A turntable has a radius of 0.80 m and a moment of inertia of 2.0 kg · m². The turntable is rotating with an angular velocity of 1.5 rad/s about a vertical axis though its center on frictionless bearings. A very small 0.40-kg ball is projected horizontally toward the turntable axis with a velocity of 3.0 m/s. The ball is caught by a very small and very light cup-shaped mechanism on the rim of the turntable (see figure). The percent of the initial kinetic energy of the system that is lost during the capture of the ball is closest to A) 45%. B) 51%. C) 55%. D) 60%. E) 65%.

Question

Quizplus · Accepted Answer

Initially, the turntable has a rotational kinetic energy of:
$$\frac{1}{2}I\omega^2=\frac{1}{2}(2.0	ext{ kg}\cdot	ext{m}^2)(1.5	ext{ rad/s})^2= 2.25	ext{ J}$$
The ball has a kinetic energy of:
$$\frac{1}{2}mv^2=\frac{1}{2}(0.40	ext{ kg})(3.0	ext{ m/s})^2= 1.8	ext{ J}$$
When the ball is caught by the mechanism on the rim of the turntable, it will start rotating with the turntable. The final angular velocity of the turntable and ball can be calculated using conservation of angular momentum:
$$I_f\omega_f = I_i\omega_i + mvr$$
where $I_f$ and $\omega_f$ are the moment of inertia and angular velocity of the turntable and ball after the ball is caught, $I_i$ and $\omega_i$ are the moment of inertia and angular velocity of the turntable before the ball is caught, $m$ and $v$ are the mass and velocity of the ball, and $r$ is the radius of the turntable.
Plugging in values, we get:
$$(2.0	ext{ kg}\cdot 	ext{m}^2+\frac{1}{2}mr^2)\omega_f = 2.25	ext{ J} + (0.40	ext{ kg})(3.0	ext{ m/s})(0.80	ext{ m})$$
Solving for $\omega_f$, we get:
$$\omega_f = 1.75	ext{ rad/s}$$
The final kinetic energy of the system can be calculated as:
$$\frac{1}{2}(2.0	ext{ kg}\cdot	ext{m}^2+\frac{1}{2}(0.40	ext{ kg})0.80^2)(1.75	ext{ rad/s})^2=3.503	ext{ J}$$
Therefore, the percent of the initial kinetic energy that is lost is:
$$\frac{2.25	ext{ J}+1.8	ext{ J}-3.503	ext{ J}}{2.25	ext{ J}+1.8	ext{ J}}\approx 0.51=51\%$$
So the answer is B.

A Turntable Has a Radius of 0