Question

A 60 kg woman is on a vertical swing of radius 20 m. The swing rotates with constant speed.
a. At what speed would she feel weightless at the top?
b. At this speed, what is her apparent weight at the bottom?

Answer 1

Apparent weight is the non-real assumed weight that corresponds to how heavy or light an object is.
Here two formulae are used $\dfrac{{m{v^2}}}{r} = mg$ and ${W_{{\text{apparent}}}} = mg + \dfrac{{m{v^2}}}{r}$, where $m =$mass of the woman, $r =$radius made by the swing, $v =$velocity, $g =$gravity of earth.

Complete step by step solution:
Given,
Weight of the woman is,
$m = {\text{60}}\,{\text{kg}}$
Radius,
$r = 20\,{\text{m}}$
We know that, acceleration due to gravity,
$g = {\text{10}}\,{\text{m/}}{{\text{s}}^{\text{2}}}$
Also, given that the swing rotates with constant speed.

For case a:

Here we need to find at what speed a woman would feel weightless at the top while she is on a vertical swing.
To find the speed for feeling weightlessness, we need to use the formula-
$\dfrac{{m{v^2}}}{r} = mg$ ...... (1)
${\text{m}}$ gets cancelled
Therefore,
$v = \sqrt {gr} \\\ = \sqrt {10 \times 20} \\\ = {\text{14}}{\text{.14}}\,{\text{m/}}{{\text{s}}^{\text{2}}} \\\$

Hence, the speed for weightlessness is ${\text{14}}{\text{.14}}\,{\text{m/}}{{\text{s}}^{\text{2}}}$.

For case b:

Here we need to find the apparent weight of the woman at the bottom when the speed is${\text{14}}{\text{.14}}\,{\text{m/}}{{\text{s}}^{\text{2}}}$.
Apparent weight usually acts under the influence of gravity.
${W_{{\text{apparent}}}} = mg + \dfrac{{m{v^2}}}{r}$ ...... (2)
Putting equation (1) in (2), we get-

${W_{{\text{apparent}}}} = mg + mg \\\ = {\text{2mg}} \\\ = 2 \times 60\,{\text{kg}} \\\ = {\text{120}}\,{\text{kg}} \\\$

Hence, the apparent weight at the bottom$= {\text{120}}\,{\text{kg}}$

Note: Apparent weight is not actual weight. The weight remains the same but the apparent weight changes due to the influence of gravity. Higher the acceleration due to gravity higher is the weight measured and vice versa.
Here $\dfrac{{m{v^2}}}{r}$ should be equal to ${\text{mg}}$ for getting weightlessness. Apparent weight is mostly found in elevators. When an elevator goes upwards, we can actually feel heavier and lighter when the elevator goes down. Here the apparent weight changes. This happens because the normal force becomes equal to apparent weight and velocity changes while going up and down, so as a result acceleration changes.