3. You are rotating a 2.0-kg stone (with the help of a very light string) in a v

Question

3. You are rotating a 2.0-kg stone (with the help of a very light string) in a vertical circle of 1.0-m radius. When the stone is at the highest position, the tension in the string is 12.4 N. a) Apply Newton's 2nd Law to calculate the acceleration and then find the speed of the stone at the highest position. B) Apply the law of conservation of energy between the highest and lowest positions and determine the speed of the stone at the lowest position. c.) Calculate the acceleration and then apply Newton's 2nd Law to find the tension in the string when the stone is at the lowest position.

Explanation / Answer

m = 2 kg

r = 1 m

T_top = mv^2 /r - mg = 12.4 N

2 v^2 / 1 - 2 * 9.81 = 12.4

v^2 = 9.81 + 6.2

v = sqrt(9.81 +6.2) = 4 m/s

(a) acceleration = a = v^2 / r = 4^2 / 1 = 16 m/s^2

Speed = v = 4 m/s

(b) T_bottom = T_top + 6 mg = 12.4 + 6 * 2 * 9.81 = 130.12 N

mv^2 / r + mg = 130.12

2*v^2 / 1 + 2*9.81 = 130.12

v^2 = 65.6 - 9.81 = 55.79

v = sqrt(55.79) = 7.47 m/s

speed at the bottom = 7.47 m/s

(c) acceleration = v^2/r = 7.47^2 / 1 = 55.8 m/s^2

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