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salantis [7]
3 years ago
13

A straight line is drawn on the surface of a 14-cm-radius turntable from the center to the perimeter. A bug crawls along this li

ne from the center outward as the turntable spins counterclockwise at a constant 45 rpm. Its walking speed relative to the turntable is a steady 3.8 cm/s. Let its initial heading be in the positive x-direction, where the x- and y-directions are fixed relative to the laboratory. As the bug reaches the edge of the turntable (still traveling at 3.8 cm/s radially, relative to the turntable) what are the x and y components of the velocity of the bug?
Physics
1 answer:
sleet_krkn [62]3 years ago
8 0

Answer:

v = \left[\begin{array}{c}0.66&0\end{array}\right]m/s

Explanation:

The position vector r of the bug with linear velocity v and angular velocity ω in the laboratory frame is given by:

\overrightarrow{r}=vtcos(\omega t)\hat{x}+vtsin(\omega t)\hat{y}

The velocity vector v is the first derivative of the position vector r with respect to time:

\overrightarrow{v}=[vcos(\omega t)-\omega vtsin(\omega t)]\hat{x}+[vsin(\omega t)+\omega vtcos(\omega t)]\hat{y}

The given values are:

t=\frac{x}{v}=\frac{14}{3.8}=3.7 s

\omega=\frac{45\times2\pi}{60s}=4.7\frac{1}{s}

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