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?

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

6/12=1/2

step-by-step explanation:

v=ir = 3x15 =45 v