The angular acceleration of the toppling pole shown in figure is given by $α = k \sin θ,$ where $θ$ is the angle between the axis of the pole and the vertical, and k is a constant. The pole starts from rest at $θ = 0 .$ Choose the correct options

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

The centripetal acceleration of the upper end of the pole is $2 k / (1 - \cos θ)$

The tangential acceleration of the upper end of the pole is $/ k \sin θ$

The centripetal acceleration of the upper end of the pole is $/ k \sin θ$

The tangential acceleration of the upper end of the pole is $2 k / (1 - \cos θ)$

answer is A, B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$α = k \sin θ$

$\frac{ω d ω}{d θ} = k \sin θ$

$\frac{ω^{2}}{2} = k (\cos θ)$

$ω^{2} = k (\cos θ - 1)$

$a_{t} = l α, a_{c} = l ω^{2}$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE