A particle of mass 2 kg is on a smooth horizontal table and moves in a circular path of radius 0.6 m. The height of the table from the ground is 0.8 m. If the angular speed of the particle is 12 rad s^-1, the magnitude of its angular momentum about a point on the ground right under the centre of the circle is :

Option 1)
8.64 kg m²s^-1

Option 2)
11.52 kg m²s^-1

Option 3)
14.4 kg m²s^-1

Option 4)
20.16 kg m²s^-1

Answers (1)

V Vakul

As we learnt

Angular momentum -

$\vec{L}=\vec{r}\times \vec{p}$

- wherein

$\vec{L}$ represent angular momentum of a moving particle about a point.

it can be calculated as $L=r_1\, P=r\, P_1$

$r_1$ = Length of perpendicular on line of motion

$P_1$ = component of momentum along perpendicualar to $r$

Angular momentum

$L_{0}=mvr sin\theta$

$\theta=90^{o}$

$L_{0}=mvrsin\;90^{o}$

After calculation we get

$L_{0}=14.4 kg\;m^{2}/sec$

Option 1)

8.64 kg m²s^-1

This is an incorrect option.

Option 2)

11.52 kg m²s^-1

This is an incorrect option.

Option 3)

14.4 kg m²s^-1

This is the correct option.

Option 4)

20.16 kg m²s^-1

This is an incorrect option.

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