A playground merry-go-round with a radius of 2.0 m and a rotational inertia of 100 kg m² is rotating at 3.0 rad/s. A child with a mass of 22 kg jumps onto the edge of the merry-go-round, traveling radially inward. What is the new angular speed of the merry-go-round? A)1.6 rad/s B)1.2 rad/s C)2.4 rad/s D)2.0 rad/s E)3.4 rad/s

Question

Quizplus · Accepted Answer

Explanation:Using the conservation of angular momentum, we have:$$I_i \omega_i = I_f \omega_f$$where:$$I_i$$ is the initial rotational inertia$$\omega_i$$ is the initial angular speed$$I_f$$ is the final rotational inertia$$\omega_f$$ is the final angular speedSubstituting the given values, we get:$$(100 	ext{ kg m}^2)(3.0 	ext{ rad/s}) = (100 	ext{ kg m}^2 + 22 	ext{ kg})(2.0 	ext{ m})^2(\omega_f)$$Solving for $$\omega_f$$, we get:$$\omega_f = 1.6 	ext{ rad/s}$$

A Playground Merry-Go-Round with a Radius of 2