1987 Specimen STEP solutions thread Watch

Dirac Spinor
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Lol, all 4 mechanics questions ftw.
Does anyone else want to do the honours of finishing off III?
I might be able to bu right now I'm feeling sleepy and don't trust my answers. A second ago I thought the area of a triangle was base*height ...
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Farhan.Hanif93
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STEP III Q9
...
Let x_i, y_i, z_i \in \mathbb{R} and \mathbf{M} =\begin{pmatrix} 1 & 0 & 0\\ x_1 &1 &0 \\ y_1 & z_1 & 1 \end{pmatrix} and \mathbf{N} = \begin{pmatrix} 1 & 0 & 0\\ x_2 &1 &0 \\ y_2 & z_2 & 1 \end{pmatrix} . Clearly \mathbf{M}, \mathbf{N} \in G

Closure
Note that \mathbf{MN} = \mathbf{M} \times \mathbf{N} = \begin{pmatrix} 1 & 0 & 0\\ x_1+x_2 &1 &0 \\ y_1+y_2 + z_1x_2 & z_1 + z_2 & 1 \end{pmatrix} \Rightarrow \forall \mathbf{M}, \mathbf{N} \in G,  \mathbf{MN} \in G

Hence G is closed under matrix multiplication. _\square

Identity

Note that \exists \mathbf{I}_3 = \begin{pmatrix} 1 & 0 & 0\\ 0 &1 &0 \\ 0 & 0 & 1 \end{pmatrix} \in G s.t. \mathbf{M}\mathbf{I}_3 = \mathbf{I}_3\mathbf{M} = \mathbf{M}, which is easily demonstrated to be unique. Hence G has a unique identity element, as required.

Inverse

Note that the cofactor matrix of \mathbf{M} is given by \mathbf{C}_M = \begin{pmatrix} 1 & -x & xz-y \\ 0 &1 &-z \\ 0 & 0 & 1 \end{pmatrix}.

Multiplying the corresponding first row elements of \mathbf{M} and \mathbf{C}_M yields that \det {\mathbf{M}}=1. This means that \mathbf{M}^{-1} = \mathbf{C}_M^T = \begin{pmatrix} 1 & 0 & 0\\ -x &1 &0 \\ xz-y & -z & 1 \end{pmatrix} \in G.

Note that all nxn matrices are invertible iff their determinant is nonzero, which is the case here. Hence all matrices in G are invertible.

Take an invertible nxn matrix \mathbf{A} such that (1) \mathbf{AB}=\mathbf{BA} = \mathbf{I}_n and (2) \mathbf{AC}=\mathbf{CA} = \mathbf{I}_n.

(1) implies that \mathbf{CI}_n = \mathbf{C}= \mathbf{C(AB)} = \mathbf{(CA)B} by the associativity of matrix multiplication and thus, by (2), \mathbf{B}=\mathbf {C}. If follows that any invertible nxn matrix has a unique multiplicative inverse.

Hence it is shown that \forall \mathbf{M} \in G, \exists !\mathbf{M}^{-1} \in G s.t. \mathbf{MM}^{-1} = \mathbf{M^{-1}M} = \mathbf{I}_3, as required. _\square




It follows that G is a group under matrix multiplication, as required. _\square.

Let H be the subset of G containing the 3x3 matrices of the above form with x,y,z \in \mathbb{Z}.

Note that x_i, y_i, z_i \in \mathbb{Z} \implies \{x_1 + x_2, z_1+z_2, y_1 + y_2 + x_1z_2\} \subset \mathbb{Z}. This means that \mathbf{MN} \in H and therefore H is closed.

Taking x=y=z=0 implies that there exists a unique identity in H.

Also x,y,z \in \mathbb{Z} \implies \{-x, -z, xz-y \} \subset \mathbb{Z} , hence \mathbf{M}^{-1} \in H, which can be shown to be unique for each element of H.

