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UPSC 2026 Maths Optional Paper 2 Q5e — Step-by-Step Solution

10 marks · Section B

Motion of rigid bodies in two dimensions · Mechanics & Fluid Dynamics · asked 5× in 14 yrs · Read the full method →

Question

A uniform heavy solid hemisphere of radius '$a$' is held at rest with its base vertical and its curved surface in contact with a horizontal plane. If the hemisphere is released when the plane is rough enough to prevent slipping, show that the angle $\theta$ that the base makes with the horizontal at time $t$, is such that

Technique

“Rough enough to prevent slipping” is a constraint, not a force to be found: the contact point has zero velocity, so friction and normal reaction do no work and energy is conserved. That makes this a one-line-in-principle problem — write $T+V=$ const with $\theta$ the single degree of freedom. The two numbers that must be right are $OG=\tfrac{3a}{8}$ and the moment of inertia; the constant $28$ in the printed denominator is $20\times\tfrac{7}{5}$ and is a direct consequence of them, so a wrong $28$ means a wrong geometry, not a wrong algebra step. The one geometric fact that unlocks everything: the centre $O$ of the flat face stays at height $a$ throughout, because the curved surface is a sphere of radius $a$ centred at $O$.

Solution

Step 1 — Set up, and say exactly what $\theta$ is.

Let $M$ be the mass, $a$ the radius, $O$ the centre of the plane (flat) face, and $G$ the centre of mass, which lies on the axis of symmetry at

(standard result for a uniform solid hemisphere), on the side of the curved surface.

$\theta$ is the angle the base (the flat face) makes with the horizontal plane. Since $OG$ is perpendicular to the base, $OG$ makes the same angle $\theta$ with the upward vertical measured from $O$ downwards — concretely, taking the vertical plane of motion as the $xy$-plane with $y$ upward, the unit vector from $O$ towards $G$ is

Check the two ends: $\theta=\tfrac{\pi}{2}$ (base vertical) gives $\hat u=(1,0)$, so $G$ is level with $O$ — the released configuration; $\theta=0$ (base horizontal, uppermost) gives $\hat u=(0,-1)$, so $G$ is directly below $O$ — the equilibrium configuration.

Initial conditions. “Held at rest with its base vertical” and then released:

(The printed formula is consistent with exactly this: it gives $\dot\theta^{2}=0$ at $\theta=\pi/2$.)

The motion is planar. The body is a solid of revolution about $OG$; at $t=0$ the vertical plane containing $OG$ is a plane of symmetry for the body, for gravity and for the horizontal plane, and the initial velocity is zero. The mirror image of the motion in that plane is therefore also a solution with the same initial data, so by uniqueness the motion coincides with its mirror image: it stays in that vertical plane. (This is worth one sentence — the rolling constraint is non-holonomic in general, and planarity is what makes the single coordinate $\theta$ sufficient.)

Step 2 — The key geometric fact: $O$ moves horizontally at height $a$.

The curved surface of the hemisphere is part of the sphere of radius $a$ centred at $O$. Contact with the horizontal plane therefore requires

and the contact point $C$ is the point of that sphere vertically below $O$:

(Consistency check: $C$ lies on the hemisphere iff $\vec{OC}\cdot\hat u\ge0$, i.e. $a\cos\theta\ge0$, true for $0\le\theta\le\pi/2$. At $\theta=\pi/2$ we get $\vec{OC}\cdot\hat u=0$: the contact is exactly on the rim of the base — the limiting configuration described in the question.)

Consequently the height of $G$ is

At $\theta=\pi/2$, $y_G=a$; at $\theta=0$, $y_G=a-\tfrac{3a}{8}=\tfrac{5a}{8}$. The drop of the centre of mass from the initial position is therefore

This is the step most candidates get wrong — the temptation is to write the drop as $c(1-\cos\theta)$ or $c\sin\theta$ by analogy with a pendulum. It is neither: $O$ does not move vertically, so the entire vertical motion of $G$ comes from the rotation of $\vec{OG}$, and at $\theta=\pi/2$ that vector is horizontal, giving the maximum height.

Step 3 — Moments of inertia.

Let the axis of rotation be the horizontal line through $O$ perpendicular to the plane of motion. Because the motion is planar and $OG$ lies in that plane, this axis lies in the plane of the flat face, i.e. it is a diameter of the base.

For a uniform solid hemisphere of mass $M$, radius $a$, about a diameter of its base through $O$: a full sphere of mass $2M$ and radius $a$ has $I=\tfrac25(2M)a^{2}=\tfrac45Ma^{2}$ about a diameter, and that diameter can be taken in the cutting plane, so by symmetry each hemisphere contributes half:

By the parallel-axis theorem (the axis through $G$ parallel to this one),

Step 4 — Kinetic energy.

