Question

A chocolate orange consists of a sphere of delicious smooth uniform chocolate of mass M and radius a, sliced into segments by planes through fixed axis. It stands on a horizontal table with this axis and it is held together by a narrow ribbon round its equator. Show that the tension in the ribbon is atleast 3/32 Mg.
(You may assume that the centre of mass of a segment of angle 2θ is at distance (3πa sinθ)/16θ from the axis.)​

Answer 1

First of all, see the 1st diagram to visualise a section of the orange.

• Now, Tension T in the ribbon is resolved into components as shown in diagram.

• Also, $T\cos(\theta)$ component gets cancelled on both sides.

Now, Mass of segment of orange :

$\rm m = \dfrac{M}{total \: angle} \times given \: angle$

$\rm \implies m = \dfrac{M}{2\pi} \times 2 \theta$

$\rm \implies m = \dfrac{M \theta}{\pi}$

Taking necessary torques :

$\rm \: \dfrac{Mg \theta}{\pi} \times \dfrac{3\pi a \sin( \theta) }{16 \theta} = 2T \sin( \theta) \times a$

$\rm \implies T = \dfrac{3Mg}{32}$

Hence proved ✔️

Answer 2

Answer:

The above diagram shows a horizontal cross-section of a segment of the orange.

Resolving the tension T in the ribbon as shown gives a horizontal component of force due to tension on the segment of 2Tsinθ towards the left of the diagram.

The volume of the segment is a fraction 2θ∕(2π) of the volume of the sphere, so the segment has mass M(2θ)∕(2π). The weight of the segment gives a force of Mgθ∕π acting downwards through the centre of mass as shown in the diagram below, which is a vertical cross-section of a segment.

Taking moments about the point of contact of the table and the segment gives

Mgθπ×3πasinθ16θ=2Tsinθ×a

which gives the required answer.