At the bottom of a loop in the vertical plane, an airplane has a horizontal velocity of 315 mi/h and is speeding up at a rate of 10 ft/s². The radius of curvature of the loop is 1 mi. The plane is being tracked by radar at O. What are the recorded values of \dot { r },\ddot { r },\dot { \theta } and \ddot { \theta } for this instant?
Question 11.169: At the bottom of a loop in the vertical plane, an airplane h...
- Horizontal velocity of the airplane: 315 mi/h
- Rate of acceleration: 10 ft/s²
- Radius of curvature of the loop: 1 mi
- Plane being tracked by radar at point O.
To solve this problem, we need to perform velocity and acceleration analysis.
Step 1:
Step 1: Velocity Analysis First, we calculate the polar coordinates of the given velocity vector. The magnitude of the velocity vector is 315 mi/h, which is equivalent to 462 ft/s. To find the angle, we use the tangent inverse function with the ratio of the y-component (1800 ft) to the x-component (2400 ft) of the velocity vector. This gives us an angle of 36.87 degrees. The magnitude of the velocity vector is 3000 ft, which we can calculate using the Pythagorean theorem.
Step 2:
Next, we break down the velocity vector into its radial and tangential components. The radial component (νr) is given by the magnitude of the velocity vector multiplied by the cosine of the angle, which gives us 369.6 ft/s. The tangential component (νθ) is given by the magnitude of the velocity vector multiplied by the sine of the angle, which gives us -277.2 ft/s.
Step 3:
Finally, we can determine the radial and tangential velocities. The radial velocity (dr/dt) is equal to the radial component of the velocity (νr), which is 369.6 ft/s. The tangential velocity (r dθ/dt) is equal to the product of the radius (3000 ft) and the tangential component of the velocity (νθ) divided by the magnitude of the velocity vector, which gives us -0.0924 rad/s.
Step 4:
Acceleration Analysis Given that the tangential acceleration (at) is 10 ft/s^2, we can calculate the normal acceleration (an) using the formula (ν^2)/ρ, where ρ is the radius of curvature. The radius of curvature can be calculated by dividing the magnitude of the velocity vector squared by the centripetal acceleration, which is (462 ft/s)^2 divided by 5280 ft/s^2. This gives us an = 40.425 ft/s^2.
Step 5:
Using the angles, we can calculate the radial acceleration (ar) and tangential acceleration (aθ). The radial acceleration is given by the product of the tangential acceleration (at) and the cosine of the angle, added to the product of the normal acceleration (an) and the sine of the angle. This gives us ar = 32.255 ft/s^2. The tangential acceleration is given by the product of the tangential acceleration (at) and the sine of the angle, subtracted from the product of the normal acceleration (an) and the cosine of the angle. This gives us aθ = 26.34 ft/s^2.
Step 6:
Finally, we can determine the radial and tangential accelerations. The radial acceleration (d^2r/dt^2) is equal to the sum of the radial acceleration (ar) and the product of the radius (3000 ft) and the square of the tangential velocity (dθ/dt)^2, which gives us 57.9 ft/s^2. The tangential acceleration (r d^2θ/dt^2) is equal to the product of the radius (3000 ft) and the second derivative of the angle (d^2θ/dt^2), divided by the magnitude of the velocity vector, which gives us 0.031 rad/s^2.
Final Answer
Geometry. The polar coordinates are
r=\sqrt{(2400)^{2}+(1800)^{2}}=3000 \mathrm{\ ft} \quad \theta=\tan ^{-1}\left\lgroup\frac{1800}{2400}\right\rgroup=36.87^{\circ}
Velocity Analysis.
\mathbf{v}=315 \mathrm{\ mi} / \mathrm{h}=462 \mathrm{\ ft} / \mathrm{s} \longrightarrow
\begin{aligned}& \nu_{r}=462 \cos \theta=369.6 \mathrm{\ ft} / \mathrm{s} \\& \nu_{\theta}=-462 \sin \theta=-277.2 \mathrm{\ ft} / \mathrm{s} \\& \nu_{r}=\dot{r} & \dot{r}=370 \mathrm{\ ft} / \mathrm{s}\blacktriangleleft\\& \nu_{\theta}=r \dot{\theta} \quad \dot{\theta}=\frac{\nu_{\theta}}{r}=-\frac{277.2}{3000}\end{aligned}
\dot{\theta}=-0.0924 \mathrm{\ rad} / \mathrm{s}\blacktriangleleft
Acceleration analysis.
\begin{aligned}& a_{t}=10 \mathrm{\ ft} / \mathrm{s}^{2} \\& a_{n}=\frac{\nu^{2}}{\rho}=\frac{(462)^{2}}{5280}=40.425 \mathrm{\ ft} / \mathrm{s}^{2}\end{aligned}
\begin{aligned}& a_{r}=a_{t} \cos \theta+a_{n} \sin \theta=10 \cos 36.87^{\circ}+40.425 \sin 36.87^{\circ}=32.255 \mathrm{~ft} / \mathrm{s}^{2} \\& a_{\theta}=-a_{t} \sin \theta+a_{n} \cos \theta=-10 \sin 36.87^{\circ}+40.425 \cos 36.87^{\circ}=26.34 \mathrm{~ft} / \mathrm{s}^{2} \\& a_{r}=\ddot{r}-r \dot{\theta}^{2} \quad\quad \ddot{r}=a_{r}+r \dot{\theta}^{2}&\ddot{r}=57.9 \mathrm{\ ft} / \mathrm{s}^{2}\blacktriangleleft \\& \ddot{r}=32.255+(3000)(-0.0924)^{2} \\& a_{\theta}=r \ddot{\theta}+2 \dot{r} \dot{\theta} \\& \ddot{\theta}=\frac{a_{\theta}}{r}-\frac{2 \dot{r} \dot{\theta}}{r} \\& =\frac{26.34}{3000}-\frac{(2)(369.6)(-0.0924)}{3000}& \ddot{\theta}=0.031\mathrm{\ rad}/\mathrm{s}^2\blacktriangleleft\end{aligned}
Related Answered Questions
First note that the vectors v and a lie in the osc...
\begin{array}{lll} R=A & \theta=2 \pi ...
The motion is projectile motion. Place the origin ...
(a) Velocity of the skier. \quad(r=500 \mat...
Hyperbolic spiral.
\begin{aligned}& r=...
Logarithmic spiral.
\begin{aligned}r & ...
We have
\mathbf{v}=\frac{d \mathbf{r}}{d t}...
For the sun,
g=274 \mathrm{~m} / \mathrm{s}...