Question

The three-dimensional motion of a particle on the surface of a right circular cylinder is described by the relations r = 2 (m) θ = πt (rad) z = sin24θ (m) Compute the velocity and acceleration of the particle at t=5 s.

Answer 1

Answer:

$V_{rex}=75.65m/s$ and $a_{res}=0$ at t=5 secs

Explanation:

We have r =2m

$\therefore \frac{dr}{dt}=0\\\\=>V_{r}=0$

Similarly

$=>V_{\theta }=\omega r\\\\\omega =\frac{d\theta }{dt}=\frac{d(\pi t)}{dt}=\pi \\\\\therefore V_{\theta }=\pi r=2\pi$

Similarly

$=>V_{z }=\frac{dz}{dt}\\\\V_{z}=\frac{dsin(24\pi t)}{dt}\\\\V_{z}=24\pi cos(24\pi t)$

Hence

at t =5s $V_{\theta}=2\pi m/s$

$V_{z}=24\pi cos(120\pi)$

$V_{z}=24\pi m/s$

$V_{res}=\sqrt{V_{\theta }^{2}+V_{z}^{2}}$

Applying values we get

$V_{res}=75.65m/s$

Similarly

$a_{\theta }=\frac{dV_{\theta }}{dt}=\frac{d(2\pi) }{dt}=0\\\\a_{z}=\frac{d^{2}(sin(24\pi t))}{dt^{2}}\\\\a_{z}=-24^{2}\pi^{2}sin(24\pi t)\\\\\therefore t=5\\a_{z}=0$