Chapter 4: Kinematics in Two Dimensions

Both hit the floor together, because gravity acts the same on both.

The dots all share the same $\omega$ (they sweep the same angle in the same time) but the one farthest out has the largest $v$ and $a_r$.

Solution: We need to analyze this motion in each dimension separately using the constant acceleration equations.

x-direction:

• Initial position: $x_i = 30$ m
• Initial velocity: $v_{xi} = -1$ m/s
• Acceleration: $a_x = 0.25$ m/s²
• Time interval: $\Delta t = 5$ s

Final position in x: $x_f = x_i + v_{xi}\Delta t + \frac{1}{2}a_x\Delta t^2$ $x_f = 30 + (-1)(5) + \frac{1}{2}(0.25)(5)^2$ $x_f = 30 - 5 + \frac{1}{2}(0.25)(25)$ $x_f = 30 - 5 + 3.125$ $x_f = 28.125$ m

y-direction:

• Initial position: $y_i = 40$ m
• Initial velocity: $v_{yi} = 1$ m/s
• Acceleration: $a_y = 0$ m/s² (no acceleration in y-direction)
• Time interval: $\Delta t = 5$ s

Final position in y: $y_f = y_i + v_{yi}\Delta t + \frac{1}{2}a_y\Delta t^2$ $y_f = 40 + (1)(5) + \frac{1}{2}(0)(5)^2$ $y_f = 40 + 5 + 0$ $y_f = 45$ m

z-direction:

• Initial position: $z_i = -10$ m
• Initial velocity: $v_{zi} = 0$ m/s (not specified, so assumed zero)
• Acceleration: $a_z = 0.33$ m/s²
• Time interval: $\Delta t = 5$ s

Final position in z: $z_f = z_i + v_{zi}\Delta t + \frac{1}{2}a_z\Delta t^2$ $z_f = -10 + (0)(5) + \frac{1}{2}(0.33)(5)^2$ $z_f = -10 + 0 + \frac{1}{2}(0.33)(25)$ $z_f = -10 + 4.125$ $z_f = -5.875$ m

Therefore, the final position vector is: $\vec{r}_f = (28.125\hat{i} + 45\hat{j} - 5.875\hat{k})$ m

Solution: We need to determine how the electron's velocity changes as it passes through the electric field between the plates.

Step 1: Determine the time the electron spends between the plates. Using the horizontal motion (x-direction):

• Initial horizontal position: $x_i = 0$ m (reference point)
• Final horizontal position: $x_f = 0.02$ m (2.0 cm)
• Initial horizontal velocity: $v_{xi} = 3.3 \times 10^7$ m/s
• Horizontal acceleration: $a_x = 0$ m/s² (no horizontal acceleration)

From the constant velocity equation: $x_f = x_i + v_{xi} \Delta t$ $0.02 = 0 + (3.3 \times 10^7) \Delta t$ $\Delta t = \frac{0.02}{3.3 \times 10^7} = 6.06 \times 10^{-10}$ s

Step 2: Calculate the vertical velocity component.

• Initial vertical velocity: $v_{yi} = 0$ m/s (starts horizontally)
• Vertical acceleration: $a_y = 2.9 \times 10^{16}$ m/s²
• Time interval: $\Delta t = 6.06 \times 10^{-10}$ s

Final vertical velocity: $v_{yf} = v_{yi} + a_y \Delta t$ $v_{yf} = 0 + (2.9 \times 10^{16})(6.06 \times 10^{-10})$ $v_{yf} = 1.76 \times 10^7$ m/s

Step 3: Calculate the magnitude and direction of the final velocity. The final horizontal velocity component is unchanged: $v_{xf} = v_{xi} = 3.3 \times 10^7$ m/s

Magnitude of the final velocity: $|v_f| = \sqrt{v_{xf}^2 + v_{yf}^2}$ $|v_f| = \sqrt{(3.3 \times 10^7)^2 + (1.76 \times 10^7)^2}$ $|v_f| = \sqrt{(10.89 \times 10^{14}) + (3.10 \times 10^{14})}$ $|v_f| = \sqrt{13.99 \times 10^{14}}$ $|v_f| = 3.74 \times 10^7$ m/s

