1.

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»

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


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