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STEP,I,II,III 1998 Mark Schemes

STEP III, 1998, Q9.

I don't know what's in the formula booklet, so I'll start by finding the M.I. of a disc radius r about an axis through its center and perpendicular to its plane.

Let

\rho

be the mass per unit area, so the mass of the disc is

\pi \rho r^2

. Divide the disc into annulae, radius x, width dx. Each annulus has mass

2\pi \rho x dx

and so moment of inertia

2\pi \rho x^3 dx

. Thus the disc has M.I. =

\int_0^r 2\pi\rho x^3 dx = \frac{\pi r^4 \rho}{2} = \frac{ma^2}{2}

Now we divide the cone up into discs of thickness dy parallel to the plane of the base. Again, set

\rho

to be the density. Then the disc at height y has radius

\frac{ay}{h}

and so has M.I.

\frac{\pi \rho}{2} (\frac{ay}{h})^4 dy

. Thus the cone has M.I.

\int_0^h \frac{\pi \rho}{2} (\frac{ay}{h})^4 dy = \frac{\pi \rho}{2} (\frac{a}{h})^4 \int_0^h y^4 dy = \frac{\pi \rho}{10}(\frac{a}{h})^4h^5 = \frac{ha^4\pi\rho}{10}

. Since the cone has mass

m = \frac{\pi\rho a^2h}{3}

, the M.I. =

\frac{3ma^2}{10}

as desired.

For the next bit, we're going to need the perpendicular and parallel axis theorems, but they are rather hard to state without diagrams and it's just bookwork. So I won't bother here.

By the perpendicular axis theorem, the M.I. of a disc of mass m and radius r about one of its diameters is

\frac{mr^2}{4}

(to use the theorem, pick two diameters at right angles).

Then by the parallel axis theorem, if you shift the axis of rotation so that instead of being a diameter, it is the same axis moved by a distance y perpendicular to the plane of the disc, then the new M.I. about that axis is

\frac{mr^2}{4}+my^2

Now again, the radius of a disc at a height y from the base is

\frac{ay}{h}

. Such a disc has mass

\pi \rho (\frac{ya}{h})^2 dy

and so it's M.I. about the axis through the apex and parallel to the base is

\pi \rho (\frac{ya}{h})^2[\frac{1}{4}(\frac{ya}{h})^2 + y^2]dy

.

So the cone's MI about that axis is:

Unparseable latex formula:

I = \displaystyle \int_0^h \pi \rho (\frac{ya}{h})^2[\frac{1}{4}(\frac{ya}{h})^2 + y^2]dy \\[br]= \frac{a^2\pi\rho}{4h^2}\int_0^h (\frac{ya}{h})^2 + y^2)y^2dy\\[br]= \frac{a^2\pi\rho}{4h^2}((\frac{a}{h})^2+4)\int_0^h y^4 dy \\[br]= \frac{a^2\pi\rho}{20h^2}((\frac{a}{h})^2+4)h^5 = \frac{a^2h\pi\rho}{20}(a^2+4h^2)

Since the cone has mass

m=\frac{\pi \rho}{3}a^2h

we can write

I = \frac{3m}{20}(a^2+4h^2)

.

For the last part, let

\theta

be the angle of the cone from the vertical, and let

L

be the distance of the c.of.g. of the cone from the apex (point of suspension). Then

I \ddot{\theta} = -mgL \sin \theta

.

Putting in our values for I, L, we have:

\displaystyle \ddot{\theta} = -\frac{3mgh}{4}\frac{20}{3m(a^2+4h^2)}\sin \theta \\ = -\frac{5gh}{a^2+4h^2}\sin \theta \approx -\frac{5gh}{a^2+4h^2}\theta

for small values of theta.

Thus we get simple harmonic motion with period

2\pi \sqrt{\frac{a^2+4h^2}{5gh}}

as required.

Comment: this question is somewhat labourious but not actually difficult. The last part is completely trivial and I imagine was only done to give you a value to check your calculated M.I. against.

Reply 61

DFranklin

18

STEP III 1998 Q13

P(k heads in n tosses) =

^nC_k \frac{1}{2^n} = \frac{n!}{k!(n-k)!2^n}

.