Direction angle: $\theta = \tan^{-1}\left(\frac{v_{yf}}{v_{xf}}\right)$ $\theta = \tan^{-1}\left(\frac{1.76 \times 10^7}{3.3 \times 10^7}\right)$ $\theta = \tan^{-1}(0.533)$ $\theta = 28.1°$

Therefore, the final velocity of the electron is $3.7 \times 10^7$ m/s at an angle of $28°$ above the horizontal.

Solution: We need to analyze this problem in two phases: the speeding up phase (when the wheel accelerates) and the slowing down phase (after being unplugged).

Step 1: Convert the initial angular velocity from rpm to rad/s. $\omega_i = 229 \text{ rpm} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = 23.98 \text{ rad/s}$

Part (a): Calculate the angle turned before being unplugged.

• Initial angular position: $\theta_i = 0$ rad
• Initial angular velocity: $\omega_i = 23.98$ rad/s
• Angular acceleration: $\alpha = 30.0$ rad/s²
• Time interval: $\Delta t = 2.00$ s

Using the angular position equation: $\theta_f = \theta_i + \omega_i\Delta t + \frac{1}{2}\alpha\Delta t^2$ $\theta_f = 0 + (23.98 \text{ rad/s})(2.00 \text{ s}) + \frac{1}{2}(30.0 \text{ rad/s}^2)(2.00 \text{ s})^2$ $\theta_f = 47.96 \text{ rad} + 60.0 \text{ rad}$ $\theta_f = 107.96 \text{ rad}$

The final angular velocity after the first 2.00 seconds: $\omega_f = \omega_i + \alpha\Delta t$ $\omega_f = 23.98 \text{ rad/s} + (30.0 \text{ rad/s}^2)(2.00 \text{ s})$ $\omega_f = 23.98 \text{ rad/s} + 60.0 \text{ rad/s}$ $\omega_f = 83.98 \text{ rad/s}$

Part (b): Calculate the time to stop after being unplugged.

• Initial angular position for slowing phase: $\theta_i = 107.96$ rad
• Final angular position: $\theta_f = 107.96 \text{ rad} + 432 \text{ rad} = 539.96$ rad
• Initial angular velocity: $\omega_i = 83.98$ rad/s
• Final angular velocity: $\omega_f = 0$ rad/s (wheel stops)

Since we know the initial and final velocities and the angular displacement, we can use: $\theta_f = \theta_i + \frac{1}{2}(\omega_i + \omega_f)\Delta t$

Rearranging to solve for $\Delta t$: $\Delta t = \frac{2(\theta_f - \theta_i)}{\omega_i + \omega_f}$ $\Delta t = \frac{2(539.96 \text{ rad} - 107.96 \text{ rad})}{83.98 \text{ rad/s} + 0 \text{ rad/s}}$ $\Delta t = \frac{2(432 \text{ rad})}{83.98 \text{ rad/s}}$ $\Delta t = \frac{864 \text{ rad}}{83.98 \text{ rad/s}}$ $\Delta t = 10.29 \text{ s}$

Part (c): Calculate the angular acceleration during slowing down.

• Initial angular velocity: $\omega_i = 83.98$ rad/s
• Final angular velocity: $\omega_f = 0$ rad/s
• Time interval: $\Delta t = 10.29$ s

Using the angular velocity equation: $\omega_f = \omega_i + \alpha\Delta t$ $0 = 83.98 \text{ rad/s} + \alpha(10.29 \text{ s})$ $\alpha = \frac{-83.98 \text{ rad/s}}{10.29 \text{ s}}$ $\alpha = -8.16 \text{ rad/s}^2$

The negative sign indicates that the angular acceleration is in the opposite direction of the angular velocity, causing the wheel to slow down.