Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2023 / Paper 2 / 7

For the stated square side length $2a$, the volume is $4a^{2}h$ and the uniform mass density is ${\rho }_1=m_1/(4a^{2}h)$. Direct integration about its central vertical axis gives the moment of inertia of a uniform square plate:Thus the PDF's coefficient $1/6$ is inconsistent with its stated side length. The coefficient $1/6$ would apply to a square of side $a$.

For the disc, use cylindrical coordinates and $m_2={\rho }_2\pi a^{2}h$:For the sphere, use spherical coordinates and $m_3=4\pi {\rho }_3{a}^3/3$:These agree with the moment of inertia of a uniform disc and the moment of inertia of a uniform solid sphere. The square's thickness does not enter its axial moment of inertia once the mass is fixed.

Measure each generalized coordinate ${\theta }_i$ from an unstressed configuration. The diagram has one torsional spring joining the frame to object 1, one joining objects 1 and 2, and one joining objects 2 and 3. The rotational kinetic energy and elastic potential energy areEach spring contribution follows by integrating its restoring torque: ${\int }_0^{\Delta \theta }{k}_{j}sds=k_j(\Delta \theta {)}^2/2$. ThusThe Euler-Lagrange equations giveThese hold for either interpretation of the square's moment of inertia; the geometry-based value from (a) is used first below.

For the stated geometry and numerical masses, the inertia and stiffness matrices areA normal mode $\mathbf{\theta }=\mathbf{v}{e}^{i\omega t}$ requires $(K-{\omega }^{2}M)\mathbf{v}=0$. Writing $z={\omega }^2$, its characteristic equation isAt the claimed $z=6$, this determinant is $-27/50\ne 0$. Hence the printed claim that $\sqrt{6}$ is a normal frequency is false for the stated data, and there is no corresponding nonzero normal mode.

The complete geometry-based answer is obtained from the three positive roots $z_j$:An exact closed form for the squared angular frequencies isThey are approximately $19.555611657$, $6.209018836$ and $1.235369507$, respectively. For any of these roots, a corresponding eigenvector can be chosen asThe last two rows of $(K-zM)\mathbf{v}=0$ give this expression directly; the first is the characteristic equation. These three modes span all motion, with arbitrary sine and cosine amplitudes. Positivity of $M$ and of the spring potential energy ensures that all squared angular frequencies are positive.

If instead one takes the PDF's printed $I_1=m_1{a}^2/6$ as a stipulated inertia, then $M=\mathrm{diag}(1/2,1/4,2/25)$ andThe corresponding angular frequencies are approximatelyThe same formula for $\mathbf{v}(z)$ applies, because its derivation used only the second and third rows. Here the determinant at $z=6$ is $-63/25\ne 0$, so following the printed inertia does not repair the frequency claim. The PDF therefore needs a further correction to its numerical data or asserted frequency; the intended correction cannot be inferred uniquely.