Question

A particle of mass M moves with constant speed V on a circular path of radius R as shown in figure the average force on it during its Motion from A to B is

Answer 1

avgF = m[ v(final) - v(initial) ]/∆t

w = angular velocity
w = 2π/T = V/R

T = time period of oscillation
T = 2πR/V

so ∆t = (2π/3)/(2π/T)
∆t = T/3
∆t = 6πR/V


to solve it in vector form
let the x-axis is from centre to A
and y-axis is perpendicular to it from centre
and let the unit vector in the direction of x&y axis are " i&j " respectively


therefore
v(initial) = v j

v(final) = v [ cos30i - sin30j ]
v(final) = v [ √3i - j ]÷2


so
=> v(final) - v(initial)
=> v [ √3/2i - 3/2j ]


so
avgF = { mv[ √3i - 3j ]÷2 }/{ 6πR/v }
= mv²[ √3i - 3j ]÷[12πR]


and by taking its mode

= mv²/12πR×[ √(3+9) ]
= √12mv²/12πR
= mv²/2√3πR


I think I made some mistakes in solving
because I'm not using pen-paper
hope so you get it that how to solve it
otherwise I have to solve it again


hope it helps you
@di

Answer 2

⭐Hey there!!

=> given that a particle of mass "m" moves with constant speed on a circular path of radius 'r'

➡ Initian velocity(u) = vj

➡ final velocitu(v) = vcos30i - vsin30j

➡ change in velocity(v') = v - u

=> v' = vcos30i - vsin30j - vj

=> v' = v√3/2 i - v/2 j - vj

=> v' = v√3/2 i -3v/2 j

> |v'| = √(3v²/4 + 9v²/4) = 2v√3/2 = v√3 m/s

for 2π (i.e 360°) rotation time taken = 2πr/v

For 1 rotation time taken = 2πr/v ÷ 2π = r/v

For 4π/3 rotation tine taken(T) = r/v × 4π/3 =4πr/3v

So, | a | = |v'| / T => v√3÷ {4πr/3v} => 3√3v²/4πr

So, Average force [ Fav] = ma

=> Fav = 3√3mv²/4πr

⭕ption '4' is correct ✔

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⭐hope it will help u