Notes: An airplane is flying in a horizontal circle at a speed of 102 m/s. The 80.0 kg pilot does not want the centripetal acceleration to exceed 6.04...

Thursday, 3 November 2016

An airplane is flying in a horizontal circle at a speed of 102 m/s. The 80.0 kg pilot does not want the centripetal acceleration to exceed 6.04...

The centripetal acceleration depends on the speed of the pilot and the radius of the path as

$\displaystyle{a}=\frac{{{v}}^{{2}}}{{r}}$

Since this acceleration cannot exceed 6.04 times free fall acceleration (acceleration due to gravity), the maximum value of the centripetal acceleration is

6.04* 9.81 = 59.25 m/s^2

Then, the minimal radius is determined by

$\displaystyle\frac{{{v}}^{{2}}}{{r}}\le{59.25}$

The given speed of the pilot is 102 m/s, so

$\displaystyle\frac{{{102}}^{{2}}}{{r}}\le{59.25}$

$\displaystyle{r}\ge\frac{{{102}}^{{2}}}{{59.25}}={175.6}$ m.

The...

The centripetal acceleration depends on the speed of the pilot and the radius of the path as

$\displaystyle{a}=\frac{{{v}}^{{2}}}{{r}}$

Since this acceleration cannot exceed 6.04 times free fall acceleration (acceleration due to gravity), the maximum value of the centripetal acceleration is

6.04* 9.81 = 59.25 m/s^2

Then, the minimal radius is determined by

$\displaystyle\frac{{{v}}^{{2}}}{{r}}\le{59.25}$

The given speed of the pilot is 102 m/s, so

$\displaystyle\frac{{{102}}^{{2}}}{{r}}\le{59.25}$

$\displaystyle{r}\ge\frac{{{102}}^{{2}}}{{59.25}}={175.6}$ m.

The minimal radius of the plane's path is 175.6 meters.

The net force on the pilot, according to the second Newton's Law, equal mass times acceleration:

$\displaystyle{F}={m}{a}={m}\frac{{{v}}^{{2}}}{{r}}$ , where m is the mass of the pilot. This net force creates the centripetal acceleration and thus maintains the circular motion.

$\displaystyle{F}={m}{a}={80}\cdot{59.25}={4},{740}$ N.

The net force on the pilot equals 4,740 N.

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