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A spherical shell of 1 \mathrm{~kg} mass and radius \mathrm{R} is rolling with angular speed \omega  on horizontal plane (as shown in figure). The magnitude of angular momentum of the shell about the origin \mathrm{O} is \frac{\mathrm{a}}{3} \mathrm{R}^{2} \omega. The value of a will be :

Option: 1

2


Option: 2

3


Option: 3

5


Option: 4

4


Answers (1)

\mathrm{\vec{L}_{0}=\vec{L}_{\mathrm{cm}}+\vec{L}_{\text {about cm }} }

        \mathrm{=m V R+\frac{2}{3} m R^{2} \omega , \quad (Using V=R \omega })

         \begin{aligned} &\mathrm{=\frac{m R^{2}}{1} \omega +\frac{2}{3} m R^{2} \omega } \\ \\&\mathrm{=\frac{5}{3} m R^{2} \omega } \end{aligned}

Hence, correct answer is Option (3).

Posted by

Kshitij

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