The angular acceleration of a particle moving along a circle is given by α=ksin⁡θ, where θ is the angle turned by particle and k is constant. The centripetal acceleration of the particle in terms of k,θ and R (radius of circle) is given by :

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

2kRsin2⁡θ

2kR(sin⁡θ)

2kR(1+cos⁡θ)

2kR(1−cos⁡θ)

answer is D.

(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=ksin⁡θ∫ωdω=∫ksin⁡θdθω22=−[kcos⁡θ]0θ⇒ω22=−k[cos⁡θ−1]ω2=2k(1−cos⁡θ)ac=ω2R=2kR(1−cos⁡θ)

Watch 3-min video & get full concept clarity