A 0.250 kg ball attached to a string s being swung in a vertical circle, like a
ID: 2122261 • Letter: A
Question
A 0.250 kg ball attached to a string s being swung in a vertical circle, like a roller coaster doing a loop-de-loop, with a radius of 1.15 m.
(a) What is the minimum speed at the top of the circle to keep the string from going slack? (At this speed, the tension in the spring is zero).
(b) If the speed of the ball increases by a factor of sqrt(5) in swinging to the bottom of the circle (v_bottom = sqrt(5) * v_top), what will be the tension in the spring at the bottom of the circle?
Explanation / Answer
a) m*v^2/r = m*g.................... ==> v = sqrt(g*r) = 3.357 m/s.......................... b) velocity at bottom, V = sqrt(5) * V_top = sqrt(5)*3.357 = 7.506 m/s............... Tension = m*v^2/r + m*g = 0.25*7.506^2/1.15 + 0.25*9.8 = 14.7 N......................
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.