A 0.60-kg rock is swung in a circular path and in a vertical plane on a 0.25-m length string. At the top of the path, the angular speed is 12.0 rad/s. What is the tension in the string at that point?Select one:a.7.9 Nb.16 Nc.6.5 Nd.22 N
Question
A 0.60-kg rock is swung in a circular path and in a vertical plane on a 0.25-m length string. At the top of the path, the angular speed is 12.0 rad/s. What is the tension in the string at that point?Select one:a.7.9 Nb.16 Nc.6.5 Nd.22 N
Solution
To solve this problem, we need to understand that the tension in the string is the sum of the centripetal force (which is what keeps the rock moving in a circle) and the force due to gravity.
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First, let's calculate the centripetal force. The formula for centripetal force is Fc = m * ω^2 * r, where m is the mass of the object, ω is the angular speed, and r is the radius of the circle.
Substituting the given values, we get Fc = 0.60 kg * (12.0 rad/s)^2 * 0.25 m = 21.6 N.
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Next, let's calculate the force due to gravity. The formula for this is Fg = m * g, where m is the mass of the object and g is the acceleration due to gravity. On Earth, g is approximately 9.8 m/s^2.
Substituting the given values, we get Fg = 0.60 kg * 9.8 m/s^2 = 5.88 N.
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Finally, let's add these two forces together to find the tension in the string. T = Fc - Fg = 21.6 N - 5.88 N = 15.72 N.
So, the tension in the string at the top of the path is approximately 15.72 N. The closest answer to this is 16 N, so the correct answer is (b) 16 N.
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