \textup{v. Show that the lateral frictional force exerted}$
$\textup{by the track on this car is equal to} \frac{mg(V^{2} - v^{2})}{\sqrt{v^{4} + r^{2}g^{2}}} \textup {newtons.}$

Paradoxica · Oct 4, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
$What do you mean? If by ``the right angled triangle with base 2 and height 1'', you mean the standard 45-45-90 triangle, then no right triangle is going to have angles that are three times the angles in this triangle.$

$$Clarification. The side length values for the larger and smaller triangles are, respectively, $(1,2,\sqrt{5}) $ and $ (2,11,5\sqrt{5})$

porcupinetree · Oct 4, 2015

Re: HSC 2015 4U Marathon

Paradoxica said:
$$Clarification. The side length values for the larger and smaller triangles are, respectively, $(1,2,\sqrt{5}) $ and $ (2,11,5\sqrt{5})$

Ah I think I have the answer, let me write up a solution

porcupinetree · Oct 4, 2015

Re: HSC 2015 4U Marathon

Finally a question from Paradoxica which I could actually do

Paradoxica · Oct 4, 2015

Re: HSC 2015 4U Marathon

porcupinetree said:
Finally a question from Paradoxica which I could actually do

that's correct, but not as quick. try thinking in the complex plane. also, in the question, I would not have stated the hypotenuse lengths were it not for the confusion.

porcupinetree · Oct 4, 2015

Re: HSC 2015 4U Marathon

Paradoxica said:
that's correct, but not as quick. try thinking in the complex plane. also, in the question, I would not have stated the hypotenuse lengths were it not for the confusion.

Ah yes I see it now. Both methods are quite efficient, in my opinion

asdfghjklqwerty · Oct 4, 2015

Re: HSC 2015 4U Marathon

A 3 metre string AB has a mass of 5kg attached at point B. The string is rotated in a horizontal circle about A and breaks as soon as it exceeds a speed of rotation of 45 revolutions per minute.
i.Find the maximum possible tension in the string.
ii. the mass at B is replaced by a 3kg mass and an additional 1kg mass is attached to the string at C, 2 metres from A (as shown below). Find the new maximum number of revolutions per minute that the string can be rotated.

i need help with part ii now

( thanks
ans: 52.5revs/minute

Drsoccerball · Oct 4, 2015

Re: HSC 2015 4U Marathon

porcupinetree said:
Here's a classic mechanics question:

$\textup{A car travelling at speed v metres per second along a banked}$
$\textup{curve of radius r metres experiences no lateral frictional force.}$
$\textup{Another car of mass m travels along the same curve at speed V}$
$\textup{metres per second, V} > \textup{v. Show that the lateral frictional force exerted}$
$\textup{by the track on this car is equal to} \frac{mg(V^{2} - v^{2})}{\sqrt{v^{4} + r^{2}g^{2}}} \textup {newtons.}$

$Firstly we use the information to calculate an expression for the angle$

$tan \theta = \frac{v^2}{rg}$

$$ LOL took me a while to see this ...$ sec^2 x = \frac{v^4}{r^2g^2} -1 (*)$

$$ Multiply vertical component by sin and horizontal by $ -cos \theta$

$F= m(V^2-v^2) \frac{cos\theta}{r}$

$$ Using (*) $ F= \frac{mg(V^2 -v^2)}{\sqrt{v^4 +r^2g^2}}$

porcupinetree · Oct 4, 2015

Re: HSC 2015 4U Marathon

Paradoxica said:
that's correct, but not as quick. try thinking in the complex plane. also, in the question, I would not have stated the hypotenuse lengths were it not for the confusion.

Drsoccerball · Oct 4, 2015

Re: HSC 2015 4U Marathon

asdfghjklqwerty said:
A 3 metre string AB has a mass of 5kg attached at point B. The string is rotated in a horizontal circle about A and breaks as soon as it exceeds a speed of rotation of 45 revolutions per minute.
i.Find the maximum possible tension in the string.
ii. the mass at B is replaced by a 3kg mass and an additional 1kg mass is attached to the string at C, 2 metres from A (as shown below). Find the new maximum number of revolutions per minute that the string can be rotated.

i need help with part ii now ( thanks
ans: 52.5revs/minute

Is i's answer 37 Newtons?

asdfghjklqwerty · Oct 4, 2015

Re: HSC 2015 4U Marathon

part i was what i just posted before. the answer for it is (135pi^2)/4=333.1 N

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