It follows that \boxed{H\leq G}. _\square

Considering the group G, note that \mathbf{MN} = \mathbf{NM} \iff [\mathbf{MN}]_{3,1} = [\mathbf{NM}]_{3,1} \iff x_1z_2 = x_2z_1. Taking x_1=z_1=0, y_1\not =0 ensures that there exists a matrix \mathbf{M} of the form \begin{pmatrix} 1 & 0 & 0 \\ 0 &1 &0 \\ y_1 & 0 & 1 \end{pmatrix} in G s.t \forall \mathbf{N}\in G, \boxed{\mathbf{MN}=\mathbf{NM}}, as required(?).

[To be honest, I'm not entirely sure what they want me to say for this last part.]

In fact, I've done very little on groups so it would be nice if anyone more knowledgeable could check through this to make sure it's right.


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Dirac Spinor
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(Original post by Farhan.Hanif93)
STEP III Q9
...
Let x_i, y_i, z_i \in \mathbb{R} and \mathbf{M} =\begin{pmatrix} 1 & 0 & 0\\ x_1 &1 &0 \\ y_1 & z_1 & 1 \end{pmatrix} and \mathbf{N} = \begin{pmatrix} 1 & 0 & 0\\ x_2 &1 &0 \\ y_2 & z_2 & 1 \end{pmatrix} . Clearly \mathbf{M}, \mathbf{N} \in G

Closure
Note that \mathbf{MN} = \mathbf{M} \times \mathbf{N} = \begin{pmatrix} 1 & 0 & 0\\ x_1+x_2 &1 &0 \\ y_1+y_2 + z_1x_2 & z_1 + z_2 & 1 \end{pmatrix} \Rightarrow \forall \mathbf{M}, \mathbf{N} \in G,  \mathbf{MN} \in G

Hence G is closed under matrix multiplication. _\square

Identity

Note that \exists \mathbf{I}_3 = \begin{pmatrix} 1 & 0 & 0\\ 0 &1 &0 \\ 0 & 0 & 1 \end{pmatrix} \in G s.t. \mathbf{M}\mathbf{I}_3 = \mathbf{I}_3\mathbf{M} = \mathbf{M}, which is easily demonstrated to be unique. Hence G has a unique identity element, as required.

Inverse

Note that the cofactor matrix of \mathbf{M} is given by \mathbf{C}_M = \begin{pmatrix} 1 & -x & xz-y \\ 0 &1 &-z \\ 0 & 0 & 1 \end{pmatrix}.

Multiplying the corresponding first row elements of \mathbf{M} and \mathbf{C}_M yields that \det {\mathbf{M}}=1. This means that \mathbf{M}^{-1} = \mathbf{C}_M^T = \begin{pmatrix} 1 & 0 & 0\\ -x &1 &0 \\ xz-y & -z & 1 \end{pmatrix} \in G.

Note that all nxn matrices are invertible iff their determinant is nonzero, which is the case here. Hence all matrices in G are invertible.

Take an invertible nxn matrix \mathbf{A} such that (1) \mathbf{AB}=\mathbf{BA} = \mathbf{I}_n and (2) \mathbf{AC}=\mathbf{CA} = \mathbf{I}_n.

(1) implies that \mathbf{CI}_n = \mathbf{C}= \mathbf{C(AB)} = \mathbf{(CA)B} by the associativity of matrix multiplication and thus, by (2), \mathbf{B}=\mathbf {C}. If follows that any invertible nxn matrix has a unique multiplicative inverse.

Hence it is shown that \forall \mathbf{M} \in G, \exists !\mathbf{M}^{-1} \in G s.t. \mathbf{MM}^{-1} = \mathbf{M^{-1}M} = \mathbf{I}_3, as required. _\square




It follows that G is a group under matrix multiplication, as required. _\square.

Let H be the subset of G containing the 3x3 matrices of the above form with x,y,z \in \mathbb{Z}.

Note that x_i, y_i, z_i \in \mathbb{Z} \implies \{x_1 + x_2, z_1+z_2, y_1 + y_2 + x_1z_2\} \subset \mathbb{Z}. This means that \mathbf{MN} \in H and therefore H is closed.

Taking x=y=z=0 implies that there exists a unique identity in H.

