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JEE MAIN - Physics (2016 (Offline) - No. 7)

A particle of mass m is moving along the side of a square of side ‘a’, with a uniform speed v in the x-y plane as shown in the figure :

JEE Main 2016 (Offline) Physics - Rotational Motion Question 178 English
Which of the following statements is false for the angular momentum $$\overrightarrow L $$ about the origin ?
$$\overrightarrow L = mv\left[ {{R \over {\sqrt 2 }} + a} \right]\widehat k$$
when the particle is moving from B to C.
$$\overrightarrow L = {{mv} \over {\sqrt 2 }}R\widehat k$$
when the particle is moving from D to A.
$$\overrightarrow L = - {{mv} \over {\sqrt 2 }}R\widehat k$$
when the particle is moving from A to B
$$\overrightarrow L = mv\left[ {{R \over {\sqrt 2 }} - a} \right]\widehat k$$
when the particle is moving from C to D.

Explanation

The angular momentum is

$$\overrightarrow L = m(\overrightarrow r \times \overrightarrow v )$$

For the particle moving from $$D \to A$$, we have

$$\overrightarrow L = m\left[ {\left( {{R \over {\sqrt 2 }}\widehat i + {R \over {\sqrt 2 }}\widehat j} \right) \times v( - \widehat j)} \right]$$

$$ = {{mvR} \over {\sqrt 2 }}( - \widehat k)$$

JEE Main 2016 (Offline) Physics - Rotational Motion Question 178 English Explanation

For the particle moving from $$A \to B$$, we have

$$\overrightarrow L = m\left[ {\left( {{R \over {\sqrt 2 }}\widehat i + {R \over {\sqrt 2 }}\widehat j} \right) \times v\widehat i} \right]$$

$$ = {{mvR} \over {\sqrt 2 }}( - \widehat k)$$

For the particle moving from $$C \to D$$, we have

$$\overrightarrow L = mv \times $$ ($$\bot$$ distance) $$\times$$ $$(\widehat k)$$

$$ = mv\left( {{R \over {\sqrt 2 }} + a} \right)(\widehat k)$$

For the particle moving from $$B \to C$$, we get

$$\overrightarrow L = mv \times $$ ($$\bot$$ distance) $$\times$$ $$(\widehat k)$$

$$ = mv\left( {{R \over {\sqrt 2 }} + a} \right)(\widehat k)$$

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