(a) We can find the velocity at the top of the circle from the law of conservation of energy:
"\\dfrac{1}{2}mv_{top}^2+mg2r=\\dfrac{1}{2}mv_{bottom}^2,""v_{top}=\\sqrt{v_{bottom}^2-4gr},""v_{top}=\\sqrt{(10\\ \\dfrac{m}{s})^2-4\\times9.8\\ \\dfrac{m}{s^2}\\times0.5\\ m}=8.97\\ \\dfrac{m}{s}."(b) We can find the tension at the top of the circle as follows:
"T_{top}=\\dfrac{mv_{top}^2}{r}-mg,""T_{top}=\\dfrac{0.2\\ kg\\times(8.97\\ \\dfrac{m}{s})^2}{0.5\\ m}-0.2\\ kg\\times9.8\\ \\dfrac{m}{s^2}=30.2\\ N."
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