is also zero, as it should be.

I found your answer insightful but I have a less complex way to construct it:

The projection of the weight of the rod onto the line of greatest slope in the plane of the ring is

W\sin{\alpha}

.

Taking the centre of the ring, O, as the pivot point, we can just take moments first in the plane of the ring and then perpendicular to it:

\left(W \sin{\alpha}\right) \left(\frac{b}{2} \sin{\theta}\right) = fa

\left(W\cos{\alpha}\right) \frac{b}{2} = Na

Replacing

f

with

\mu N

and dividing the two equations gives the same answer as above.

(edited 11 years ago)

Reply 264

ryanboi

Can someone here please explain the last part for Question 5, STEP 1, 1987 paper. I dont understand the solution. Thanks in advance

Reply 265

josephtsui

Is there a mistake in the solution for Question 10, Step 1, 1987 paper? When solving for the modulus of elasticity, the author divided the extension by 2*pi, it should be 2arcsin(1/2) instead of arcsin(1/2).

Thanks in advance.

Reply 266

brianeverit

Original post by josephtsui

I do agree that there is an error in the solution.Here is my amended solution (without the diagram)
Resolving vertically,

2T\cos\theta=mg

and

\sin\theta=\frac{1}{x}

\cos\theta=\frac{\sqrt(x^2-1)}{x}

Length of unstretched string=

\pi+2\arcsin\frac{1}{x}+2\sqrt(x^2-1)

So by Hooke's law,

T=\frac{\lambda(2\theta+2\sqrt(x^2-1)-\pi)}{2\pi}

\mathrm{i.e.}\frac{mg}{2\cos \theta}=\frac{ \lambda}{\pi}\left(\theta+\sqrt(x^2-1)- \frac {\pi}{2}\right)

x=2\rightarrow\cos\theta=\frac{\sqrt3}{2}\rightarrow\theta=\frac{\pi}{6}

\frac{mg}{\sqrt3}=\frac{\lambda}{\pi}\left(\frac{\pi}{6}+\sqrt3-\frac{\pi}{2}\right)=\frac {\lambda}{\pi}\left(\sqrt3-\frac{\pi}{3}\right)

\mathrm{hence} \lambda= \frac{mg}{\sqrt3}\left(\sqrt3-\frac{\pi}{3}\right) =\frac{3mg}{9-\pi}

Reply 267

abra-cad-abra

Original post by coffeym

STEP I Q3

Our differential equation, referred to from here on as (*) is:

\displaystyle x^3\frac{dv}{dx}+x^2v=\frac{1+x^2v^2}{(1+x^2)v}

Let

\displaystyle y=xv \implies v=\frac{y}{x}

\displaystyle \implies \frac{dv}{dx}=\frac{x\frac{dy}{dx}-y}{x^2}

Hence in (*):

\displaystyle x^2\frac{dy}{dx}=\frac{x(1+y^2)}{y(1+x^2)}

After cancelling an

x

and rearranging, (*) reduces to:

\displaystyle \frac{y}{1+y^2}\frac{dy}{dx}=\frac{1}{x(1+x^2)}

Hence, after dealing with partial fractions (which should be routine)

\displaystyle \int{\frac{y}{1+y^2}}\,dy=\int{\frac{1}{x}-\frac{x}{1+x^2}}\,dx

Thus

\displaystyle \frac12 \ln{(1+y^2)}=\ln{|x|}-\frac12 \ln{(1+x^2)}+c

Now

x=1,v=1 \implies y=1 \implies c=\ln{2}

\displaystyle \therefore \ln{(1+y^2)}=\ln{\left(\frac{4x^2}{1+x^2}\right)}

\displaystyle \implies 1+x^2v^2=\frac{4x^2}{1+x^2}

\displaystyle \implies x^2v^2=\frac{3x^2-1}{1+x^2}

\displaystyle \implies v^2=\frac{3}{1+x^2}-\frac{1}{x^2(1+x^2)}

Hence as

x \to \infty, v \to 0

This completes the question, although I would like someone to have a quick scan for errors as I did this quite roughly.

