A) 12.5
B) 40
C) 32.6
D) 15.6
Correct Answer: A
Solution :
From equation of motion \[0={{\omega }_{0}}-\alpha t\] \[\alpha =\frac{{{\omega }_{0}}}{t}\] \[=\frac{(100\times 2\pi )/60}{15}\] \[=0.6\,\,rad{{s}^{-2}}\] Now, angle rotated before coming to rest \[\theta =\frac{\omega _{0}^{2}}{2\alpha }\] or \[\theta =\frac{{{\left( \frac{100\times 2\pi }{60} \right)}^{2}}}{2\times 0.7}=78.25\,rad\] Number of rotations \[n=\frac{\theta }{2\pi }=12.5\]
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