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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