Let $\omega$ be the angular speed of the body. Since the orientation of $\hat u$ is $\theta$-determined, $\dfrac{d\hat u}{dt}=\dot\theta(\cos\theta,\sin\theta)$, while for a rigid body $\dfrac{d\hat u}{dt}=\vec\omega\times\hat u$; with $\vec\omega=\omega\hat k$ this is $\omega(\cos\theta,\sin\theta)$. Hence

Velocity of $G$. Rolling without slipping means $\vec v_C=0$, so $C$ is the instantaneous centre of rotation and

From $\vec{CG}=\vec{CO}+\vec{OG}=(0,a)+c(\sin\theta,-\cos\theta)=\bigl(c\sin\theta,\ a-c\cos\theta\bigr)$,

(Equivalently, without invoking the instantaneous centre: $\vec v_O=-a\dot\theta\,\hat i$ from the rolling condition, and $\vec v_G=\vec v_O+\vec\omega\times\vec{OG}=\dot\theta\bigl(c\cos\theta-a,\ c\sin\theta\bigr)$, whose square is again (4).)

Therefore, by König’s theorem (translation of $G$ plus rotation about $G$),

Insert $c=\tfrac{3a}{8}$ and (3), and divide the bracket by $Ma^{2}$:

since $448/320=7/5$.

(Sanity check on (6) at $\theta=0$: there $|CG|=a-c$ and $T=\tfrac12\dot\theta^{2}\bigl[M(a-c)^{2}+I_G\bigr]=\tfrac12\dot\theta^{2}Ma^{2}\bigl[\tfrac{25}{64}+\tfrac{83}{320}\bigr]=\tfrac12\dot\theta^{2}Ma^{2}\cdot\tfrac{125+83}{320}=\tfrac12\dot\theta^{2}Ma^{2}\cdot\tfrac{13}{20}$, and (6) gives $\tfrac75-\tfrac34=\tfrac{28-15}{20}=\tfrac{13}{20}$ ✓.)

Step 5 — Energy conservation.

The forces acting are gravity (conservative) and the reaction of the plane at $C$ — normal reaction and friction. Because the contact is rolling, the material point of the body at $C$ has zero velocity, so the reaction does no work. Hence the total mechanical energy is conserved:

Evaluating the constant at $t=0$ ($\theta=\pi/2$, $\dot\theta=0$, $V=Mga$):

which is just $T=Mg\,h(\theta)$ with the drop (2), as it must be.

Step 6 — Solve for $\dot\theta^{2}$.

Cancel $M$ and one factor $a$:

Multiply throughout by $20$:

Step 7 — Reading the result (cheap marks, one line each).

• At $\theta=\pi/2$: $\dot\theta^{2}=0$ ✓ — the released-from-rest condition is built in.
• Denominator: $28-15\cos\theta\ge13>0$ for all $\theta$, so $\dot\theta^{2}\ge0$ throughout $|\theta|\le\pi/2$ and the motion is genuine; $\dot\theta^{2}$ is even in $\theta$, so the body performs finite oscillations about $\theta=0$ between $\theta=+\pi/2$ and $\theta=-\pi/2$.
• Maximum angular speed at the equilibrium position $\theta=0$: $\dot\theta^{2}_{\max}=\dfrac{15g}{13a}$.
• The result is independent of $M$ (energy equation homogeneous in $M$) and of the coefficient of friction, provided only that rolling is maintained — which is what “rough enough to prevent slipping” grants. The reaction components can be recovered afterwards from $M\ddot{\vec r}_G=\vec N+\vec F+M\vec g$ if required; they are not needed for the result, and that is precisely the economy the energy method buys.
• Related standard result (different initial data). The Lagrange equation of motion behind (6) is

so for small-amplitude oscillations about $\theta=0$ (not the large release of this question, where $\dot\theta^{2}$ is not small at $\theta=0$) the linearisation is $\ddot\theta\approx-\dfrac{15g}{26a}\theta$, giving the classical period $2\pi\sqrt{\dfrac{26a}{15g}}$ for a rocking hemisphere. Worth one line if time allows; do not confuse it with the present motion.

Answer

obtained from $T+V=$ const with $OG=\dfrac{3a}{8}$, $I_G=\dfrac{83}{320}Ma^{2}$, drop of $G$ $=\dfrac{3a}{8}\cos\theta$, and $2T=Ma^{2}\dot\theta^{2}\Bigl(\dfrac75-\dfrac34\cos\theta\Bigr)$.