A) 0.02
B) 0.28
C) 0.5
D) 0.3
Correct Answer: C
Solution :
Angular impulse = change in angular momentum \[\tau \,\Delta t=\Delta L\] \[mg\frac{\ell }{2}\times .01=\frac{m{{\ell }^{2}}}{3}\omega \] \[\omega =\frac{3g\times 0.01}{2\ell }\] \[=\frac{3\times 10\times .01}{2\times 0.3}\] \[=\frac{1}{2}=0.5rad/s\] time taken by rod to hit the ground \[t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 5}{10}}=1\sec .\] in this time angle rotate by rod \[\theta =\omega t=0.5\times 1=0.5\] radian
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