SATHEE: Rotational Motion Question 171

Rotational Motion Question 171

Question: 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

Options:

A) $14.4 k g m^{2} s^{- 1}$

B) $8.64 k g m^{2} s^{- 1}$

C) $20.16 k g m^{2} s^{- 1}$

D) $11.52 k g m^{2} s^{- 1}$

Show Answer

Answer:

Correct Answer: A

Solution:

[a] Angular momentum,

$L_{0} = m v r \sin 90^{o}$

$= 2 \times 0.6 \times 12 \times 1 \times 1$

$A s V = r ω, S i n 90^{o} = 1$

$S o, L_{0} = 14.4 k g m^{2} / s$

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Rotational Motion Question 171

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