A) top of the circle
B) bottom of the circle
C) half way down
D) None of the above
Correct Answer: A
Solution :
\[mg=1\times 10=10N\] \[\frac{m{{v}^{2}}}{r}=\frac{1\times {{(4)}^{2}}}{1}=16\] Tension at the top of circle \[=\frac{m{{v}^{2}}}{r}-mg=6N\] Tension at the bottom of circle \[=\frac{m{{v}^{2}}}{r}+mg=26N\]You need to login to perform this action.
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