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

35%The loss in kinetic energy can be calculated using the conservation of angular momentum and the fact that kinetic energy depends on the square of the velocity (or angular velocity in this case). Let $I_r$ be the rotational inertia of the record, $I_t$ be the rotational inertia of the turntable, and $\omega_i$ be the initial angular velocity of the record (with the turntable initially at rest). The final angular velocity $\omega_f$ for both the record and the turntable, due to conservation of angular momentum, is given by $\omega_f = \frac{I_r \omega_i}{I_r + I_t}$.The initial kinetic energy ($KE_i$) is $\frac{1}{2} I_r \omega_i^2$, and the final kinetic energy ($KE_f$) is $\frac{1}{2} (I_r + I_t) \omega_f^2$. Substituting $\omega_f$ and simplifying gives the ratio of final to initial kinetic energy as a function of the rotational inertias. Given $I_r = 0.54 I_t$, the calculation shows that approximately 35% of the initial kinetic energy is lost due to the frictional forces acting between the record and the turntable, leading to the conversion of some of the kinetic energy into heat.

A Record Is Dropped Vertically onto a Freely Rotating (Undriven)turntable