A) at the top of the circle
B) at the bottom of the circle
C) halfway down
D) None of the above
Correct Answer: B
Solution :
Given, \[m=2kg.\,\,,\,v=4m/s\] \[r=1m,\] \[T=52N\] \[T-mg\,\cos \theta =\frac{m{{v}^{2}}}{r}\] \[52-2\times 10\cos \theta =\frac{2\times 4\times 4}{1}\] \[\cos \theta =\frac{20}{20}=1\] \[=\cos \,{{0}^{o}}\] \[\therefore \] \[\theta ={{0}^{o}}\] \[\theta ={{0}^{o}}\]at the bottom of the circle.
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