Question

A mass of 30.0 kg sits on the London Eye whirled in a vertical circle of radius 10.0 m at a constant speed of 5.0 m s¹. If the mass breaks off at the lowest point of the vertical circle, how far would the mass move in the horizontal direction from the lowest point in 5.0 s? (Neglect air resistance)​

Answer 1

Step-by-step explanation:

The horizontal distance moved by the mass in 5.0 s is 25 m. The formula for the horizontal distance moved by a mass in a vertical circle is: Horizontal distance = (2πr/T) × t where r = radius of the vertical circle, T = period of the circle, and t = time. Given: r = 10.0 m, T = 2π√(r/g) = 2π√(10.0/9.8) = 6.25 s, t = 5.0 s Therefore, Horizontal distance = (2πr/T) × t = (2π × 10.0/6.25) × 5.0 = 25 m