Let

P_n(k)

= P(k heads in n tosses).

P_n(1) = \frac{n!}{(n-1)!2^n} = \frac{n}{2^n}

. Then set

r_n = P_{n+1}(1)/P_{n}(1) = \frac{n+1}{2^{n+1}} \Big/ \frac{n}{2^n} = \frac{n+1}{2n}

Then

r_1 = 1, r_n < 1\, \forall n > 1

. So since

P_n(1) > 0 \,\forall n

, we have

P_1(1)=P_2(1), P_n(1) > P_{n+1}(1) \,\forall n > 1

. Thus we have the MLE's n=1, n=2.

More generally, we have:

P_n(k) = \frac{n!}{k!(n-k)!2^n}

. Again, set

Unparseable latex formula:

r_n = P_{n+1}(k)/P_{n}(k) = \frac{(n+1)!}{k!(n+1-k)!2^{n+1}} \Big/ \frac{n!}{k!(n-k)!2^n} = \frac{n+1}{2(n+1-k)} \\[br]= \frac{2(n+1-k)+2k-n+1}{2(n+1-k)} = 1 + \frac{n-(2k-1)}{2(n+1-k)}

Thus we find:

Unparseable latex formula:

\begin{displaymath}[br]\left\{\begin{array}{ll}[br]r_n > 1 & n < 2k-1 \\[br]r_n = 1 & n = 2k-1 \\[br]r_n < 1 & n > 2k-1 \\[br]\end{array}\right.[br]\end{displaymath}

and we deduce:

Unparseable latex formula:

\begin{displaymath}[br]\left\{\begin{array}{ll}[br]P_n(k) < P_n(k+1) & n < 2k-1 \\[br]P_n(k) = P_n(k+1) & n = 2k-1 \\[br]P_n(k) > P_n(k+1) & n > 2k-1 \\[br]\end{array}\right.[br]\end{displaymath}

In other words,

P_n(k)

increases until

n = 2k-1, \, P_{2k-1}(k)=P_{2k}(k)

and

P_n(k)

decreases thereafter. Thus we find n =2k-1 and n=2k are the two MLEs.

For the last part, our equation for

P_n(k)

becomes

P_n(k) = \frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}

. Playing the same game, we find

r_n = P_{n+1}(k)/P_{n}(k) = \frac{(n+1)!}{k!(n+1-k)!}p^k(1-p)^{n+1-k} \Big/[br]\frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}

=\frac{(n+1)(1-p)}{n+1-k}=\frac{(n+1)(1-p)+(n+1)p-(n+1)p+k-k}{n+1-k}

=\frac{(n+1-k)+k-(n+1)p}{n+1-k} = 1 - \frac{k-p(n+1)}{n+1-k}

Again we deduce

Unparseable latex formula:

\begin{displaymath}[br]\left\{\begin{array}{ll}[br]P_n(k) < P_n(k+1) & p(n+1) < k \text{ i.e. } n < k/p - 1\\[br]P_n(k) = P_n(k+1) & p(n+1) = k \text{ i.e. } n = k/p - 1\\[br]P_n(k) > P_n(k+1) & p(n+1) > k \text{ i.e. } n > k/p - 1\\[br]\end{array}\right.[br]\end{displaymath}

So if k/p is an integer, we have the 2 MLEs n = k/p -1, n=k/p.
If k/p is not an integer, then if n is the integer part of k/p we have n-1 < k/p-1 and n > kp-1 and so

P_{n-1}(k) < P_n(k) > P_{n+1}(k)

and so n is our unique MLE. Explicitly, the MLE is unique if k/p is not an integer.

Reply 62

justinsh

4

I don't agree with datr for problem 4 on STEP 98 with the two last lines. Indeed, cos(n*pi) depends on the parity of n...

datr

\displaystyle I_n - I_{n-1} = -\frac{2}{n^2}[cos(nx)]_0^\pi = \frac{4}{n^2}

\displaystyle I_n = \sum_{r=1}^{n} \frac{4}{r^2}

Justin

Reply 63

datr

12

Good catch. Obviously it should be:

\displaystyle I_n - I_{n-1} = \frac{2}{n^2}(1 - cosn\pi)

Then I'm not sure if it can simplified much further, which would leave the sum as.

