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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