Question

A 1.0 m string with a 5 g (.005 kg) stopper on the end is whirled in a vertical circle. The speed of the stopper is 8 m/s at the top of the circle.

What is the speed of the stopper at the bottom of the circle (Use energy conservation principle s)(answer 10.2 m/s), what is the tension in the string when the stopper is at at the top of the circle (.27N), What is the tension in the string when the stopper is at the bottom of the circle (.57N)?

Can you show step by step please and formulas used? Thanks in advance

Answer 1

From conservation of energy 1/2 mv_b^r = 1/2 mv_t^2 + mgh 1/2 times v_b^2 = 1/2 mv_t^2 + mgh 1/2 times v_b^2 = 1/2 times 8^2 + 9.8 times 2 v_b^2 = 8^2+ 4 times 9.8 v_b = 10.2 m/s at top mv_t^2/r = t_top + mg 5 times 10^-3 times 8^2/1 - 5 times 10^-3 9.8 = t_top t_top = 0.27 n at bottom t_ bottom = mv_b^2/r + mg t_bottom = 5 times 10^3 times 10.2^2 + 5 times 10^-3 times 9.8 t_bottom = 0.57 N