Rotation 2 Question 11

11. A disc of mass $M$ and radius $R$ is rolling with angular speed $\omega$ on a horizontal plane as shown. The magnitude of angular momentum of the disc about the origin $O$ is

(1999, 2M)

(a) $\frac{1}{2} M R^{2} \omega$

(b) $M R^{2} \omega$

(c) $\frac{3}{2} M R^{2} \omega$

(d) $2 M R^{2} \omega$

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Answer:

Correct Answer: 11. (c)

Solution:

  1. From the theorem

(a)

(b)

$$ \mathbf{L} _0=\mathbf{L} _{CM}+M(\mathbf{r} \times \mathbf{v}) $$

We may write

Angular momentum about $O=$ Angular momentum about $CM+$ Angular momentum of $CM$ about origin

$$ \begin{aligned} \therefore \quad L _0 & =I \omega+M R v \\ & =\frac{1}{2} M R^{2} \omega+M R(R \omega)=\frac{3}{2} M R^{2} \omega \end{aligned} $$

NOTE That in this case [ Figure (a) ] both the terms in Eq. (i), i.e. $L _{C M}$ and $M(r \times v)$ have the same direction $\ddot{A}$. That is why, we have used $L _0=/ \omega+M R v$. We will use $L _0=\mid \omega \sim M R v$ if they are in opposite direction as shown in figure (b).



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