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UPSC 2023 Maths Optional Paper 1 Q7b — Step-by-Step Solution

15 marks · Section B

Principle of virtual work · Dynamics & Statics · asked 6× in 13 yrs · Read the full method →

Question

A solid hemisphere is supported by a string fixed to a point on its rim and to a point on a smooth vertical wall with which the curved surface is in contact. If $\theta$ is the angle of inclination of the string with vertical and $\phi$ is the angle of inclination of the plane base of the hemisphere to the vertical, then find the value of $(\tan\phi-\tan\theta)$.

Technique

2D cross-section through plane of symmetry; place wall-contact at origin and centre of hemisphere at $(a,0)$; identify CG and rim attachment using the inclined axis; force balance + moments about wall-contact give the result.

Solution

Setup. Hemisphere of radius $a$, weight $W$. Curved surface touches a smooth vertical wall at one point. A string ties from a point on the rim to a point on the wall.

Place the wall along $x=0$. Let the contact point with the wall be the origin. The hemisphere’s centre $O=(a,0)$ (since at the contact, the radius from the centre is horizontal, length $a$).

Base plane inclined at angle $\phi$ to vertical → axis of hemisphere (perpendicular to base) inclined at angle $\phi$ to horizontal. Choose the orientation: axis points from $O$ toward the wall (the curved side), with the rim above the contact point.

In a 2D cross-section through the symmetry plane: axis direction = $(-\cos\phi,-\sin\phi)$ (the centre of mass is between $O$ and the wall, slightly below the horizontal line through $O$).

Centre of gravity of solid hemisphere is at distance $3a/8$ from the centre of the flat base, along the axis (toward the curved side). So

Rim attachment point $R$: on the rim, in our 2D cross-section. The rim is a circle in the base plane (centred at $O$, radius $a$). In 2D cross-section, the rim has two points; pick the upper one (where string is naturally attached):

(The rotation: axis $=(-\cos\phi,-\sin\phi)$; rotate $90°$ CCW: $(\sin\phi,-\cos\phi)$ or CW $(-\sin\phi,\cos\phi)$. Take the one with positive $y$-component → $(-\sin\phi,\cos\phi)$.)

Step 1 — Force balance

Forces on hemisphere:

1. Weight $W$ downward at $G$.
2. Wall reaction $R_w$ horizontal at the wall-contact $(0,0)$, pointing $+x$ (perpendicular to smooth wall).
3. String tension $T$ at $R$, along direction toward the wall point. The string makes angle $\theta$ with vertical, with the wall on the $-x$ side, so direction $=(-\sin\theta,\cos\theta)$ (toward wall and upward).

Vertical equilibrium: $T\cos\theta=W$, so $T=W/\cos\theta$.

Horizontal equilibrium: $R_w-T\sin\theta=0$, so $R_w=T\sin\theta=W\tan\theta$.

Step 2 — Take moments about wall-contact $(0,0)$

Wall reaction $R_w$ acts at the moment axis: zero moment.

Weight $W$ at $G=(a(1-3\cos\phi/8),\,-3a\sin\phi/8)$: moment $z$-component $=x_G\cdot(-W)-y_G\cdot 0=-Wa(1-3\cos\phi/8)\cdot 1$. Hmm we want the convention: moment of $\vec F=(0,-W)$ at position $\vec r=G$: moment $z=x_G\cdot F_y-y_G\cdot F_x=x_G(-W)-y_G\cdot 0=-Wx_G$.

So $M_W=-W\cdot a(1-3\cos\phi/8)=-Wa(8-3\cos\phi)/8$. (Negative = clockwise.)

String tension $T(-\sin\theta,\cos\theta)$ at $R=(a(1-\sin\phi),a\cos\phi)$: moment $z$-component $=x_R\cdot F_y-y_R\cdot F_x=a(1-\sin\phi)\cdot T\cos\theta-a\cos\phi\cdot T(-\sin\theta)=aT[(1-\sin\phi)\cos\theta+\cos\phi\sin\theta]$. (Positive = counter-clockwise.)

Sum to zero:

Substitute $T=W/\cos\theta$:

Divide by $\cos\phi$:

Answer