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6.System of Particles and Rotational Motion
medium
$V _{ CM }=2\; m / s , m =2\;kg , R =4 \;m$ જ્યારે રીંગ સંપૂર્ણ ગબડે ત્યારે તેનું કોણીય વેગમાન ઉદગમબિંદુને અનુલક્ષીને ($kgm ^{2} / s$ માં)

A
$32$
B
$24$
C
$16$
D
$8$
(AIIMS-2019)
Solution
The moment of inertia about centre of mass is,
$I_{C M}=M R^{2}$
The angular velocity of the wheel is,
$\omega=\frac{V_{C M}}{R}$
$\omega=\frac{2}{4}=0.5 rad / s$
The angular momentum of the ring about origin is,
$\vec{L}=I_{C M} \omega+m V_{C M} R$
$=2 \times 16 \times 0.5+2 \times 2 \times 4$
$=16+16$
$=32 \;kg \cdot m ^{2} / s$
Standard 11
Physics
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