Question

A bucket of mass 1.90 kg is whirled in a vertical circle of radius 1.00 m . at the lowest point of its motion the tension in the rope supporting the bucket is 26.0 n .how fast must the bucket move at the top of the circle so that the rope does not go slack?

Answer 1

To prevent the rope from going slack, the bucket must move at a minimum speed of 3.13 m/s at the top of the vertical circle, calculated using the equation for centripetal force.

Step-by-step explanation:

To determine how fast the bucket must move at the top of the circle to ensure the rope does not go slack, we need to calculate the minimum speed that provides the necessary centripetal force to balance the gravitational force on the bucket. At the top of the circle, the gravitational force and the centripetal force required to keep the bucket in circular motion must be equal.

At the top of the circle, the tension in the rope (T) is zero if it's on the verge of going slack. Therefore, the only force acting on the bucket is the gravitational force (mg), which must provide the necessary centripetal force (mv2/r) to keep it in a circular path. Setting these equal gives us:

mg = mv2/r

Solving for v gives us:

v = √(gr)

Where g is the acceleration due to gravity (9.80 m/s2) and r is the radius of the circle (1.00 m).

Substituting the values, we find:

v = √(9.80 m/s2 × 1.00 m) = √(9.80 m2/s2) = 3.13 m/s

The bucket must be moving at least 3.13 m/s at the top of the circle to ensure the rope remains taut.