Answer:
559.5 N at the bottom and 519.6 N at the top of the wheel
Step-by-step explanation:
If it completes 1 revolution (or 2π rad) per 28s then its angular speed is
![\omega = 2\pi/28 = 0.224 rad/s](https://img.qammunity.org/2021/formulas/physics/college/lvxk13h67rfg1i1mkm6527r132ptrlko48.png)
The centripetal acceleration would be:
![a_c = \omega^2 R = 0.224^2*7.2 = 0.363 m/s^2](https://img.qammunity.org/2021/formulas/physics/college/l8nuwxhhy2slguwjftdhdzcn2zjxuylb00.png)
Let gravitational acceleration g = 9.81 m/s2.
At the bottom of the wheel the net acceleration would be g plus the centripetal acceleration:
![a_b = a_c + g = 9.81 + 0.363 = 10.17 m/s^2](https://img.qammunity.org/2021/formulas/physics/college/87foy9zfzlvhpjncugg6zif0huzljbhj71.png)
So the weight at the bottom of the wheel would be
![W_b = a_b*m = 10.17*55 = 559.5 N](https://img.qammunity.org/2021/formulas/physics/college/cgfa5mvwhg6ge9j93ngd95amtdc09um623.png)
Similarly at the top of the wheel the net acceleration is g subtracted by the centripetal acceleration:
![a_t = g - _a_c = 9.81 - 0363 = 9.45 m/s^2](https://img.qammunity.org/2021/formulas/physics/college/cgx2385xh7j1asgelmnhuvpbjewul8z2m7.png)
And the weight at the top is
![W_t = a_t*m = 9.45*55 = 519.6 N](https://img.qammunity.org/2021/formulas/physics/college/qlcjcbxpg3nxy2u8rpejsnjz99qpwkmexl.png)