A particle of mass $15 \mathrm{~kg}$ is moving with a uniform speed $8 \mathrm{~ms}^{-1}$ in $x y$-plane…

A particle of mass $15 \mathrm{~kg}$ is moving with a uniform speed $8 \mathrm{~ms}^{-1}$ in $x y$-plane along the line $3 y=4 x+10$, then the magnitude of its angular momentum about the origin in $\mathrm{kg} \cdot \mathrm{m}^2 \mathrm{~s}^{-1}$ is $\ldots\left(\sin 53^{\circ}=\frac{4}{5}\right)$
  1. 240
  2. 80
  3. 120
  4. 280

Solution


Angular momentum $=m v r_1$ $ =15 \times 8 \times\left(\frac{10}{\sqrt{3^2+4^2}}\right)=15 \times 8 \times 2=240 \mathrm{~kg}-\mathrm{m}^2 \mathrm{~s}^{-1} $

Asked in: AP EAMCET 2018 (23 Apr Shift 2)

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