wow thats somne mad academics

Reply 268

brianeverit

Original post by abra-cad-abra

wow thats somne mad academics

Looks o.k. if you correct the latex errors in lines 7 and 9, though I think you should give

v=\frac{1}{x} \sqrt { \frac{4x^2}{1+x^2}-1}= \frac{1}{x} \sqrt{ \frac{3x^2-1}{1+x^2} }

since question asks for the solution v(x)

Reply 269

abra-cad-abra

Original post by kabbers

Q I/8:

don't blame me for any mistakes, i've learnt c3&4 myself in the past month and a bit, still a little slow in some areas! :wink:

please check for any!

part one : [thanks nota bene!]: x = 1/t will not work when x = 0, which is sure to happen since the integration is between bounds of -1 & 1.

Part two part one:

Integrate

\displaystyle\int^1_{-1} \frac{1}{(1+x^2)^2} \, \mathrm{d}x

Let's try a substitution of

x

with

tan\theta

Thus

x = tan\theta

\frac{\mathrm{d}x}{\mathrm{d}\theta} = sec^2 \theta

\mathrm{d}x = sec^2 \theta \cdot \mathrm{d}\theta

Find the bounds of the new integral:

1 = tan\theta

, so the top one will be

\pi/4

-1 = tan\theta

, so the bottom one will be

-\pi/4

Perform the substitution:

=\displaystyle\int^{\pi/4}_{-\pi/4} \frac{sec^2 \theta}{(1+tan^2 \theta)^2} \, \mathrm{d}\theta

tan^2 \theta + 1 = sec^2 \theta

, so:

=\displaystyle\int^{\pi/4}_{-\pi/4} \frac{sec^2 \theta}{sec^4 \theta} \, \mathrm{d}\theta

=\displaystyle\int^{\pi/4}_{-\pi/4} \frac{1}{sec^2 \theta} \, \mathrm{d}\theta

=\displaystyle\int^{\pi/4}_{-\pi/4} cos^2 \theta \, \mathrm{d}\theta

Use the identity

cos 2\theta = cos^2 \theta - sin^2 \theta

cos^2 \theta = 0.5(cos (2\theta) + 1) = 0.5cos (2\theta) + 0.5

=\displaystyle\int^{\pi/4}_{-\pi/4} 0.5cos (2\theta) + 0.5 \, \mathrm{d}\theta

=[0.25sin (2\theta) + 0.5\theta]^{\pi/4}_{-\pi/4}

= 0.5 + \pi/4

Part two part two:

Integrate

\displaystyle\int^1_{-1} \frac{-t^2}{(1+t^2)^2} \, \mathrm{d}t

Let's try (again) a substitution of

t

with

tan\theta

Thus

t = tan\theta

\frac{\mathrm{d}t}{\mathrm{d}\theta} = sec^2 \theta

\mathrm{d}t = sec^2 \theta \cdot \mathrm{d}\theta

Find the bounds of the new integral:

t = tan\theta

1 = tan\theta

, so the top one will be

\pi/4

-1 = tan\theta

, so the bottom one will be

-\pi/4

Perform the substitution:

=\displaystyle\int^{1}_{-1} \frac{-tan^2 \theta}{(1+tan^2 \theta)^2} \, \mathrm{d}t

=\displaystyle\int^{\pi/4}_{-\pi/4} \frac{-tan^2 \theta}{sec^4 \theta} \cdot sec^2\theta \, \mathrm{d}\theta

=\displaystyle\int^{\pi/4}_{-\pi/4} \frac{-tan^2 \theta}{sec^2 \theta} \, \mathrm{d}\theta

=\displaystyle\int^{\pi/4}_{-\pi/4} -tan^2\theta \cdot cos^2 \theta \, \mathrm{d}\theta

=\displaystyle\int^{\pi/4}_{-\pi/4} -sin^2 \theta \, \mathrm{d}\theta

Use the identity

cos 2\theta = cos^2 \theta - sin^2 \theta

sin^2 \theta = 0.5(1 - cos (2\theta)) = - 0.5cos (2\theta) + 0.5

=\displaystyle\int^{\pi/4}_{-\pi/4} -(-0.5cos (2\theta) + 0.5) \, \mathrm{d}\theta