Also x,y,z \in \mathbb{Z} \implies \{-x, -z, xz-y \} \subset \mathbb{Z} , hence \mathbf{M}^{-1} \in H, which can be shown to be unique for each element of H.

It follows that \boxed{H\leq G}. _\square

Considering the group G, note that \mathbf{MN} = \mathbf{NM} \iff [\mathbf{MN}]_{3,1} = [\mathbf{NM}]_{3,1} \iff x_1z_2 = x_2z_1. Taking x_1=z_1=0, y_1\not =0 ensures that there exists a matrix \mathbf{M} of the form \begin{pmatrix} 1 & 0 & 0 \\ 0 &1 &0 \\ y_1 & 0 & 1 \end{pmatrix} in G s.t \forall \mathbf{N}\in G, \boxed{\mathbf{MN}=\mathbf{NM}}, as required(?).

[To be honest, I'm not entirely sure what they want me to say for this last part.]

In fact, I've done very little on groups so it would be nice if anyone more knowledgeable could check through this to make sure it's right.


Unfortunately I'm not one of those knowledgeable people but I don't understand your notation in the last part. What does [\mathbf{MN}]_{3,1} mean?

What I did for the last part was a bit simple but I simply multiplied two matrices in this form and then inverted the order and equated entries. This showed that one of the entries didn't matter so if you let the other two be 0 but this other one non zero then commutativity would hold for all B. I don't know whether it's valid...
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DFranklin
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What either of you have done is fine for "justfiy your answers". Justify here means "you're not going to get marks for just writing 'yes'".

The only other thing I'll say is that there are "minimal" axioms for a group: The closure + associativity axioms are the same, but the other two become:

\exists e \in G : ea = a \forall a \in G (i.e. there's a left identity)

\forall b \in G, \exists a\in G :  ab = e (i.e. there's a left inverse)

It can be helpful to state that G is a group if these conditions apply and then you have less to verify.
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Farhan.Hanif93
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(Original post by ben-smith)
Unfortunately I'm not one of those knowledgeable people but I don't understand your notation in the last part. What does [\mathbf{MN}]_{3,1} mean?
In an nxn matrix M, [\mathbf{M}]_{ij} is the ij-th entry i.e. the element in the i-th row, j-th column.

What I did for the last part was a bit simple but I simply multiplied two matrices in this form and then inverted the order and equated entries. This showed that one of the entries didn't matter so if you let the other two be 0 but this other one non zero then commutativity would hold for all B. I don't know whether it's valid...
This sounds similar to what I did.

Question did feel a bit pointless, though. I was expecting a catch at the end i.e. either H wasn't a subgroup or G wasn't commutative under matrix multiplication but that wasn't the case.

Along with the fact that a lot of people have done nothing on groups by the time they get on to STEP, I can understand why they got rid of them.
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Farhan.Hanif93
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(Original post by DFranklin)
What either of you have done is fine for "justfiy your answers". Justify here means "you're not going to get marks for just writing 'yes'".

The only other thing I'll say is that there are "minimal" axioms for a group: The closure + associativity axioms are the same, but the other two become:

\exists e \in G : ea = a \forall a \in G (i.e. there's a left identity)

\forall b \in G, \exists a\in G :  ab = e (i.e. there's a left inverse)

It can be helpful to state that G is a group if these conditions apply and then you have less to verify.
Never knew about this. Thanks.
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Dirac Spinor
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(Original post by DFranklin)
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Can you have a look at the last bit of question 13 please. I haven't really convinced myself there...
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DFranklin
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(Original post by ben-smith)
Can you have a look at the last bit of question 13 please. I haven't really convinced myself there...
I think I would have to do the question myself to be give a "proper answer", but I'd be surprised if the correct answer isn't something along the lines of "at this point, the y' equals the acceleration due to gravity, and it follows that the reaction (which is acting against gravity) must be 0. After this point, the hoop will leave the table".
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Dirac Spinor
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STEP I Q2


PB=\sqrt{x^2+h^2},PC=\sqrt{x^2+(  d+h)^2}
Cosine rule:
cos\theta=\dfrac{d^2-x^2-h^2-x^2-(h+d)^2}{-2\sqrt{(x^2+h^2)(x^2+(d+h)^2)}}

\Rightarrow Tan\theta=\frac{\sqrt{1-cos^2\theta}}{cos\theta}

=\dfrac{\sqrt{(x^2+h^2)(x^2+(d+h  )^2)-(x^2+hd+x^2)^2}}{x^2+hd+h^2}}

=\dfrac{\sqrt{x^4+x^2h^2+2x^2hd+  x^2d^2+h^2x^2+h^4+2h^3d+h^2d^2-x^4-2x^2hd-2x^2h^2-h^2d^2-2h^3d-h^4}}{x^2+hd+h^2}

=\dfrac{xd}{x^2+hd+h^2}
Note that max[\theta] \Rightarrow max[Tan\theta]
so we require an x such that \frac{d}{dx}[Tan\theta]=0. (obviously will be a max because min occurs a x=infinity).
\frac{d}{dx}[Tan\theta]=\frac{d(h^2+hd-x^2)}{(x^2+hd+h^2)^2}

\Rightarrow x^2=hd+h^2 \Rightarrow x=\sqrt{h(h+d)}
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Farhan.Hanif93
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(Original post by ben-smith)
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Which paper are you working from? :p:
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Dirac Spinor
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STEP I Q3
\frac{d^2y}{dx^2}+2\frac{dy}{dx}-3y=2e^x

\Rightarrow \frac{d^2y}{dx^2}-\frac{dy}{dx}+3(\frac{dy}{dx}-y)=2e^x

\Rightarrow \frac{dz}{dx}+3z=2e^x

\Rightarrow \frac{dz}{dx}e^{3x}+3e^{3x}z=2e^  {4x}

\Rightarrow \frac{d}{dx}[ze^{3x}]=2e^{4x} \Rightarrow z=\frac{e^x}{2}+C_1e^{-3x}
Using the initial conditions z(0)=1 so C_1=1/2. Now, to solve for y:
\frac{dy}{dx}-y=\frac{e^x}{2}+C_1e^{-3x}

\Rightarrow \frac{d}{dx}[ye^{-x}]=1/2+C_1e^{-4x}

\Rightarrow y=\frac{xe^x}{2}-C_1e^{-3x}+C_2e^x

y(0)=1 \Rightarrow C_2=9/8
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Dirac Spinor
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(Original post by Farhan.Hanif93)
Which paper are you working from? :p:
I managed to acquire the papers missing. I'll upload in the OP.
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Dirac Spinor
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STEP I Q4
I=\displaystyle \int^{\infty}_{1} \dfrac{1}{(x+1)\sqrt{x^2+2x-2}}dx
Let:
u^2=(x+1)^2-3 \Rightarrow x=\sqrt{u^2+3}-1 \Rightarrow \frac{dx}{du}=\frac{u}{\sqrt{u^2  +3}}

\therefore I=\displaystyle \int^{\infty}_{1} \frac{u}{\sqrt{u^2+3}u\sqrt{u^2+  3}}du

=\displaystyle \int^{\infty}_{1} \frac{1}{u^2+3}du=\frac{1}{\sqrt  3}\displaystyle \int^{\infty}_{1} \frac{1}{\sqrt3}\frac{1}{(u/\sqrt3)^2+1}du=\frac{1}{\sqrt3}[arctan(u/\sqrt3)]^{\infty}_{1}

=\dfrac{\pi}{3\sqrt3}
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Farhan.Hanif93
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STEP I - Q7
|1+e^{2i\alpha}| = \sqrt{(cos2\alpha+1)^2 + \sin ^22\alpha} = \sqrt{2(\cos 2\alpha +1)} = 2\cos \alpha

\arg (1+e^{2i\alpha}) = \tan^{-1} \dfrac{\sin 2\alpha}{\cos 2\alpha +1} = \tan^{-1} (\tan \alpha) = \alpha.

Hence 1+e^{2i\alpha} = 2\cos \alpha e^{i\alpha}.

