91Ó°ÊÓ

We know that the radius of the ferries wheel is $r=15\mathrm{m}$, and the period of one rotation is $T=25\mathrm{s}$. We have to calculate the speed and the magnitude of acceleration.

The speed of an object in circular motion is given by

$v=\mathrm{Ó\neg }r$

where $\mathrm{Ó\neg }$is the angular velocity and is given by $\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{T}$. Hence, the speed is given by

$v=\frac{2\mathrm{Ï€}}{T}r$

Substituting the values we have

$v=\frac{2\mathrm{Ï€}}{(25\mathrm{s})}(15\mathrm{m})$

$=3.77m/s$

And the acceleration here will be centripetal acceleration, which is given by

$a={\mathrm{Ó\neg }}^{2}r=\frac{4{\mathrm{Ï€}}^2}{T^2}r$

Substituting the values we have

$=0.947m/s^2$

My speed is $3.77\mathrm{m}/\mathrm{s}$and the magnitude of acceleration is$0.947\mathrm{m}/{\mathrm{s}}^2$.

We know that the acceleration of the gravity is $g=9.81\mathrm{m}/{\mathrm{s}}^2$.

Consider my mass is $m$. Hence, my weight standing at the ground is

While at the top, the centrifugal force acts vertically upward and the weight acts vertically downward. Hence, net weight at the top is given by

Hence, the ratio is

Substituting the values we have

$\frac{w_1}{w}=\frac{\left[\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)-\left(0.947\mathrm{m}/{\mathrm{s}}^2\right)\right]}{\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)}$

$=0.903$

The ratio of the weight at the top of the ferries wheels to the weight while standing at the ground is $0.903$.

We know that the acceleration of the gravity is $g=9.81\mathrm{m}/{\mathrm{s}}^2$

At the bottom, the centrifugal force and the weight, both are directed vertically downwards. Hence, the effect will add up. Hence, the net weight of me will be

Hence, the ratio is

Substituting the values we have

$\frac{w_2}{w}=\frac{\left[\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)+\left(0.947\mathrm{m}/{\mathrm{s}}^2\right)\right]}{\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)}$

The ratio of your weight at the bottom of the ride to your weight while standing on the ground is$1.09$.