Rotation 2 Question 6
6. A small mass $m$ is attached to a massless string whose other end is fixed at $P$ as shown in the figure. The mass is undergoing circular motion in the $x-y$ plane with centre at $O$ and constant angular speed $\omega$. If the angular momentum of the system, calculated about $O$ and $P$ are denoted by $\mathbf{L} _O$ and $\mathbf{L} _P$ respectively, then
(a) $\mathbf{L} _O$ and $\mathbf{L} _P$ do not vary with time
(b) $\mathbf{L} _O$ varies with time while $\mathbf{L} _P$ remains constant
(c) $\mathbf{L} _O$ remains constant while $\mathbf{L} _P$ varies with time
(d) $\mathbf{L} _O$ and $\mathbf{L} _P$ both vary with time
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Answer:
Correct Answer: 6. (c)
Solution:
- Angular momentum of a particle about a point is given by $\mathbf{L}=\mathbf{r} \times \mathbf{p}=m(\mathbf{r} \times \mathbf{v})$
For $\mathbf{L} _O$
$|\mathbf{L}|=(m v r \sin \theta)=m(R \omega)(R) \sin 90^{\circ}=$ constant Direction of $\mathbf{L} _O$ is always upwards. Therefore, complete $\mathbf{L} _O$ is constant, both in magnitude as well as direction.
For $\mathbf{L} _P$
$\left|\mathbf{L} _P\right|=(m v r \sin \theta)=(m)(R \omega)(l) \sin 90^{\circ}=(m R l \omega)$
Magnitude of $\mathbf{L} _P$ will remain constant but direction of $\mathbf{L} _P$ keeps on changing.
