(D) The centripetal acceleration of a particle moving in a circular path is given by the formula $a_{cp} = \frac{v^2}{r}$.Initially,the speed is $v$,s

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(D) The centripetal acceleration of a particle moving in a circular path is given by the formula $a_{cp} = \frac{v^2}{r}$.Initially,the speed is $v$,so the initial centripetal acceleration is $a_1 = \frac{v^2}{r}$.Finally,the speed is $2v$,so the final centripetal acceleration is $a_2 = \frac{(2v)^2}{r} = \frac{4v^2}{r}$.Comparing the two,we get $a_2 = 4 \times \left(\frac{v^2}{r}\right) = 4a_1$.Therefore,the centripetal acceleration has changed by a factor of $4$.

Class 11 Physics 3-2.Motion in Plane Non-uniform Circular Motion (Centrifugal/Centripetal and tangential Accelaration)

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$A$ particle moves in a circular path of radius $r$ with speed $v.$ It then increases its speed to $2v$ while travelling along the same circular path. The centripetal acceleration of the particle has changed by a factor of

A
$0.25$
B
$0.5$
C
$2$
D
$4$

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3-2.Motion in Plane

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Non-uniform Circular Motion (Centrifugal/Centripetal and tangential Accelaration)

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$A$ car moves with a speed of $30 \, m/s$ on a circular path of radius $500 \, m$. If its tangential acceleration is $2 \, m/s^2$,what is the magnitude of the total acceleration of the car in $m/s^2$?

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$A$ particle of mass $5 \ kg$ is moving on a circle of radius $1 \ m$ with an angular velocity of $2 \ rad \ s^{-1}$. The centripetal force is (in $N$)

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$A$ particle moves in a circular path of radius $r$ with speed $v.$ It then increases its speed to $2v$ while travelling along the same circular path. The centripetal acceleration of the particle has changed by a factor of

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