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  • A cubical block of mass '  ' and density <img alt="1000 \mathrm{~kg} / \mathr

A cubical block of mass ' 1 \mathrm{~kg} ' and density 1000 \mathrm{~kg} / \mathrm{m}^3 is moving with velocity of 12 \mathrm{~m} / \mathrm{s} on a frictionless plane as shown in the figure. Moving forward, it hits a ridge, then the angular speed of the block just after hitting the ridge will be:-

Option: 1

15 rad/s


Option: 2

30 rad/s


Option: 3

60 rad/s


Option: 4

90 rad/s


Answers (1)

best_answer

\begin{aligned} & v=\frac{m}{\rho}=\frac{1}{1000} \mathrm{~m}^3 \\ & v=10^{-3} \times 10^6 \mathrm{~cm}^3 \\ & v=10^3 \mathrm{~cm}^3 \\ & a^3=10^3 \mathrm{~cm}^3 \\ & a=10 \mathrm{~cm} \end{aligned}

conserving the angular momentum about '0'

\begin{aligned} 1 \times 12 \times \frac{5}{100} & =\left(\frac{1 \times 10^2 \times 10^{-4}}{6}+\frac{1 \times}{10^4}\left(\frac{10}{\sqrt{2}}\right)^2\right) \omega \\ \frac{12 \times 5}{100} & =\left(\frac{10^{-2}}{6}+\frac{100}{2} \times 10^{-4}\right) \omega \\ \frac{12 \times 5}{100} & =\frac{1}{100}\left(\frac{1}{6}+\frac{1}{2}\right) \omega \\ 60 & =\left(\frac{2+6}{12}\right) \omega \\ \omega & =\frac{12 \times 60}{8} \\ \omega & =90 \mathrm{rad} / \mathrm{s} \end{aligned}

Posted by

himanshu.meshram

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