Question

What linear speed must a 0.0507-kg hula hoop have if its total kinetic energy is to be 0.160 J? Assume the hoop rolls on the ground without slipping.

Answer 1

total kinetic energy = linear kinetic energy + rotational kinetic energy

linear kinetic energy = 0.5 * mv^2

rotational knetic energy = 0.5 * moment of inertia * angular velocity^2

angular velocity = linear velocity / radius

rotational knetic energy = 0.5 * moment of inertia * (linear velocity / radius)^2

moment of inertia of a loop = mass * radius^2

rotational knetic energy = 0.5 * mass * radius^2 * (linear velocity / radius)^2

rotational knetic energy = 0.5 * mass * linear velocity^2

total kinetic energy = 0.5 * mv^2 + 0.5 * mv^2

0.16 = 0.5 * 0.0507 * v^2 + 0.5 * 0.0507 * v^2

v = 1.7764 m/s

linear speed of the loop = 1.7764 m/s