You swing a bucket full of water in a vertical circle at the end of a rope. The mass of the bucket plus the water is 2.4 kg. The
center of mass of the bucket plus the water moves in a circle of radius 1.0 m. At the instant that the bucket is at the top of the circle, the speed of the bucket is 7.8 m/s. What is the tension in the rope at this instant?
1 answer:
Answer:
122.5 N
Explanation:
The free body diagram in the bucket at the top of the circle is shown in the attached image.
The force balance hence, is
T + mg = mv²/r
where T = tension in the rope = ?
mg = weight of the bucket system
mv²/r = force keeping the bucket in circular motion.
v = velocity of the bucket = 7.8 m/s
r = radius of the circular path = 1.0 m
T + mg = mv²/r
T + (2.4×9.8) = (2.4×7.8²)/1
T = 146.016 - 23.52 = 122.496 N
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