SATHEE: Rotation 2 Question 12

Rotation 2 Question 12

12. A mass $m$ is moving with a constant velocity along a line parallel to the $X$ -axis, away from the origin. Its angular momentum with respect to the origin

(1997C, 1M)

(a) is zero

(b) remains constant

(d) goes on decreasing

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Answer:

Correct Answer: 12. (b)

Solution:

$| v | = v =$ constant and $| r | = r$ (say)

Angular momentum of the particle about origin $O$ will be given by

$\begin{array}{ll} L = r \times p = m (r \times v) \\ or & | L | = L = m r v \sin θ = m v (r \sin θ) = m v h \end{array}$

Now, $m, v$ and $h$ all are constants.

Therefore, angular momentum of particle about origin will remain constant. The direction of $r \times v$ also remains the same (negative $z$ ).

NOTE Angular momentum of a particle moving with constant velocity about any point is always constant. e.g. Angular momentum of the particle shown in figure about origin $O$ will be

$L = m v h =$ constant

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Rotation 2 Question 12

12. A mass m is moving with a constant velocity along a line parallel to the X-axis, away from the origin. Its angular momentum with respect to the origin

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12. A mass $m$ is moving with a constant velocity along a line parallel to the $X$ -axis, away from the origin. Its angular momentum with respect to the origin