A turntable has a moment of inertia of 3.0 x 10^-2 kg*m² and spins freely on a frictionless bearing at 25 rev/min.A 0.30-kg ball of putty is dropped vertically on the turntable and sticks at a point 0.10 m from the center.By what factor does the kinetic energy of the system change after the putty is dropped onto the turntable? A)0.91 B)1.0 C)0.82 D)1.5

Question

Quizplus · Accepted Answer

Explanation:The initial kinetic energy of the system is:$$K_i = \frac{1}{2}I\omega_i^2 = \frac{1}{2}(3.0 	imes 10^{-2} 	ext{ kg m}^2)(25 	ext{ rev/min})^2(2\pi 	ext{ rad/rev})^2$$$$K_i = 1.88 	ext{ J}$$The final kinetic energy of the system is:$$K_f = \frac{1}{2}(I + mr^2)\omega_f^2$$where $m$ is the mass of the putty, $r$ is the distance from the center of the turntable to the point where the putty sticks, and $\omega_f$ is the final angular velocity of the system.We can use the conservation of angular momentum to find $\omega_f$:$$I\omega_i = (I + mr^2)\omega_f$$$$\omega_f = \frac{I\omega_i}{I + mr^2}$$$$\omega_f = \frac{(3.0 	imes 10^{-2} 	ext{ kg m}^2)(25 	ext{ rev/min})^2(2\pi 	ext{ rad/rev})^2}{3.0 	imes 10^{-2} 	ext{ kg m}^2 + (0.30 	ext{ kg})(0.10 	ext{ m})^2}$$$$\omega_f = 22.5 	ext{ rev/min}$$Now we can calculate $K_f$:$$K_f = \frac{1}{2}(3.0 	imes 10^{-2} 	ext{ kg m}^2 + 0.30 	ext{ kg}(0.10 	ext{ m})^2)(22.5 	ext{ rev/min})^2(2\pi 	ext{ rad/rev})^2$$$$K_f = 1.71 	ext{ J}$$The factor by which the kinetic energy of the system changes is:$$\frac{K_f}{K_i} = \frac{1.71 	ext{ J}}{1.88 	ext{ J}} = 0.91$$Therefore, the correct answer is A.

A Turntable Has a Moment of Inertia of 3