A merry-go-round with a moment of inertia of 6.78 $\times$ 10³ kg · m² is coasting at 2.20 rad/s.When a 72.6-kg man steps onto the rim,the angular velocity decreases to 2.0 rad/s.The radius of the merry-go-round is A)3.06 m B)3.66 m C)4.27 m D)4.88 m E)5.49 m

Question

Quizplus · Accepted Answer

Explanation:The angular momentum of the system is conserved, so:$$I_i\omega_i = I_f\omega_f$$$$(6.78	imes10^3 	ext{ kg$\cdot$m}^2)(2.20 	ext{ rad/s}) = (6.78	imes10^3 	ext{ kg$\cdot$m}^2 + 72.6 	ext{ kg}r^2)(2.00 	ext{ rad/s})$$$$r = 3.06 	ext{ m}$$

A Merry-Go-Round with a Moment of Inertia of 6 ×\times× 103 Kg · M2 Is Coasting at 2

A Merry-Go-Round with a Moment of Inertia of 6 $\times$ 10³ Kg · M² Is Coasting at 2