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Physics-

General

Easy

Question

A particle is moving in a circle of radius $R$ in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at $t equals 0 blank i s blank v subscript 0 end subscript$ , the time taken to complete the first revolution is

$\frac{R}{v_{0}}$
$\frac{R}{v_{0}} (1 - e^{- 2 π})$
$\frac{R}{v_{0}} e^{- 2 π}$
$\frac{2 π R}{v_{0}}$

The correct answer is: $fraction numerator R over denominator v subscript 0 end subscript end fraction left parenthesis 1 minus e to the power of negative 2 pi end exponent right parenthesis$

When a particle moves in a circular motion, it is acted upon by centripetal force directed towards the centre. Hence, centripetal acceleration is
$a subscript N end subscript equals fraction numerator d v over denominator d t end fraction equals fraction numerator v to the power of 2 end exponent over denominator R end fraction$
$o r blank not stretchy integral subscript 0 end subscript superscript t end superscript fraction numerator d t over denominator R end fraction equals not stretchy integral subscript v subscript 0 end subscript end subscript superscript v end superscript fraction numerator d v over denominator v to the power of 2 end exponent end fraction$
$o r blank t equals negative R open square brackets fraction numerator 1 over denominator v end fraction close square brackets subscript v subscript 0 end subscript end subscript superscript v end superscript$
$v equals fraction numerator v subscript 0 end subscript R over denominator R minus v subscript 0 end subscript t end fraction$
$A l s o blank fraction numerator d r over denominator d t end fraction equals fraction numerator v subscript 0 end subscript R over denominator open parentheses R minus v subscript 0 end subscript t close parentheses end fraction$
$not stretchy integral subscript 0 end subscript superscript 2 pi blank R end superscript d r equals v subscript 0 end subscript R blank not stretchy integral subscript 0 end subscript superscript T end superscript fraction numerator d t over denominator R minus v subscript 0 end subscript t end fraction$
$⟹ blank T equals fraction numerator R over denominator v subscript 0 end subscript end fraction open parentheses 1 minus 1 minus e to the power of negative 2 pi end exponent close parentheses$

Related Questions to study

A particle of mass $m$ is rotating in a horizontal circle of radius $R$ and is attached to a hanging mass $M$ as shown in the figure. The speed of rotation required by the mass $m$ keep $M$ steady is

A particle of mass $m$ is rotating in a horizontal circle of radius $R$ and is attached to a hanging mass $M$ as shown in the figure. The speed of rotation required by the mass $m$ keep $M$ steady is
Physics-General

A constant force $stack F with rightwards arrow on top$ = 3 $stack i with hat on top$ + 2 $stack j with hat on top$ – 4 $stack k with hat on top$ is applied at point A (1, –1, 2). Then vector moment of $stack F with rightwards arrow on top$ about point B (2,–1, 3) is -

A constant force $stack F with rightwards arrow on top$ = 3 $stack i with hat on top$ + 2 $stack j with hat on top$ – 4 $stack k with hat on top$ is applied at point A (1, –1, 2). Then vector moment of $stack F with rightwards arrow on top$ about point B (2,–1, 3) is -

The vectors $stack i with hat on top$ + 2 $stack j with hat on top$ +3 $stack k with hat on top$ ,  $stack i with hat on top$ + 4 $stack j with hat on top$ + 7 $stack k with hat on top$ &–3 $stack i with hat on top$ –2 $stack j with hat on top$ –5 $stack k with hat on top$ are collinear, if  =

The vectors $stack i with hat on top$ + 2 $stack j with hat on top$ +3 $stack k with hat on top$ ,  $stack i with hat on top$ + 4 $stack j with hat on top$ + 7 $stack k with hat on top$ &–3 $stack i with hat on top$ –2 $stack j with hat on top$ –5 $stack k with hat on top$ are collinear, if  =

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If $stack u with rightwards arrow on top$ , $stack v with rightwards arrow on top$ , $stack w with rightwards arrow on top$ are non-coplanar vectors then
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