System of Particles and Rotational Motion - Result Question 48
51. The angular momentum of a body with mass (m), moment of inertia (I) and angular velocity $(\omega) rad / sec$ is equal to
[1996]
(a) $I \omega$
(b) $I \omega^{2}$
(c) $\frac{I}{\omega}$
(d) $\frac{I}{\omega^{2}}$
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Answer:
Correct Answer: 51. (a)
Solution:
- (a) Let body contain $m_1, m_2, m_3 \ldots \ldots . m_n$ masses at distance $r_1, r_2, r_3 \ldots \ldots . . r_n$ from axis $O A$.
Angular momentum of body
$=m_1 v_1 r_1+m_2 v_2 r_2 \ldots . .+m_n v_n r_n$
$=m_1(\omega r_1) r_1+m_2(\omega r_2) r_2 \ldots \ldots+m_n(\omega r_n) r_n$
$=(m_1 r_1^{2}) \omega+(m_2 r_2^{2}) \omega \ldots \ldots .+(m_n r_n^{2}) \omega$
$=(\sum _{i=1}^{n} m_i r_i^{2}) \omega=I \omega$
