Answer:
we know that,
for linear
Force=m×a
for angular
Force=m×alpha
we know that a=v^2/r
i.e for angular acceleration alpha=(omega)^2r
Force (centripetal)=m(omega)^2r
But the force acts towards the centre of cirucular path so,
Fc.p=-m(omega)^2r
Copyright © 2024 EHUB.TIPS team's - All rights reserved.
Answers & Comments
Answer:
we know that,
for linear
Force=m×a
for angular
Force=m×alpha
we know that a=v^2/r
i.e for angular acceleration alpha=(omega)^2r
Force (centripetal)=m(omega)^2r
But the force acts towards the centre of cirucular path so,
Fc.p=-m(omega)^2r