Question

A .001 kg bead slides without friction around a loop-the-loop. The bead is released from a height of 20.6 m from the bottom of the loop-the-loop which has a radius 7 m. The acceleration of gravity is 9.8 m/s [photo] What is its speed at point A ? Answer in units of m/s. Your answer must be within ± 5.0%

Answer 1

Answer:

11.37 m/s

Explanation:

We are given;

Mass; m = 0.001 kg

Initial velocity; v1 = 0 m/s (since it was released from rest)

Initial height; h1 = 20.6m

From the attached image of the bead in the loop, we can see that height at A; h2 = twice the radius = 2R = 2 × 7 = 14 m

g = 9.8

To solve for the speed at point, we will use formula for conservation of energy. Thus;

½m(v1)² + mgh1 = ½m(v2)² + mgh2

Divide through by m to get;

½(v1)² + gh1 = ½(v2)² + gh2

Plugging in the relevant values;

½(0²) + (9.8 × 20.6) = ½(v2)² + (9.8 × 14)

Rearranging, we have;

½(v2)² = ½(0²) + (9.8 × 20.6) - (9.8 × 14)

Simplifying, we have;

½(v2)² = 64.68

Multiply both sides by 2 to get;

(v2)² = 2 × 64.68

(v2)² = 129.36

v2 = √129.36

v2 = 11.37 m/s