\displaystyle I_n = 2\sum_{r=1}^{n} \frac{1}{r^2}(1 - cosr\pi)

Reply 64

DFranklin

18

Well, in that case, you are just summing over "r odd", which you can rearrange into a sum looking like

4\sum \frac{1}{(2r-1)^2}

, although you'll need to take some care over the limits.

Reply 65

dr_98_98

STEP III Qn 11:

Referring to my attached diagram:

For the first part use the conservation of energy:

PE at A = KE at B

PE at A=

mgl(cos\theta - cos\gamma)

KE at B =

\frac{1}{2}mv^2

As the particle describes a circle, so we can make use of

v =l\omega

where

\omega

is the angular velocity i.e.

\omega = \frac{d\theta}{dt}

So KE at B =

\frac{1}{2}ml^2(\frac{d\theta}{dt})^2

Substituting in and cancelling an

m

and and

l

gives

\frac{1}{2}l(\frac{d\theta}{dt})^2 = g(cos\theta - cos\gamma)

The period P is the time taken for the pendulum to move from an angular displacement of

\gamma

radians to

-\gamma

radians and all the way back to

\gamma

radians (two swings). Thus P/4 is the time taken for the pendulum to move from an angular displacement of

\gamma

radians to 0 radians

Substituting

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

and

cos\gamma = 1 - 2sin^2(\frac{\gamma}{2})

gives

\frac{1}{2}l(\frac{d\theta}{dt})^2 = 2g(sin^2(\frac{\gamma}{2}) - sin^2(\frac{\theta}{2}))

By separating variables and square rooting

\frac{d\theta}{dt} = 2(\frac{g}{l}(sin^2(\frac{\gamma}{2}) - sin^2(\frac{\theta}{2}))^{0.5}

So

Unparseable latex formula:

2\displaystyle\int^{P/4}_0 dt \, \dt = \sqrt \frac{l}[br]{g} \displaystyle\int^{\gamma}_0 (sin^2(\frac{\gamma}{2})^2 - sin^2(\frac{\theta}{2}))^{-0.5} d{\theta}\, \dt

Integrating the left hand side to get

P/2

gives

Unparseable latex formula:

P = 2\sqrt \frac{l}[br]{g} \displaystyle\int^{\gamma}_0 (sin^2(\frac{\gamma}{2}) - sin^2(\frac{\theta}{2}))^{-0.5} d{\theta}\, \dt

Using the suggested subsitution gives

d\theta = 2sin(\frac{\gamma}{2})sec(\frac{\theta}{2})cos\phi d\phi

the limits change from

\gamma

and

0

to

\frac{\pi}{2}

and

0

(sin^2(\frac{\gamma}{2}) - sin^2(\frac{\theta}{2}))^{-0.5}

becomes

cosec(\frac{\gamma}{2})sec\phi

so that

Unparseable latex formula:

P = 4\sqrt \frac{l}[br]{g} \displaystyle\int^{\pi/2}_0 sec(\frac{\theta}{2}) d{\phi}\, \dt

but

sec(\frac{\theta}{2}) = (1 - sin^2(\frac{\theta}{2}))^{-0.5} = (1 - sin^2(\frac{\gamma}{2})sin^2\phi)^{-0.5}

So

Unparseable latex formula:

P = 4\sqrt \frac{l}{g} \displaystyle\int^{\pi/2}_0 (1 - sin^2(\frac{\gamma}{2})sin^2\phi)^{-0.5} d{\phi}\, \dt

Expanding

(1 - sin^2(\frac{\gamma}{2})sin^2\phi)^{-0.5}

for the first two terms gives

Unparseable latex formula:

P \approx 4\sqrt \frac{l}{g} \displaystyle\int^{\pi/2}_0 (1 + \frac{1}{2} sin^2(\frac{\gamma}{2})sin^2\phi) d{\phi}\, \dt

Substituting

cos2\phi = 1 - 2sin^2(\phi)

Unparseable latex formula:

P \approx 4\sqrt \frac{l}{g} \displaystyle\int^{\pi/2}_0 (1 + \frac{1}{4} sin^2(\frac{\gamma}{2}) - \frac{1}{4}sin^2(\frac{\gamma}{2})cos(2\phi)) d{\phi}\, \dt

