Question

A uniform rod AB of length 3r remains in equillibrium on a hemispherical bowl of radius r as shown .

Answer 1

since there is equilibrium between rid and bowl
First draw complete circle to the bowl.
Now make the diameter , through lower end , weight force line and one of the normal line at
othe end meet at a point on the circle D (let)
Now
by geometry rod divides angle made by rod with
horizontal and angle made by rod with diameter made by lower end and joining common
point equally.

After this we get an equation

2Rcos2@=1.5Rcos@
ie
2cos2@=1.5cos@
After solving this we get
cos@=0.92

Please mark me the brainliest!
Suhani
xx

Answer 2

Answer: cos⁻¹(0.92)

Explanation:

The forces acting on the rod are:

(i) Weight W of the rod acting vertically downwards from centre of gravity.

(ii) Reaction R at A acting normally at A i.e. along AO.

(iii) Reaction R’ at C acting at right angles to rod. For equilibrium of rod the three forces will be concurrent. By geometry, ∠OCA = ∠OAC = ∠GDC = θ

In Triangle GDC,

tan θ = GC/DC = (AC-AG)/DC or sinθ / cosθ = (2r cosθ - 1.5r)/2r sinθ

2 sin²θ = 2 cos²θ - 1.5 cosθ or

1 - cos²θ = cos²θ - 0.75 cosθ or

2 cos²θ - 0.75 cosθ - 1 = 0

on solving,

cosθ = 0.92 (using the +ve answer)