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Physics-

General

Easy

Question

A particle moves in a circle with a uniform speed. When it goes from a point $A$ to a diametrically opposite point $B$ , the momentum of the particle changes by $stack p with rightwards arrow on top subscript A end subscript minus stack p with rightwards arrow on top subscript B end subscript equals 2 blank k g blank m s to the power of negative 1 end exponent left parenthesis stack j with hat on top right parenthesis$ and the centripetal fore acting on it changes by $stack F with rightwards arrow on top subscript A end subscript minus stack F with rightwards arrow on top subscript B end subscript equals 8$ N $open parentheses stack i with hat on top close parentheses$ where $stack i with hat on top$ and $stack j with hat on top$ are unit vectors along $X$ and $Y$ axes respectively. The angular velocity of the particle is

Dependent of its mass
4 $r a d s^{- 1}$
$\frac{2}{π} r a d s^{- 1}$
$16 μ$ rad $s^{- 1}$

The correct answer is: 4 $r a d blank s to the power of negative 1 end exponent$

$stack p with rightwards arrow on top subscript B end subscript minus stack p with rightwards arrow on top subscript A end subscript equals m open parentheses stack v with rightwards arrow on top subscript B end subscript minus stack v with rightwards arrow on top subscript A end subscript close parentheses equals m v left parenthesis stack j with hat on top plus stack i with hat on top right parenthesis$
$equals 2 m v stack j with hat on top equals 2 k g blank m s to the power of negative 1 end exponent$ (i)
$stack F with rightwards arrow on top subscript B end subscript minus stack F with rightwards arrow on top subscript A end subscript equals fraction numerator m v to the power of 2 end exponent over denominator R end fraction open parentheses negative stack i with hat on top close parentheses minus fraction numerator m v to the power of 2 end exponent over denominator R end fraction left parenthesis plus stack i with hat on top right parenthesis$
$equals fraction numerator 2 m v to the power of 2 end exponent over denominator R end fraction open parentheses negative stack i with hat on top close parentheses equals 8 N$ (ii)
Dividing Eq. (ii) by Eq. (i), $fraction numerator v over denominator R end fraction equals omega equals 4$ rad $s to the power of negative 1 end exponent$

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