Question

The airplane on the amusement park ride moves along a path defined by the equations r = 4 m, θ = (0.2t)rad, and z = (0.5cosθ)m, where t is in seconds.

A)Determine the cylindrical components of the velocity of the airplane when t = 6s.

B)Determine the cylindrical components of the acceleration of the airplane when t = 6s.

Answer 1

Start with the velocity.

\(v_{r}=\frac{\mathrm{dr}}{\mathrm{d} t}=\frac{d}{\mathrm{~d} t}(4)=0\)

\(v_{\theta}=r * \frac{d \theta}{\mathrm{d} t}=(4)\left(\frac{d}{\mathrm{dt}} 0.2 t\right)=4 * .2=0.8 \mathrm{~m} / \mathrm{s}\)

\(v_{z}=\frac{\mathrm{d} z}{\mathrm{~d} \mathrm{t}}=-0.5 \sin (\theta) \frac{d \theta}{\mathrm{d} t}\)

The angle that the airplane is swinging through at \(t=6 \mathrm{sec}\) is:

\(\theta(6)=1.2 \mathrm{rad}\)

\(v_{z}=-0.5 \sin (1.2) * 0.2=-0.0932 \mathrm{~m} / \mathrm{s}\)

Now do acceleration.

\(a_{r}=\frac{d^{2} r}{\mathrm{~d} \mathrm{t}^{2}}=0 \quad a_{r}=\frac{d^{2} r}{\mathrm{dt}^{2}}-r\left(\frac{d \theta}{\mathrm{d} \mathrm{t}}\right)^{2}=0-4 *(0.2)^{2}=-0.16 \mathrm{~m} / \mathrm{s}^{2}\)

\(a_{\theta}=r \frac{d^{2} \theta}{\mathrm{dt}^{2}}+2 \frac{\mathrm{dr}}{\mathrm{dt}} \frac{d \theta}{\mathrm{dt}}=0+0=0\)

\(a_{z}=\frac{d^{2} z}{\mathrm{~d} \mathrm{t}^{2}}=\left[-0.5 \cos (\theta)\left(\frac{d \theta}{\mathrm{d} \mathrm{t}}\right)^{2}-0.5 \sin \theta \frac{d^{2} \theta}{\mathrm{d} t}\right]_{<-\mathrm{Again}, \text { time derivatives. Take care here. }}\)

\(=-0.5 \cos (1.2) * 0.2^{2}-0.5 \sin (1.2) * 0\)

\(=-7.25 * 10^{-3} \frac{\mathrm{m}}{\mathrm{s}^{2}}\)