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description\tshow all you work, including units, on separate paper. follow the \problem solving method\.
\tthis final exam is cumulative and covers material from the entire course.
instructions\tfrom the list of choices, select the one best answer.
multiple attempts\tnot allowed. this test can only be taken once.
force completion\tthis test can be saved and resumed later.
\tyour answers are saved automatically.

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question 52
a turntable in the shape of a flat uniform disk has a radius 0.500 m and a mass of 0.120 kg. the turntable is initially rotating at 4.00 rad/s about a vertical axis through its center. suddenly, a 2.50 x 10⁻² kg spider crawls onto the turntable at a point near the outer edge. treat the sprider as a particle. what is the angular speed of the turntable after the spider crawls on it?
○ 1.21 rad/s
○ 1.80 rad/s
○ 2.82 rad/s
○ 3.12 rad/s

Explanation:

Step1: Recall Angular Momentum Conservation

Angular momentum \( L = I\omega \) is conserved (no external torque). For a disk, \( I_{\text{disk}} = \frac{1}{2}MR^2 \); for a particle, \( I_{\text{particle}} = mr^2 \) (here \( r = R \)).
Initial \( L_1 = I_{\text{disk}}\omega_1 \), final \( L_2 = (I_{\text{disk}} + I_{\text{particle}})\omega_2 \). Set \( L_1 = L_2 \).

Step2: Calculate Initial Moment of Inertia

\( M = 0.120\,\text{kg} \), \( R = 0.500\,\text{m} \), \( \omega_1 = 4.00\,\text{rad/s} \).
\( I_{\text{disk}} = \frac{1}{2}(0.120)(0.500)^2 = 0.015\,\text{kg·m}^2 \).

Step3: Calculate Final Moment of Inertia

Spider mass \( m = 2.50 \times 10^{-2}\,\text{kg} \), \( r = R = 0.500\,\text{m} \).
\( I_{\text{particle}} = (2.50 \times 10^{-2})(0.500)^2 = 0.00625\,\text{kg·m}^2 \).
\( I_{\text{total}} = 0.015 + 0.00625 = 0.02125\,\text{kg·m}^2 \).

Step4: Solve for Final Angular Speed

From \( I_{\text{disk}}\omega_1 = I_{\text{total}}\omega_2 \):
\( \omega_2 = \frac{I_{\text{disk}}\omega_1}{I_{\text{total}}} = \frac{0.015 \times 4.00}{0.02125} \approx 2.82\,\text{rad/s} \).

Answer:

2.82 rad/s (corresponding to the option "2.82 rad/s")