For a particle in uniform circular motion the acceleration →a at a point P(R,�

Question

For a particle in uniform circular motion the acceleration

\to a

at a point

P (R, θ)

on the circle of radius R is (here

θ

is measured from the x-axis)

Options:

A . −v2Rcosθ^i+v2Rsinθ^j

B . −v2Rsinθ^i+v2Rcosθ^j

C . −v2Rcosθ^i+v2Rsinθ^j

D . v2R+v2R^j

Answer: Option C
:
C

\to r = R c o s θ^j

∴ \frac{d \to r}{d t} = - R s i n θ \frac{d θ}{d t}^i + R c o s θ \frac{d θ}{d t}^j

\Rightarrow \to v = - R ω s i n θ^i + R ω c o s θ^j

(a s ω = \frac{d θ}{d t} a n d \frac{d \to r}{d t} = \to v)

\Rightarrow \frac{d \to v}{d t} = \to a = R ω c o s θ \frac{d θ}{d t}^i + R ω (- s i n θ) \frac{d θ}{d t}^j

\Rightarrow \frac{d s}{d t} \equiv \frac{d θ r}{d t} = r \frac{d θ}{d t} = r ω

∴ ω o r \frac{d θ}{d t} i s c o n s t a n t

∴ \to v = - R ω^{2} c o s θ^i - R ω^{2} s i n θ^j

Substituting

ω = \frac{v}{R}

\to a = \frac{- v^{2}}{R} c o s θ^i - \frac{- v^{2}}{R} s i n θ^j

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