Conservation of angular momentum: A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turntable to a common angular speed. If the rotational inertia of the record is 0.54 times that of the turntable, what percentage of the initial kinetic energy is lost?
A) 35%
B) 18%
C) 46%
D) 60%

Question

Quizplus · Accepted Answer

Explanation:The initial kinetic energy is:$$K_i = \frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2$$where $I_r$ and $\omega_r$ are the rotational inertia and angular speed of the record, and $I_t$ and $\omega_t$ are the rotational inertia and angular speed of the turntable.The final kinetic energy is:$$K_f = \frac{1}{2}(I_r + I_t)\omega_f^2$$where $\omega_f$ is the final angular speed of the system.The percentage of the initial kinetic energy that is lost is:$$\frac{K_i - K_f}{K_i} \times 100\% = \frac{\frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2 - \frac{1}{2}(I_r + I_t)\omega_f^2}{\frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2} \times 100\%$$$$= \frac{\frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2 - \frac{1}{2}(I_r + I_t)\omega_f^2}{\frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2} \times 100\%$$$$= \frac{\frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2 - \frac{1}{2}(I_r + I_t)\left(\frac{I_r\omega_r + I_t\omega_t}{I_r + I_t}\right)^2}{\frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2} \times 100\%$$$$= \frac{\frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2 - \frac{1}{2}(I_r + I_t)\left(\frac{I_r\omega_r^2 + I_t\omega_t^2 + 2I_rI_t\omega_r\omega_t}{I_r + I_t}\right)}{\frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2} \times 100\%$$$$= \frac{\frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2 - \frac{1}{2}(I_r + I_t)\left(\frac{I_r\omega_r^2 + I_t\omega_t^2}{I_r + I_t}\right) - I_rI_t\omega_r\omega_t}{\frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2} \times 100\%$$$$= \frac{\frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2 - \frac{1}{2}(I_r\omega_r^2 + I_t\omega_t^2) - I_rI_t\omega_r\omega_t}{\frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2} \times 100\%$$$$= \frac{I_rI_t\omega_r\omega_t}{\frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2} \times 100\%$$$$= \frac{2I_rI_t\omega_r\omega_t}{I_r\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{2(0.54I_t)I_t\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\

Conservation of Angular Momentum: a Record Is Dropped Vertically onto a Freely