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THE K.E OF A PARTICLE MOVING ALONG A CIRCLE OF RADIUS R DEPENDS ON THE DISTANCE COVERED S. IT IS GIVEN BY K=as WHAT IS THE FORCE ACTING ON THE PARTICLE?

nik_gup

since K.E.=as

1/2 m.v^2=a.s

mv^2=2.a.s............(1)

now net force on body = mv^2/R(since body is performing

circular motion of radius R)

therefore ,F=2.a.s/R.....(from(1))

please rate if correct

rhd92781

K=as = mv^2/2

v=^]

(2as/m)

acc, A=dv/dt = ^]

(2a/m) * 1/2^]

s *v

A=

(a/2ms) *

(2as/m)

A=a/m

so, F_t=mA = a

F_c=mv^2/R = 2as/R

Resultant F=

(a²+4(as/R)²)

krishna.gopal

Perfect answer by Rhd92781. nik you missed the tangential acc. Its not uniform circular motion

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