=\displaystyle\int^{\pi/4}_{-\pi/4} 0.5cos (2\theta) - 0.5 \, \mathrm{d}\theta

=[0.25sin (2\theta) - 0.5\theta]^{\pi/4}_{-\pi/4}

= 0.5 - \pi/4

for part 1 of the question can you just explain t is undefined at x=0 which is in the range . or ddo you have to say this and algebraicaly show that the first integral does become the second itnegral but still because of the substituon undefined at x=0 it wont work

Reply 270

abra-cad-abra

Original post by brianeverit

STEP I numbers 13-16
Again if anyone could check them I would be most grateful.

For q13 you have to draw an x=t graph.

how can you have negative time on yours?

and for x<0 surely you cant use the initial conditions for x=0.

(edited 10 years ago)

Reply 271

abra-cad-abra

Original post by brianeverit

STEP I numbers 13-16
Again if anyone could check them I would be most grateful.

you say the pdf is 2/pi and then go on to use 1/pi? for q 13

im afraid i disagree in my approach to the question. i said X=sqrt(2-2x) and x is unform and has pdf 1/2 on [-1,1] which got me to E(X)=4/3

your approach looks fine and mine looks suitable to me. where have i gone wrong?

(edited 10 years ago)

Reply 272

abra-cad-abra

Original post by brianeverit

1987 STEP I number 11
Just spotted a mistake in my solution in post number 166 so here is the revised version.

\text{Resolving radially for circular motion }mg\cos \theta-R=\dfrac{mv^2}{r}

\text{By conservation of energy }\dfrac{1}{2} mv^2=mgr(1- \cos \theta) \text{ so }mg \cos \theta-R=2mg(1- \cos \theta)

\Rightarrow R=mg(3 \cos \theta-2)

\text{particle loses contact when }R=0 \text{ i.e. } \theta=\text{arc}\cos \dfrac{2}{3} \text{ which is the angle of the velocity to the horizontal}

\text{From start, particle now falls a distance }2r \text{ so loss of potential energy is }2mgr

\text{horizontal component of velocity is }v \cos \theta=\dfrac{2v}{3} \text{ so if vertical component is }V \text{ then gain in kinetic energy is }

\dfrac{1}{2}m\left(\dfrac{4v^2}{9}+V^2 \right)\text{ and } v^2=2gr(1- \cos \theta)=\dfrac{2}{3}gr

\text{hence }\dfrac{4v^2}{9}+V^2=4gr \Rightarrow V^2=4 \left(gr-\dfrac{v^2}{9} \right) \text{ so }V^2=4gr\left(1-\dfrac{2}{27} \right)=\dfrac{100gr}{27} \Rightarrow V=\dfrac{10}{3} \sqrt{ \dfrac{gr}{3}}

Unparseable latex formula:

\text{so vertical component of momentu at floor is }\dfrac{10m}{3}\sqrt \dfrac{gr}{3}}

why do you consider gpe from the start rather than the point at whcih the ball loses contact as it is there that it has that horizontal component of velocity

Reply 273

abra-cad-abra

Original post by brianeverit

1987 STEP I number 13
Just spotted a mistake in my solution in post number 167 so here is the revised version.

\text{For }x \geq 0, m\dfrac{d^2x}{dt^2}=-m \mu^2x \Rightarrow \dfrac{d^2x}{dt^2}+\mu^2x=0 \Rightarrow x=A \cos \mu t+B \sin |mu t

\text{at }t=0,x=0 \text{ and } \dfrac{dx}{dt}=v_0 \text{ so }A=0 \text{ and }B=\dfrac{v_0}{\mu} \text{ i.e. }x= \dfrac{v_0}{\mu} \sin \mu t

