A small mass attached to a string rotates on frictionless table top as shown. If the tension is the string is increased by pulling the string causing the radius of the circular motion to decrease by a factor of 2 , the kinetic energy of the mass will

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Correct answer: C. increase by a factor of 4

From the law of conservation of angular momentum So, aligned m v r & =m v^ (r)/(2) y^ & =2 y (k_1)/(k_2)& =((1)/(2) m v^2)/((1)/(2) m v^2) (k_1)/(k_2)& = y^2 v^ 2 & =(y^2)/((2 v)^2) (k_1)/(k_2)& =(1)/(4) k_1 & =(1)/(4)k_2 k_2 & =4 k_1 aligned

A small mass attached to a string rotates on frictionless table top...

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