Year: 2015
Paper: 3
Question Number: 10
Course: UFM Mechanics
Section: Variable Force
No solution available for this problem.
A very similar number of candidates to 2014 once again ensured that all questions received a decent number of attempts, with seven questions being very popular rather than five being so in 2014, but the most popular questions were attempted by percentages in the 80s rather than 90s. All but one question was answered perfectly at least once, the one exception receiving a number of very close to perfect solutions. About 70% attempted at least six questions, and in those cases where more than six were attempted, the extra attempts were usually fairly superficial.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
A light rod of length $2a$ has a particle of mass $m$ attached to each end and it moves in a vertical plane. The midpoint of the rod has coordinates $(x,y)$, where the $x$-axis is horizontal (within the plane of motion) and $y$ is the height above a horizontal table. Initially, the rod is vertical, and at time $t$ later it is inclined at an angle $\theta$ to the vertical.
Show that the velocity of one particle can be written in the form
\[
\begin{pmatrix}
\dot x + a \dot\theta \cos\theta \\
\dot y - a \dot\theta \sin\theta
\end{pmatrix}
\]
and that
\[
m\begin{pmatrix}
\ddot x + a\ddot\theta \cos\theta - a \dot\theta^2 \sin\theta
\\
\ddot y- a\ddot\theta \sin\theta - a \dot\theta^2 \cos\theta
\end{pmatrix}
=-T\begin{pmatrix}
\sin\theta
\\
\cos\theta
\end{pmatrix}
-mg
\begin{pmatrix}
0 \\ 1
\end{pmatrix}
\]
where the dots denote differentiation with respect to time $t$ and $T$ is the tension in the rod. Obtain the corresponding equations for the other particle.
Deduce that $\ddot x =0$, $\ddot y = -g$ and $\ddot\theta =0$.
Initially, the midpoint of the rod is a height $h$ above the table, the velocity of the higher particle is $\Big(\begin{matrix} \, u \, \\ v \end{matrix}\Big)$, and the velocity of the lower particle is $\Big(\begin{matrix}\, 0 \, \\ v\end{matrix}\Big)$. Given that the two particles hit the table for the first time simultaneously, when the rod has rotated by $\frac12\pi$, show that
\[
2hu^2 = \pi^2a^2 g - 2\pi uva \,.
\]
Whilst this was the least popular question with just over 8% attempting it, it was only slightly less successfully attempted than question 5, though those making substantial attempts at it invariably scored half to two thirds of the marks comfortably. Most successfully wrote the position vector of one of the particles and then differentiated with respect to time to obtain the velocity correctly, though a few succeeded by adding velocities. The second displayed equation was almost always correctly derived, though many did far too much work obtaining the corresponding equations for the other particle when it could just be written down. Deducing certain quantities was fine, but θ frequently wasn't. At this point, finding initial values caused some issues, if it was realised that these were needed, and although many wrote the uniform acceleration equation for the displacement of the midpoint of the rod, the final result eluded many.