Answer
(a) Please see the work below.
(b) $0.52KN$ at the top and $0.56KN$ at the bottom
Work Step by Step
(a) We know that the apparent weight is greater at the bottom than at the top because at the bottom the normal force not only supports the weight but also provides the upward centripetal acceleration.
(b) We can calculate the apparent weight at the top as follows:
$W_{app}=m\omega^2r-mg$
$\implies W_{app}=m(\frac{2\pi}{T})^2r-mg$
We plug in the known values to obtain:
$W_{app}=(55Kg)(\frac{2\times3.1416}{28s})^2\times 7.2m-(55Kg\times 9.8m/s^2)$
$W_{app}=520N=0.52KN$
We can calculate the apparent weight at the bottom as follows:
$W_{app}=m\omega^2r+mg$
$\implies W_{app}=m(\frac{2\pi}{T})^2r+mg$
We plug in the known values to obtain:
$W_{app}=(55Kg)(\frac{2\times3.1416}{28s})^2\times 7.2m+(55Kg\times 9.8m/s^2)$
$W_{app}=560N=0.56KN$
