Question

A spaceman in training is rotated in a seat at the end of a horizontal arm of length 5 m. If he can withstand acceleration up to 9 g, then what is the maximum number of revolutions per second permissible? (Take g =10\,ms^ (-2))

Answer 1

Hint:To find the necessary solution let us first understand the concept of centripetal force in circular motion. A centripetal force is defined as a force that makes a body follow a curved path. Any object traveling along a circular path of radius r with velocity v experiences an acceleration directed toward the centre of its path.

Complete step by step answer:
Let us denote centripetal force by ${F_{_{_C}}}$
So ${F_{_{_C}}} = \dfrac{{m{v^2}}}{r}$
We know that $v = w \times r$
So ${F_c} = m{w^2}r$
We know that $w = 2\pi n$ where n is the number of revolutions per second. Also centripetal force to the man is provided by the effective weight of man. So let us equate these both
$m \times 9g = mr{w^2}$
Substitute the value of w
$m \times 9g$$ = mr{(2\pi n)^2} = mr{n^2}4{\pi ^2}$
After rearranging the terms we have
$n = \sqrt {\dfrac{{9g}}{{4{\pi ^2}r}}} $
Use the given values and solving we get
$n = \sqrt {\dfrac{{9 \times 10}}{{4 \times {{(3.14)}^2} \times 5}}} $
$\therefore n = 0.6756\,Hz$

So the number of revolutions per second will be 0.675.

Note:Centripetal force is used to refer to the force experienced by an object traveling in a circle. It's necessary to have a centripetal force to maintain a circular motion because if there is no resultant force acting on an object, then the object travels with uniform motion in a straight line, or stays at rest. Centripetal force is always perpendicular to the path and pointing to the centre of curvature.