A) at \[{{P}_{4}}\]
B) at \[{{P}_{1}}\]
C) at \[{{P}_{2}}\]
D) at \[{{P}_{3}}\]
Correct Answer: A
Solution :
Since angular momentum is conserved Therefore, \[{{I}_{1}}{{\omega }_{1}}={{I}_{2}}{{\omega }_{2}}\] or \[MR_{1}^{2}{{\omega }_{1}}=MR_{2}^{2}{{\omega }_{2}}\] or \[M{{R}_{1}}{{\upsilon }_{1}}=M{{R}_{2}}{{\upsilon }_{2}}\] At \[{{P}_{4}}\] the value of R is minimum Hence the velocity is maximum or kinetic energy will be maximum.
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