A 60 cm rope is tied to the handle of a bucket which is then whirled in a vertical circle. The mass of the bucket is 3 kg.What is the critical speed below which the rope would become slack when the bucket reaches the highest point in the circle?
Question
A 60 cm rope is tied to the handle of a bucket which is then whirled in a vertical circle. The mass of the bucket is 3 kg.What is the critical speed below which the rope would become slack when the bucket reaches the highest point in the circle?
Solution
To solve this problem, we need to use the concept of centripetal force and gravitational force.
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At the highest point of the circle, the bucket is held up by the tension in the rope and the gravitational force. The tension in the rope becomes zero when the bucket is at the verge of making the rope slack. So, the gravitational force provides the necessary centripetal force for circular motion at this point.
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The centripetal force is given by the equation F = mv^2/r, where m is the mass of the bucket, v is the velocity (or speed) of the bucket, and r is the radius of the circle.
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The gravitational force is given by the equation F = mg, where m is the mass of the bucket and g is the acceleration due to gravity.
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Setting these two equations equal to each other gives us mv^2/r = mg.
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We can simplify this to v^2 = rg.
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Plugging in the given values (r = 0.6 m, g = 9.8 m/s^2), we get v^2 = 0.6 m * 9.8 m/s^2 = 5.88 m^2/s^2.
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Taking the square root of both sides gives us v = sqrt(5.88 m^2/s^2) = 2.43 m/s.
So, the critical speed below which the rope would become slack when the bucket reaches the highest point in the circle is 2.43 m/s.
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