(A) The formula for centripetal acceleration $a_c$ is given by $a_c = r \omega^2$,where $r$ is the radius and $\omega$ is the angular speed.Given: $r

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(A) The formula for centripetal acceleration $a_c$ is given by $a_c = r \omega^2$,where $r$ is the radius and $\omega$ is the angular speed.Given: $r = 20 \, cm$ and $a_c = 980 \, cm \, s^{-2}$.Substituting the values into the formula:$980 = 20 \times \omega^2$$\omega^2 = \frac{980}{20} = 49$$\omega = \sqrt{49} = 7 \, rad \, s^{-1}$.

Class 11 Physics 3-2.Motion in Plane Angular Variables and Basic of Uniform Circular Motion

Easy

$A$ heavy stone is tied to the end of a string and whirled in a horizontal circular path of radius $20 \, cm$ with a constant angular speed. If the centripetal acceleration is $980 \, cm \, s^{-2}$,what is its angular speed?

A
$7 \, rad \, s^{-1}$
B
$14 \, rad \, s^{-1}$
C
$21 \, rad \, s^{-1}$
D
$49 \, rad \, s^{-1}$

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3-2.Motion in Plane

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Angular Variables and Basic of Uniform Circular Motion

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MediumAIPMT 1995

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DifficultJEE MAIN 2023

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For a particle in uniform circular motion,the acceleration $\vec{a}$ at a point $P(R, \theta)$ on the circle of radius $R$ is (Here $\theta$ is measured from the $x$-axis):

MediumAIEEE 2010

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$A$ particle is moving on a circular path of radius $r$ with uniform speed $v$. The change in velocity when the particle moves from $P$ to $Q$ is $(\angle POQ = 40^\circ)$.

Easy

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$A$ heavy stone is tied to the end of a string and whirled in a horizontal circular path of radius $20 \, cm$ with a constant angular speed. If the centripetal acceleration is $980 \, cm \, s^{-2}$,what is its angular speed?

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