\text{For }x<0, \dfrac{d^2x}{dt^2}=-\kappa \dfrac{dx}{dt} \Rightarrow \dfrac{d^2x}{dt^2}+\kappa \dfrac{dx}{dt}=0 \Rightarrow x=C+D\text{e}^{-\kappa t}

Unparseable latex formula:

\text{at }t= \dfrac{\pi}{\mu}, x=o \text{ and } \dfrac{dx}{dt}=-v_0 \text{ so } C+D \text{e}^{-\frac{\kappa \pi}{\mu}}=0\text{ and }- \kappa D\text{e}^{-\frac{\kappa \pi}{\mu} }=-v_0 [br]\text{ so }D=\dfrac{v_0}{\kappa}\text{e}^[\frac{\kappa \pi}{\mu} }\text { and }C=-\dfrac{v_0}{\kappa}

\text{and so }x= \dfrac{v_0}{\kappa} \left[ \text{e}^{\kappa(\frac{\pi}{\mu}-1)}-1\right]

Original post by brianeverit

1987 STEP I number 13
Just spotted a mistake in my solution in post number 167 so here is the revised version.

\text{For }x \geq 0, m\dfrac{d^2x}{dt^2}=-m \mu^2x \Rightarrow \dfrac{d^2x}{dt^2}+\mu^2x=0 \Rightarrow x=A \cos \mu t+B \sin |mu t

\text{at }t=0,x=0 \text{ and } \dfrac{dx}{dt}=v_0 \text{ so }A=0 \text{ and }B=\dfrac{v_0}{\mu} \text{ i.e. }x= \dfrac{v_0}{\mu} \sin \mu t

\text{For }x<0, \dfrac{d^2x}{dt^2}=-\kappa \dfrac{dx}{dt} \Rightarrow \dfrac{d^2x}{dt^2}+\kappa \dfrac{dx}{dt}=0 \Rightarrow x=C+D\text{e}^{-\kappa t}

Unparseable latex formula:

\text{and so }x= \dfrac{v_0}{\kappa} \left[ \text{e}^{\kappa(\frac{\pi}{\mu}-1)}-1\right]

just seen this. agree with redone conditions

but again how have you drawn the graph. negative time???

and doesnt it fluctuate between x>=0 and x<0 every pi/u and kpi/u so you would have to draw a section of the graph?

Reply 274

brianeverit

Original post by abra-cad-abra

For q13 you have to draw an x=t graph.

how can you have negative time on yours?

and for x<0 surely you cant use the initial conditions for x=0.

The particle was moving before the time started to be recorded so I believe

Reply 275

brianeverit

Original post by abra-cad-abra

How do you get X=sqrt(2-2x)? What is x? If it is a trig function then it is not uniform.

Reply 276

abra-cad-abra

Original post by brianeverit

How do you get X=sqrt(2-2x)? What is x? If it is a trig function then it is not uniform.

no i said the point on x axis has uniform distribution. the y coord for each x is sqrt(1-x^2) or the negative of this but as the point from which we are measuring distance on lies on x axis we can just consider y>0

then i used pythagoras to get a general function for distance from (1,0) to (x,sqrt(x^2-1)

it gave some pretty close answers to urs but clearly very different

Reply 277

abra-cad-abra

Original post by brianeverit

The particle was moving before the time started to be recorded so I believe

what is on your x axis and what on your y axis

Reply 278

brianeverit

Original post by abra-cad-abra

The distance measured on the x axis does NOT have a uniform distribution.
It is the distance measured along the curve that has one.

Reply 279

abra-cad-abra

Original post by brianeverit

The distance measured on the x axis does NOT have a uniform distribution.
It is the distance measured along the curve that has one.

What do you mean by the distance measured on the x axis?

as the question says a point is chosen randomly. meaning their x coorcdinate is chosen randomly on [-1,1] so there is an equal probability for all values on this domain which is the whole domain considered int he question.

so x has uniform distribution 1/2 on [-1,1] . im not saying the distance has uniform distribution, rather the x-coord of the point has uniform distribution. which is surely correct no, thats just what you have done with theta

from this g(x)=sqrt(2-2x). and we want E[g(x)] and Var[g(x)]