If a yo yo has a speed of 3 m/s at the top of the circle, whose radius is 80 cm, What is its speed at the bottom

Answer 1

the kinetic energy at the bottom is the kinetic energy at the top plus the gravitational potential across the circle

.5 m (Vb)^2 = [.5 m (Vt)^2] + [m g h]

dividing by .5 m

(Vb)^2 = (Vt)^2 + 2 g h

v^2 = 3^2 + [2 * 9.8 * (2 * .8)]