Two light elastic springs each have natural length \(a\). One end of each spring is attached to a particle \(P\) of weight \(W\). The other ends of the springs are attached to the end-points, \(B\) and \(C\), of a fixed horizontal bar \(BC\) of length \(2a\). The moduli of elasticity of the springs \(PB\) and \(PC\) are \(s_1 W\) and \(s_2 W\) respectively; these values are such that the particle \(P\) hangs in equilibrium with angle \(BPC\) equal to \(90^\circ\).
- Let angle \(PBC = \theta\). Show that \(s_1 = \dfrac{\sin\theta}{2\cos\theta - 1}\) and find \(s_2\) in terms of \(\theta\).
- Take the zero level of gravitational potential energy to be the horizontal bar \(BC\) and let the total potential energy of the system be \(-paW\). Show that \(p\) satisfies
\[ \frac{1}{2}\sqrt{2} \geqslant p > \frac{1}{4}(1+\sqrt{3}) \]
and hence that \(p = 0.7\), correct to one significant figure.
A particle \(P\) of mass \(m\) moves freely and without friction on a wire circle of radius \(a\), whose axis is horizontal. The highest point of the circle is \(H\), the lowest point of the circle is \(L\) and angle \(PHL = \theta\). A light spring of modulus of elasticity \(\lambda\) is attached to \(P\) and to \(H\). The natural length of the spring is \(l\), which is less than the diameter of the circle.
- Show that, if there is an equilibrium position of the particle at \(\theta = \alpha\), where \(\alpha > 0\), then \(\cos\alpha = \dfrac{\lambda l}{2(a\lambda - mgl)}\).
Show also that there will only be such an equilibrium position if \(\lambda > \dfrac{2mgl}{2a - l}\).
When the particle is at the lowest point \(L\) of the circular wire, it has speed \(u\).
- Show that, if the particle comes to rest before reaching \(H\), it does so when \(\theta = \beta\), where \(\cos\beta\) satisfies
\[(\cos\alpha - \cos\beta)^2 = (1 - \cos\alpha)^2 + \frac{mu^2}{2a\lambda}\cos\alpha,\]
where \(\cos\alpha = \dfrac{\lambda l}{2(a\lambda - mgl)}\).
Show also that this will only occur if \(u^2 < \dfrac{2a\lambda}{m}(2 - \sec\alpha)\).
A straight road leading to my house consists of two sections. The first section is inclined downwards at a constant angle \(\alpha\) to the horizontal and ends in traffic lights; the second section is inclined upwards at an angle \(\beta\) to the horizontal and ends at my house. The distance between the traffic lights and my house is \(d\).
I have a go-kart which I start from rest, pointing downhill, a distance \(x\) from the traffic lights on the downward-sloping section. The go-kart is not powered in any way, all resistance forces are negligible, and there is no sudden change of speed as I pass the traffic lights. Given that I reach my house, show that \(x \sin \alpha\ge d \sin\beta\,\).
Let \(T\) be the total time taken to reach my house.
Show that
\[
\left(\frac{g\sin\alpha}2 \right)^{\!\frac12} T =
(1+k) \sqrt{x} - \sqrt{k^2 x -kd\;}
\,,
\]
where \(k = \dfrac{\sin\alpha}{\sin\beta}\,\).
Hence determine, in terms of \(d\) and \(k\), the value of \(x\) which minimises \(T\).
[You need not justify the fact that the stationary value is a minimum.]
