A small particle of mass $m$ is given an initial high velocity in the horizontal plane and winds its cord around the fixed vertical shaft of radius $a$ . All motion occurs essentially in horizontal plane. If the angular velocity of the cord is $ω_{0}$ , when the distance from the particle to the tangency point is $r_{0}$ , then the angular velocity of the cord $ω$ after it has turned through an angle $θ$ is :

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

$ω = ω_{0}$

$ω = \frac{a}{r_{0}} ω_{0}$

$ω = (\frac{ω_{0}}{1 - \frac{a}{r_{0}} θ})$

$ω = ω_{0} θ$

answer is C.

(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

$r = r_{0} - a θ$

In the process $v$ remains constant, so

$ω_{0} r_{0} = ω r \to ω = \frac{ω_{0} r_{0}}{r_{0} - a θ} = \frac{ω_{0}}{(1 - \frac{a θ}{r})}$

JEE

NEET

Foundation JEE

Foundation NEET

CBSE