Note that \displaystyle\sum _{r=0}^n \begin{pmatrix} n \\ r \end{pmatrix} \sin (2r+1)\alpha \right]= Im\left[e^{i\alpha}\displaystyle\sum _{r=0}^n \begin{pmatrix} n \\ r \end{pmatrix} (e^{2i\alpha})^r\right]
=Im(e^{i\alpha} (1+e^{2i\alpha})^n)
 \implies \boxed{\displaystyle\sum _{r=0}^n \begin{pmatrix} n\\ r \end{pmatrix} \sin (2r+1)\alpha=2^n\cos ^n\alpha \sin(n+1)\alpha}.
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Dirac Spinor
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STEP I Q7
 z=1+e^{2i\alpha}=1+cos2\alpha+is  in2\alpha

\Rightarrow |z|=\sqrt{(1+cos2\alpha)^2+(sin2  \alpha)^2}=\sqrt{2+2cos\alpha}=2  cos\alpha

arg(z)=arctan(\dfrac{sin2\alpha}  {1+cos2 \alpha})=arctan(Tan(2\alpha/2))= \alpha

\displaystyle \Sigma^{n}_{r=0} \displaystyle \binom{n}{r} sin(2r+1)=\Im[\displaystyle \Sigma \displaystyle \binom{n}{r} e^{i(2r+1)\alpha}]=\Im[e^{i\alpha} \displaystyle \Sigma \displaystyle \binom{n}{r} e^{i2r\alpha}]

=\Im[e^{i\alpha}(1+e^i\alpha)^n]=\Im[e^{i\alpha}(2cos\alpha e^{i\alpha})^n]

=2^n cos^n\alpha\Im[e^{i(n+1)\alpha}]=2^n cos^n\alpha sin(n+1)\alpha
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Dirac Spinor
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STEP I Q6
Consider the sector formed by the points A,D and E and let Q be the centre of the whole coin. We want to find the area of the segment subtended on A and the area of the triangle QDE. Let QD=QE=x and 2y=DE:
Sin(\pi/14)=y/a,Sin(\pi/7)=y/x

\Rightarrow x=a\dfrac{sin(\pi/14)}{sin(\pi/7)}
Adding the two areas we mentioned earlier we get:


a^2/2(\pi/7-sin(\pi/7))+a^2/2 \dfrac{sin^2(\pi/14)sin(2\pi/7)}{sin^2(\pi/7)}

=a^2/2(\frac{\pi}{7}-sin(\pi/7)+\dfrac{sin^2(\pi/14)2sin(\pi/7)cos(\pi/7)}{sin^2(\pi/7)}

=a^2/2(\frac{\pi}{7}+\frac{2sin^2(\pi/7)cos(\pi/7)-sin^2(\pi/7)}{sin(\pi/7)})

=a^2/2(\frac{\pi}{7}+\frac{cos(\pi/7)-1}{sin(\pi/7)})



=a^2/2(\frac{\pi}{7}-Tan(\pi/14)

Multiplying by 7 gives the required result.
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Farhan.Hanif93
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STEP I - Q11
Let the cross section of the cylinder have centre O and radius a. Let the contact force exerted by the ramp on the cylinder have magnitude R. Note that there is no reaction force exerted by the horizontal surface as the cylinder is on the point of losing contact with it. Friction must be limiting on the vertical surface. On the point of rolling:

\Re (\mathrm{up} \mathrm{the} \mathrm{ramp}): \dfrac{P}{\sqrt 2} - \dfrac{\mu P}{\sqrt 2} - \dfrac{Mg}{\sqrt 2} - \nu R =0 (1)

\Re (\mathrm{perpendicular} \mathrm{to} \mathrm{ramp}): R - \dfrac{Mg}{\sqrt 2} - \dfrac{P}{\sqrt 2} - \dfrac{\mu P}{\sqrt 2} = 0 (2)

Taking moments about O: a\nu R = a\mu P (3)

Subbing (3) into (1): P(1-\mu -\sqrt 2 \mu) - Mg=0 \implies \boxed{P=\dfrac{Mg}{1-\mu(1+\sqrt 2)}}, as required.