Show Solution
Applying conservation of energy, since there are no external forces (other than gravity) the condition to reach the house (with any speed) is the initial GPE is larger than the final GPE, ie:
\begin{align*}
&& m g x \sin \alpha &\geq m g d \sin \beta \\
\Rightarrow && x \sin \alpha &\geq d \sin \beta
\end{align*}
Let \(T_1\) be the time taken on the downward section, and \(T_2\) the time taken on the upward section, then:
\begin{align*}
&& s &= ut + \frac12 a t^2 \\
\Rightarrow && x &= \frac12 g \sin \alpha T_1^2 \\
\Rightarrow && T_1^2 &= \frac{2x}{g \sin \alpha} \\
&& v &= u + at \\
\Rightarrow && v &= T_1 g \sin \alpha \\
&& mg x \sin \alpha &= mg d \sin \beta + \frac12 m w^2 \\
\Rightarrow && w &= \sqrt{2(x \sin \alpha - d \sin \beta)} \\
&& w &= v - g \sin \beta T_2 \\
\Rightarrow && T_2 &= \frac{v - w}{g \sin \beta} \\
\Rightarrow && T &= T_1 + T_2 \\
&&&= \sqrt{\frac{2x}{g \sin \alpha}} + \frac{\sqrt{\frac{2x}{g \sin \alpha}} g \sin \alpha- \sqrt{2(x \sin \alpha - d \sin \beta)}}{g \sin \beta} \\
&&&= \left ( \frac{2}{g \sin \alpha} \right)^{\tfrac12} \left ( \sqrt{x} + \sqrt{x}k - \sqrt{k^2x-kd}\right)
\end{align*}
Differentiating wrt to \(x\), we obtain:
\begin{align*}
&& \frac{\d T}{\d x} &= C(-(1+k)x^{-1/2}+k^2(k^2 x - kd)^{-1/2}) \\
\text{set to }0: && 0 &= k^2(k^2 x - kd)^{-1/2} - (1+k)x^{-1/2} \\
\Rightarrow && \sqrt{x} k^2 &= \sqrt{k^2x - kd} (1+k) \\
\Rightarrow && x k^4 &= (k^2x-kd)(1+k)^2 \\
\Rightarrow && x(k^4-k^2(1+k)^2) &= -kd(1+k)^2 \\
\Rightarrow && x(2k^2+k) &= d \\
\Rightarrow && x &= \frac{d}{(2k^2+k)}
\end{align*}
In this question take \(g = 10 ms^{-2}.\)
The point \(A\) lies on a fixed rough plane inclined at \(30^{\circ}\) to the horizontal and \(\ell\) is the line of greatest slope through \(A\). A particle \(P\) is projected up \(\ell\) from \(A\) with initial speed \(6\)ms\(^{-1}\). A time \(T\) seconds later, a particle \(Q\) is projected from \(A\) up \(\ell\), also with speed \(6\)ms\(^{-1}\). The coefficient of friction between
each particle and the plane is \(1/(5\sqrt{3})\,\) and the mass of each particle is \(4\)kg.
- Given that \(T<1+\sqrt{3/2}\), show
that the particles collide at a time \((3-\sqrt6)T+1\) seconds after \(P\) is projected.
- In the case \(T=1+\sqrt{2/3}\,\),
determine the energy lost due to friction from the instant at which \(P\) is projected to the time of the collision.
Show Solution
Since the particles are identical and are projected with the same speed, the only way they can reach the same point \(x\) at the same time, is if \(A\) has reached it's apex and started descending.
Considering \(P\), we must have (setting the level of \(A\) to be the \(0\) G.P.E. level), suppose it travels a distance \(x\) before becoming stationary:
\begin{align*}
\text{N2}(\nwarrow): && R - 4g \cos(30^\circ) &= 0 \\
\Rightarrow && R &= 20\sqrt{3} \\
\Rightarrow && \mu R &= \frac1{5 \sqrt{3}} (20 \sqrt{3}) \\
&&&= 4 \\
\end{align*}
Therefore in the two phases of the journey the particle is being accelerated down the slope by either \(6\) or \(4\).
\(v^2 = u^2 + 2as \Rightarrow 0 = 36 - 12s \Rightarrow s = 3\). \(v = u + at \Rightarrow t = 1\). Therefore after \(1\) second \(P\) reaches its highest point having travelled \(3\) metres. It will pass back to the start in \(s = ut + \frac12 a t^2 \Rightarrow 3 = \frac12 4 t^2 \Rightarrow t = \sqrt{3/2}\) seconds, ie the constraint is that the particle hasn't already past \(Q\) before the collision.