Integrating gives

P \approx 4\sqrt \frac{l}{g} ((1+\frac{1}{4} sin^2(\frac{\gamma}{2}))\phi - \frac{1}{8}sin^2(\frac{\gamma}{2})sin(2\phi))

and putting in the limits

P \approx 4\sqrt \frac{l}{g} (1+\frac{1}{4} sin^2(\frac{\gamma}{2})(\frac{\pi}{2})

P \approx 2\pi\sqrt \frac{l}{g} (1+\frac{1}{4} sin^2(\frac{\gamma}{2}))

For small values of

\frac{\gamma}{2}

sin(\frac{\gamma}{2}) \approx [br]\frac{\gamma}{2}

so

sin^2(\frac{\gamma}{2}) \approx [br]\frac{\gamma^2}{4}

Hence

P \approx 2\pi\sqrt \frac{l}{g} (1+\frac{\gamma^2}{16})

I normally stay away from the mechanics questions but I'll have a go at STEP II Q11 as it looks easy enough.

R(&#8593

Initial Velcoity = usinα
Acceleration = -9.8

y = ut\sin\alpha - 4.9t^2

R(&#8594

Initial Velocity = ucosα

x = ut\cos\alpha

Notice that tanβ will vary in the same way as tanθ but in the opposite direction. Therefore:

\frac{d}{dt}\tan\beta = -\frac{d}{dt}\tan\theta

\tan\beta = \frac{y}{x} = \frac{ut\sin\alpha - 4.9t^2}{ut\cos\alpha} = \tan\alpha - \frac{4.9t}{u}\sec\alpha

\frac{d}{dt}\tan\beta = -\frac{4.9}{u}\sec\alpha

\frac{d}{dt}\tan\theta = \frac{4.9}{u}\sec\alpha

Which is clearly constant.

For the second part of the question change the frame of reference so that the girl is once again stationary which adds an extra horizontal component v to the speed of the ball.

R(&#8594 :wink:

Initial Velocity = ucosα + v

x = (u\cos\alpha + v)t

Now we use the same argument again.

\tan\beta = \frac{y}{x} = \frac{ut\sin\alpha - 4.9t^2}{(u\cos\alpha + v)t} = \frac{u\sin\alpha - 4.9t}{u\cos\alpha + v}

\frac{d}{dt}\tan\beta = -\frac{4.9}{u\cos\alpha + v}

\frac{d}{dt}\tan\theta = \frac{4.9}{u\cos\alpha + v}

Which again is constant.

Reply 67

Swayum

18

DFranklin

Paper I, Q6:

Unparseable latex formula:

a_1 = \cos x, b_1 = 1\\[br]a_2 = \frac{1}{2}(1 + \cos x) = \cos^2(x/2), b_2 = \sqrt{a_2} = \cos(x/2)\\

Sorry but how did you make the jump from 1/2(1 + cosx) to cos^2(x/2) so easily? Is there some identity I'm missing?

Reply 68

insparato

15

Yes its very useful in C4 and FP1-3.

cos2x = cos^2x - sin^2x = 2cos^2x - 1

1 + cos2x = 2cos^2x

cos^2x = \frac{1}{2}(1+cos2x)

cos^2\frac{x}{2} = \frac{1}{2}(1+cosx)

Reply 69

Swayum

18

Oh ok thanks, I'll try to keep that in mind.

Reply 70

v-zero

14

I really have a problem with the STEP 1 question 3 part iii) answer. I believe it to be true, since for any given value a power series can be constructed to a finite degree which approximates cos(theta) to a degree of accuracy better than six decimal places, since the taylor series to infinity holds for all values of theta exactly.

I can see where the ukgea is coming from with his answer, but I think the question is too ambiguous to avoid this problem.

Reply 71

Dystopia

10

v-zero

I really have a problem with the STEP 1 question 3 part iii) answer. I believe it to be true, since for any given value a power series can be constructed to a finite degree which approximates cos(pheta) to a degree of accuracy better than six decimal places, since the taylor series to infinity holds for all values of pheta exactly.

