A) \[8\,\,rad/\sec \]
B) \[4\,\,rad/\sec \]
C) \[2\,\,rad/\sec \]
D) \[1\,\,rad/\sec \]
Correct Answer: B
Solution :
Using the relation Force = breaking stress\[\times \]area \[=(4.8\times {{10}^{7}})({{10}^{-6}})=48\,\,N\] ... (1) Using the relation \[F=Mr{{\omega }^{2}}\] ... (2) \[Mr{{\omega }^{2}}=48\] \[10\times 0.3\times {{\omega }^{2}}=48\] \[{{\omega }^{2}}=\frac{48}{10\times 0.3}=16\] or \[{{\omega }^{2}}=16\] \[\omega =4\,\,rad/\sec \]You need to login to perform this action.
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