The collision will occur when \(s = 6t - \frac12 6 t^2\) and \(s =3 - \frac12 4 (t+T-1)^2\) coincide, ie:
\begin{align*}
&& 6t - 3t^2 &= 3 - 2(t+T-1)^2 \\
&& 0 &= 3 -2(T-1)^2 -(4(T-1)+6)t + t^2 \\
&& 0 &= 3 -2(T-1)^2 -(4T+2)t + t^2 \\
\Rightarrow && t &= \frac{4T+2 \pm \sqrt{(4T+2)^2 - 4(3-2(T-1)^2)}}{2} \\
&&&= \frac{4T+2 \pm \sqrt{24T^2}}{2} \\
&&&= 2T + 1 \pm \sqrt{6} T \\
&&&= (2 \pm \sqrt{6})T + 1
\end{align*}
we must take the smaller root, ie \((2-\sqrt{6})T + 1\).
In the case the collision occurs exactly at the start, the particle \(P\) has traveled \(6\) meters, against a force of \(4\) newtons of friction, ie work done is \(24\) Joules.
In a certain race, runners run 5\(\,\)km in a straight line to a fixed point and then turn and run back to the starting point. A steady wind of 3\(\,\text{ms}^{-1}\) is blowing from the start to the turning point. At steady racing pace, a certain runner expends energy at a constant rate of 300\(\,\)W. Two resistive forces act. One is of constant magnitude \(50\,\text{N}\). The other, arising from air resistance, is of magnitude \(2w\,\mathrm{N}\), where \(w\,\text{ms}^{-1}\) is the runner's speed relative to the air. Give a careful argument to derive formulae from which the runner's steady speed in each half of the race may be found. Calculate, to the nearest second, the time the runner will take for the whole race.
\textit{Effects due to acceleration and deceleration at the start and turn
may be ignored.}
The runner may use alternative tactics, expending the same total energy during the race as a whole, but applying different constant powers, \(x_{1}\,\)W in the outward trip, and \(x_{2}\,\)W on the return trip. Prove that, with the wind as above, if the outward and return speeds are \(v_{1}\,\)ms\(^{-1}\) and \(v_{2}\,\)ms\(^{-1}\) respectively, then \(v_{1}+v_{2}\) is independent of the choices for \(x_{1}\) and \(x_{2}\).
Hence show that these alternative tactics allow the runner to run the whole race approximately 15 seconds faster.
Show Solution
Note that \(P = Fv\). Since he is running at a steady pace, we can say that \(F\) must be equal to the resistive forces (as net force is \(0\)).
Therefore \(F = 50 + 2(v+3)\) on the way out.
ie, \(300 = (2v + 56)v \Rightarrow 150 = v^2 + 28v \Rightarrow v = \sqrt{346}-14\)
On the way back, \(F = 50 + 2(v-3)\), ie \(300 = (2v+44)v \Rightarrow 150 = v^2 +22v \Rightarrow v = \sqrt{271}-11\)
Therefore the total time will be \(\frac{5000}{\sqrt{346}-150} + \frac{5000}{\sqrt{271}-11} \approx 2002\), or 33 minutes, 22 seconds. Very respectable!
The total energy in this first run is \(E = Pt = 2002 \cdot 300\).
Now suppose we apply two different powers as in the question, then we must have:
\begin{align*}
&& x_1 &= 2v_1^2 + 56v_1 \\
&& x_2 &= 2v_2^2 + 44v_2 \\
&& E &= x_1 \frac{5000}{v_1} + x_2 \frac{5000}{v_2} \\
&&&= 5000 \left ( \frac{x_1}{v_1} + \frac{x_2}{v_2} \right) \\
\Rightarrow && \frac{x_1}{v_1} &= 2v_1 + 56 \\
&& \frac{x_2}{v_2} &= 2v_2 + 44 \\
\Rightarrow && \frac{E}{5000} &= 2(v_1+v_2) + 100 \\
\Rightarrow && v_1+v_2 &\text{ is independent of the choices for }x_i
\end{align*}
We wish to minimize
\begin{align*}
&& \frac{5000}{v_1} + \frac{5000}{v_2} &\underbrace{\geq}_{AM-HM} 10\,000 \cdot \frac{2}{v_1+v_2} \\
&&&= 10\,000 \cdot \frac{2}{\sqrt{346}-14+\sqrt{271}-11} \\
&&&\approx 1987
\end{align*}
ie they can go 15 seconds quicker with better strategy.