# In a garden a boy is swinging in the standing position how will the time period of the swing be affected if the boy sites down on the swing?

Mar 2, 2018

Time period will increase

#### Explanation:

During this swinging, angular momentum of the whole system will be constant.

So,angular momentum = $I \omega$ (where, $I$ is the moment of inertia of the system and $\omega$ is the angular velocity)

Now,when the boy sits down,moment of inertia increases,well that's because,here we calculate the moment of inertia along that axis,from where the swing is swinging to and fro.

So,initially,his head was closer to the point of swing than his legs,so his body mass was distributed along the radius,but on sitting,the whole mass went towards the cicumferance,so $I$ increased.

Now,as angular momentum is conserved,so on increasing $I$, $\omega$ must decrease

Now,we know, $T = \frac{2 \pi}{\omega}$

So, if $\omega$ decreases,time period will increase.

Mar 3, 2018

The boy sitting down will make the time period increase unless he is really, really tall!

#### Explanation:

To understand this problem, we will make a few semi-realistic assumptions. We will model the boy standing straight on the swing by a straight uniform rod of mass $m$. We will also model the boy sitting down on the swing by a point mass (not very realistic, I know - but this should convey the general idea). We will also ignore the weight and the moment of inertia of the swing compared to the boy (this, of course, is a much more realistic approximation).

Let us denote the height of the boy and the length of the swing by $h$ and $l$, respectively.

For a compound pendulum of moment of inertia $I$, mass $m$, and for which the center of gravity is at a distance $L$ from the point of suspension, the restoring torque when the angular displacement from the equilibrium position is $\theta$ is given by $- m g L \sin \theta$. Thus the equation of motion is

$I \frac{{d}^{2} \theta}{\mathrm{dt}} ^ 2 = - m g L \sin \theta \approx - m g L \theta$

for small oscillations. So, the time period is given by

$T = \frac{1}{2 \pi} \sqrt{\frac{I}{m g L}}$

For the boy standing up, his center of gravity is at a distance $L = l - \frac{h}{2}$ from the point of suspension. The moment of inertia of the boy about his center of mass is given by $\frac{1}{12} m {h}^{2}$. What we need is the moment of inertia about the point of suspension. Using the parallel axis theorem, this turns out to be simply $m \left[\frac{1}{12} {h}^{2} + {\left(l - \frac{h}{2}\right)}^{2}\right] = m \left({l}^{2} - h l + \frac{1}{3} {h}^{2}\right)$. So, the time period in this case is given by

${T}_{1} = \frac{1}{2 \pi} \sqrt{\frac{1}{g} \frac{{l}^{2} - h l + \frac{1}{3} {h}^{2}}{l - \frac{1}{2} h}}$

On the other hand, for the boy sitting down, the moment of inertia is $m {l}^{2}$ and the distance of the center of gravity from the point of suspension is given by $l$ - leading to the familiar result

${T}_{2} = \frac{1}{2 \pi} \sqrt{\frac{l}{g}}$

Sitting down will increase the time period, ${T}_{2} > {T}_{1}$ if

$l > \frac{{l}^{2} - h l + \frac{1}{3} {h}^{2}}{l - \frac{1}{2} h} \implies {l}^{2} - \frac{1}{2} h l > {l}^{2} - h l + \frac{1}{3} {h}^{2}$
$\implies \frac{1}{2} h l > \frac{1}{3} {h}^{2} \implies h < \frac{2}{3} l$

Thus the boy sitting down will make the time period increase unless he is really, really tall!