Note that P>0 \implies 1-\mu (1+\sqrt 2) > 0 \implies \boxed{\mu < \dfrac{1}{\sqrt 2 + 1} = \sqrt 2 -1}, as required.

Furthermore, in order for the cylinder to be rolling up the slope, we require the clockwise moment to be greater than or equal to the anti-clockwise moment.

By (2): R = \dfrac{Mg(2- \sqrt 2 \mu)}{\sqrt 2(1-\mu (1+\sqrt 2))}

We require \nu R \geq \mu P

\implies \dfrac{\nu (2-\sqrt 2\mu)}{\sqrt 2(1-\mu (1+\sqrt 2))} \geq \dfrac{\mu}{1-\mu (1+\sqrt 2)}

\implies  \nu (\sqrt 2 - \mu)  \geq  \mu

By the earlier result, \sqrt 2 - \mu > 1 so it's safe to divide through by it to obtain the required result \boxed{\nu \geq \dfrac{\mu}{\sqrt 2 - \mu}}.
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Farhan.Hanif93
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STEP I - Q12
Let A have initial speed U and immediately before the collision, speed v. Immediately after the collision, let A have speed w and let B have speed x.

By the conservation of energy:
\Delta E_k = \mathrm{w.d.} \mathrm{by}  \mathrm{friction}

\implies \dfrac{1}{2}mx^2 = \mu mgd

\implies \boxed{x=\sqrt{2\mu gd}}, as required.

Whilst sliding from O to disc B, disc A is subject to a constant deceleration of \mu g. Hence v^2=u^2 +2as \implies v = \sqrt{U^2 -2\mu g d}

Also by conservation of momentum: mv = mw + mx \implies \sqrt{U^2 -2\mu g d} = w+\sqrt{2\mu g d} (1)

By Newton's Law of Restitution: ev = x-w \implies e\sqrt{U^2 -2\mu g d} = \sqrt{2\mu gd} - w. (2)

(1)+(2): (1+e)\sqrt{U^2 -2\mu g d} = 2\sqrt{2\mu g d}

\implies \boxed{U = \dfrac{\sqrt{2\mu g d}}{{1+e}}(5+2e+e^2)^{\frac{1}{2  }}}, as required.

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Farhan.Hanif93
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STEP I - Q13
Let the string have modulus \lambda. Whilst in equilibrium, note that \dfrac{\lambda a}{a} - mg = 0 \implies \lambda = mg.

Whilst the particle moves in the described horizontal circle, let the string have length a+x; be inclined at an angle \theta to the downwards vertical and exert a tension T on the particle. It follows that the circle must have radius r=(a+x)\sin \theta.

\Re (\mathrm{inwards}): T\sin \theta = mr\omega ^2 (1)
\Re (\uparrow): T\cos \theta - mg =0 (2)

Note that from (2),  \dfrac{mgx}{a}\cos \theta =mg \implies \cos \theta = \dfrac{a}{x}. Furthermore, \cos \theta < 1 \implies \boxed{x>a} i.e. In order for the particle to be moving in such a circle, the string must stretched to a longer length than what it was when hanging in equilibrium, as required.

From (1): \dfrac{gx}{a}\sin \theta = r\omega ^2 = (a+x)\sin\theta \omega ^2

\implies \omega ^2 = \dfrac{g}{a} - \dfrac{g}{a+x}.

Observe that x>a \implies \dfrac{g}{a} - \dfrac{g}{a+x} >  \dfrac{g}{a}-\dfrac{g}{2a} = \dfrac{g}{2a} and that \dfrac{g}{a+x} > 0 \implies \dfrac{g}{a} - \dfrac{g}{a+x} < \dfrac{g}{a}.

Hence the required result \dfrac{g}{2a} <\omega ^2 < \dfrac{g}{a} follows.
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oh_1993
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STEP I Q1

\displaystyle y\leq-2 or \displaystyle y\geq-\frac{6}{5}

Coefficient of x^n = \displaystyle2^{-(n+1)}+\frac{\frac{3}{4}(n+1)}{2  ^n}+(-1)^n

Expansion valid for: |x|<1
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