Math  /  Algebra

Question48. (II) A bucket of mass 2.00 kg is whirled in a vertical circle of radius 1.10 m . At the lowest point of its motion the tension in the rope supporting the bucket is 25.0 N . (a) Find the speed of the bucket. (b) How fast must the bucket move at the top of the circle so that the rope does not go slack?

Studdy Solution
Calculate the minimum speed at the top of the circle.
The condition for the rope not going slack is:
mg=mv2r mg = \frac{mv^2}{r}
Canceling m m from both sides gives:
g=v2r g = \frac{v^2}{r}
Rearrange to solve for v v :
v2=gr v^2 = gr
v=gr v = \sqrt{gr}
Substitute the known values:
v=9.81×1.10 v = \sqrt{9.81 \times 1.10}
v=10.791 v = \sqrt{10.791}
v3.28m/s v \approx 3.28 \, \text{m/s}
The speed of the bucket at the lowest point is approximately 1.72m/s 1.72 \, \text{m/s} , and the minimum speed at the top of the circle to prevent the rope from going slack is approximately 3.28m/s 3.28 \, \text{m/s} .

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