Question

show all you work, including units, on separate paper. follow the \problem solving method\. this final exam is cumulative and covers material from the entire course. from the list of choices, select the one best answer. multiple attempts not allowed. this test can only be taken once. force completion this test can be saved and resumed later. your answers are saved automatically. question completion status: 53 54 55 56 57 58 59 60 61 62 question 42 of 62 1.6 points save answer an ice skater, spinning on the ice, increases her angular velocity from 2.00 rev/s to 6.00 rev/s while rotating through one - half a revolution. how much time does this take? 0.0625 s 0.250 s 0.125 s 0.0833 s

Explanation:

Step1: Recall rotational kinematic formula

For rotational motion with constant angular acceleration, the formula relating angular displacement ($\Delta\theta$), initial angular velocity ($\omega_0$), final angular velocity ($\omega$), and time ($t$) is $\Delta\theta=\frac{\omega_0 + \omega}{2}t$.

Step2: Convert angular displacement to revolutions

The skater rotates through one - half a revolution, so $\Delta\theta = 0.5\space rev$. The initial angular velocity $\omega_0=2.00\space rev/s$ and the final angular velocity $\omega = 6.00\space rev/s$.

Step3: Solve for time $t$

Rearrange the formula $\Delta\theta=\frac{\omega_0+\omega}{2}t$ to solve for $t$: $t=\frac{2\Delta\theta}{\omega_0 + \omega}$.

Substitute $\Delta\theta = 0.5\space rev$, $\omega_0 = 2.00\space rev/s$, and $\omega=6.00\space rev/s$ into the formula:

$t=\frac{2\times0.5}{2.00 + 6.00}=\frac{1}{8.00}=0.125\space s$? Wait, no, wait. Wait, let's check again. Wait, $\Delta\theta = 0.5\space rev$, $\omega_0 = 2\space rev/s$, $\omega = 6\space rev/s$.

$t=\frac{2\Delta\theta}{\omega_0+\omega}=\frac{2\times0.5}{2 + 6}=\frac{1}{8}=0.125\space s$? But wait, maybe I made a mistake. Wait, no, let's re - derive the formula. The average angular velocity $\omega_{avg}=\frac{\omega_0+\omega}{2}$, and $\Delta\theta=\omega_{avg}t$, so $t = \frac{\Delta\theta}{\omega_{avg}}=\frac{\Delta\theta}{\frac{\omega_0+\omega}{2}}=\frac{2\Delta\theta}{\omega_0+\omega}$.

Substituting the values: $\Delta\theta = 0.5\space rev$, $\omega_0 = 2\space rev/s$, $\omega = 6\space rev/s$.

$t=\frac{2\times0.5}{2 + 6}=\frac{1}{8}=0.125\space s$? Wait, but let's check the answer options. Wait, no, wait, maybe I messed up the angular displacement. Wait, no, one - half revolution is $0.5\space rev$. Wait, but let's recalculate:

Wait, $\omega_{avg}=\frac{2 + 6}{2}=4\space rev/s$. $\Delta\theta = 0.5\space rev$. Then $t=\frac{\Delta\theta}{\omega_{avg}}=\frac{0.5}{4}=0.125\space s$? Wait, but the options are 0.0625 s, 0.250 s, 0.125 s, 0.0833 s. Wait, 0.125 s is one of the options. Wait, but let's check again. Wait, maybe the angular displacement is in radians? No, the velocities are in rev/s, so using revolutions is okay.

Wait, no, wait a second. Wait, the formula is correct. Let's do the calculation again:

$\Delta\theta = 0.5\space rev$, $\omega_0 = 2\space rev/s$, $\omega = 6\space rev/s$.

$t=\frac{2\times0.5}{2 + 6}=\frac{1}{8}=0.125\space s$. So the time taken is $0.125\space s$.

Wait, but let's check the options. The options are 0.0625 s, 0.250 s, 0.125 s, 0.0833 s. So 0.125 s is an option.

Wait, maybe I made a mistake in the formula? Let's think about rotational kinematics. The equation for angular displacement with constant angular acceleration is also $\Delta\theta=\omega_0t+\frac{1}{2}\alpha t^2$, and $\omega=\omega_0+\alpha t$. But since we can use the average velocity formula (because angular acceleration is constant, average angular velocity is the average of initial and final), the first formula is valid.

So, using $\Delta\theta=\frac{\omega_0+\omega}{2}t$, we have $t=\frac{2\Delta\theta}{\omega_0+\omega}$. Plugging in the numbers: $\Delta\theta = 0.5\space rev$, $\omega_0 = 2\space rev/s$, $\omega = 6\space rev/s$.

$t=\frac{2\times0.5}{2 + 6}=\frac{1}{8}=0.125\space s$. So the answer should be 0.125 s.

Answer:

0.125 s