I can see where the ukgea is coming from with his answer, but I think the question is too ambiguous to avoid this problem.

I do not think the question is ambiguous. There does not exist a polynomial which satisfies

|P(\theta) - \cos\theta| \leq 10^{-6} \; \forall \; \theta

as

\displaystyle \lim_{\theta \to \infty} |P(\theta)| = \infty

Given a certain value of

\theta

, a polynomial satisfying the condition can be found, but for any polynomial a value of

\theta

can be chosen such that the condition no longer holds.

Reply 72

Swayum

18

A polynomial has to have a finite number of terms though. Otherwise, it isn't polynomial.

Reply 73

v-zero

14

Swayam

A polynomial has to have a finite number of terms though. Otherwise, it isn't polynomial.

That was not my point, simply that for any numerical value of theta there is a polynomial that satisfies the identity. However, I can see both sides of the argument and do not have a problem accepting the answer given as an alternative to my own argument - it is the only one of those questions that I think was badly worded.

Reply 74

Zhen Lin

12

A minor nitpick, but you do know it's theta right?

Reply 75

v-zero

14

Yes sorry, I don't know why I wrote it like an idiot, fixing...

Reply 76

DFranklin

18

v-zero

That was not my point, simply that for any numerical value of theta there is a polynomial that satisfies the identity. However, I can see both sides of the argument and do not have a problem accepting the answer given as an alternative to my own argument - it is the only one of those questions that I think was badly worded.

I'm afraid I don't think there's any ambiguity here. Your argument:

for any numerical value of theta there is a polynomial that satisfies the identity

would be valid if they had said:

"For all real

\theta

, there exists a polynomial P such that

|P(\theta)-\cos \theta| \le 10^{-6}

",

but this is a very different statement from

"There exists a polynomial P such that

|P(\theta)-\cos \theta| \le 10^{-6}

, for all real

\theta

",

which is what they actually said.

Reply 77

Glutamic Acid

17

I think I've got a rather different method for I/4.

Call the lengths of the rectangle x and y, and the radius of the circle r. Since if the rectangle's vertices didn't lie on the circle it could simply be enlarged, the vertices will lie on the circle. The diagonal cannot be longer than the diameter, so

\sqrt{x^2+y^2} = 2r \Rightarrow x^2 + y^2 = 4r^2

We want to maximize Z = x + y (Z being the perimeter).
So

y = \sqrt{4r^2-x^2}

Substituting this in, we get

Z = x + \sqrt{4r^2-x^2}

.
Differentiating, we get

\dfrac{dZ}{dx} = 1 - \dfrac{x}{\sqrt{4r^2-x^2}}

\sqrt{4r^2-x^2} = x \Rightarrow x^2 = r^2 - x^2 \Rightarrow x^2 = 2r^2 \Rightarrow x = \sqrt{2}r.

Substituting back in, we get

y = \sqrt{2}r

x = y, hence it is a square.

Considering the square of x and y, Z = x^2 + y^2 and x^2 + y^2 = 4r^2. Z = 4r^2, independent of x and y.

Considering the cube of x and y.

Z = x^3 + y^3, x^2 + y^2 = 4r^2.

Once again,

y = \sqrt{4r^2-x^2}

Therefore

Z = x^3 + (\sqrt{4r^2-x^2})^3 = x^3 + (4r^2-x^2)\sqrt{4r^2-x^2}

. Differentiating this, we get

\dfrac{dZ}{dx} = (-3x)(\sqrt{4r^2-x^2}) = 0.

. Remembering x = 0 as a solution, 4r^2 - x^2 = 0. Hence x^2 = 4r^2, and x = 2r. But this leads to y = 0. Instead, we can say as x or y tends to 2r and hence x or y tends to 0, the sum of the cubes of the sides will increase.

Hope this is right!

Reply 78

lordcrusade9

1

I have a question about question 7 STEP III

I have trouble with the very last part. surely K varies as T changes.

but K = -6 (1+1/2^4 + 1/3^4 + 1/4^4 ... )

which is a constant.

Reply 79

DFranklin

18

No, the whole point of the question is that K is to be a constant independent of T.

You need to post your working.

STEP,I,II,III 1998 Mark Schemes

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