A student swings a 100 g ball on a 60 cm long string in a vertical circle about

Question

A student swings a 100 g ball on a 60 cm long string in a vertical circle about a point 200 cm above the floor. The tension in the string when the ball is at the very bottom of the circle is 5.0 N. Another sneaky student decides to spoil all the fun by suddenly cutting the string directly below the point of support. Draw the free body diagram for the ball when it is at the very bottom of the circle. Write Newton's 2nd Law along the two axes in your free body diagram. Where does the ball hit the floor? (2.625153057)

Explanation / Answer

C) Well, we know at the bottom, T = mg +mv^2/r so 5.0 N = .1*9.8+ .1*v^2/(.6) Solving for v, we get v = 4.9112116631234699819722222348404 m/s. Now finding how long it takes for the ball to hit earth/ground, we get 1.40 = .5gt^2 so t = 0.53452248315669566417045030903797. It is 1.40 because while the center is at 2 m, the object is extended .6 below the center at the bottom. Multiplying this by the velocity, we should get x = 2.6251530534808822837745514478569m. For part B) along x axis, Fx = 0. And along y axis: T - mg = mv^2/r. Unfortunately I cannot do part a on here but I can describe it as a simple box with an arrow T going up and another arrow mg going down. If you want, you can show mv^2/r going up as a resultant in a different colored pencil or with a tiny note next to it saying "resultant"

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